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<div class="breadcrumb"><a href="index.html">Home</a> / Dynamic Programming</div>
<h1>Dynamic Programming</h1>
<p>Break hard problems into simple subproblems. DP is just smart recursion -- and it is way less scary than it sounds.</p>
</div>
<!-- Table of Contents -->
<div class="toc">
<h4>On This Page</h4>
<a href="#what-is-dp">1. What is Dynamic Programming?</a>
<a href="#top-down-vs-bottom-up">2. Top-Down vs Bottom-Up</a>
<a href="#framework">3. The DP Problem-Solving Framework</a>
<a href="#fibonacci">4. Fibonacci (The Classic Intro)</a>
<a href="#climbing-stairs">5. Climbing Stairs</a>
<a href="#1d-dp">6. 1D DP Problems</a>
<a href="#2d-dp">7. 2D DP Problems</a>
<a href="#categories">8. DP Categories Cheat Sheet</a>
<a href="#leetcode">9. LeetCode Problems by Category</a>
<a href="#tips">10. Tips for DP</a>
<a href="#quiz">11. Practice Quiz</a>
</div>
<!-- ======================= SECTION 1 ======================= -->
<section id="what-is-dp">
<h2>1. What is Dynamic Programming?</h2>
<p>
Dynamic Programming (DP) is a technique for solving problems by <strong>breaking them into overlapping subproblems</strong> and storing the results so you never solve the same subproblem twice. That is literally all it is.
</p>
<p>For a problem to be solvable with DP, it needs two key properties:</p>
<ol>
<li><strong>Optimal substructure</strong> -- the optimal solution to the problem contains optimal solutions to its subproblems. If you can build the answer from answers to smaller versions of the same problem, you have this.</li>
<li><strong>Overlapping subproblems</strong> -- the same subproblems are solved multiple times. This is why storing results helps. If each subproblem is unique (like in merge sort), DP won't help -- that is just divide and conquer.</li>
</ol>
<div class="formula-box">
DP = Recursion + Memoization (or Bottom-Up Tabulation)
</div>
<div class="formula-box">
<strong>"Is This a DP Problem?" -- Checklist:</strong><br><br>
1. Can the problem be broken into overlapping subproblems? (same subproblem solved multiple times)<br>
2. Does the problem have optimal substructure? (optimal solution uses optimal solutions to subproblems)<br>
3. Is the problem asking for: count, min/max, or feasibility?<br><br>
If YES to all three → try DP.<br><br>
<strong>Complexity rule:</strong> Time = (number of distinct states) × (work per state)<br>
Space = number of states stored simultaneously
</div>
<div class="tip-box">
<div class="label">Don't Panic</div>
<p>DP is NOT as scary as people make it sound. If you can write a recursive solution, you can convert it to DP. The hard part is defining what your subproblem is -- the actual coding is usually straightforward. Think of DP as "remembering your work so you don't redo it."</p>
</div>
<h3>Why naive recursion is slow</h3>
<p>Consider computing Fibonacci(5). A naive recursive approach recalculates the same values over and over:</p>
<pre><code><span class="comment">/*
fib(5)
/ \
fib(4) fib(3)
/ \ / \
fib(3) fib(2) fib(2) fib(1)
/ \ / \ / \
fib(2) fib(1) ... ...
/ \
fib(1) fib(0)
fib(3) is computed 2 times
fib(2) is computed 3 times
This EXPLODES exponentially as n grows
*/</span></code></pre>
<p>With DP, you compute each value <strong>once</strong>, store it, and look it up next time. That takes you from O(2^n) to O(n).</p>
</section>
<!-- ======================= SECTION 2 ======================= -->
<section id="top-down-vs-bottom-up">
<h2>2. Top-Down (Memoization) vs Bottom-Up (Tabulation)</h2>
<p>There are two ways to implement DP. Both give the same results -- they just approach the problem from different directions.</p>
<h3>Top-Down (Memoization)</h3>
<ul>
<li>Start from the original problem and recurse downward</li>
<li>Cache (memoize) results of subproblems as you go</li>
<li>Easier to think about -- just write the recursive solution and add a cache</li>
<li>May have recursion overhead (stack frames)</li>
</ul>
<h3>Bottom-Up (Tabulation)</h3>
<ul>
<li>Start from the smallest subproblems and build up iteratively</li>
<li>Fill a table (array) from base cases to final answer</li>
<li>No recursion overhead -- pure iteration</li>
<li>Can be more efficient in practice</li>
<li>Sometimes easier to optimize space</li>
</ul>
<table>
<thead>
<tr>
<th>Aspect</th>
<th>Top-Down (Memo)</th>
<th>Bottom-Up (Tab)</th>
</tr>
</thead>
<tbody>
<tr><td>Direction</td><td>Big problem down to base cases</td><td>Base cases up to big problem</td></tr>
<tr><td>Implementation</td><td>Recursion + cache</td><td>Iterative + table</td></tr>
<tr><td>Ease of thought</td><td>Often easier to derive</td><td>Requires knowing computation order</td></tr>
<tr><td>Stack overflow risk</td><td>Yes, for large inputs</td><td>No</td></tr>
<tr><td>Space optimization</td><td>Harder</td><td>Easier (can drop old rows/values)</td></tr>
<tr><td>Computes unused states?</td><td>No (only what's needed)</td><td>Sometimes yes</td></tr>
</tbody>
</table>
<div class="tip-box">
<div class="label">Recommendation</div>
<p>In interviews, start with top-down -- it is the natural extension of your recursive brute force. Once that works, mention you could convert to bottom-up for better space/performance. We will show both approaches for every problem below.</p>
</div>
</section>
<!-- ======================= SECTION 3 ======================= -->
<section id="framework">
<h2>3. The DP Problem-Solving Framework</h2>
<p>Use this 5-step framework for every DP problem. Seriously -- every single one.</p>
<div class="roadmap">
<div class="roadmap-step">
<div class="step-num">1</div>
<span class="step-label">Define the state</span>
<span class="step-desc">What does dp[i] (or dp[i][j]) represent?</span>
</div>
<div class="roadmap-step">
<div class="step-num">2</div>
<span class="step-label">Find the recurrence</span>
<span class="step-desc">dp[i] = f(dp[i-1], dp[i-2], ...)</span>
</div>
<div class="roadmap-step">
<div class="step-num">3</div>
<span class="step-label">Set base cases</span>
<span class="step-desc">dp[0] = ?, dp[1] = ?</span>
</div>
<div class="roadmap-step">
<div class="step-num">4</div>
<span class="step-label">Determine computation order</span>
<span class="step-desc">Left to right? Bottom to top?</span>
</div>
<div class="roadmap-step">
<div class="step-num">5</div>
<span class="step-label">Optimize space</span>
<span class="step-desc">Do you really need the whole table?</span>
</div>
</div>
<div class="example-box">
<div class="label">Example: Applying the framework</div>
<p>Problem: "How many ways to reach step n if you can take 1 or 2 steps?"</p>
<ol>
<li><strong>State:</strong> dp[i] = number of ways to reach step i</li>
<li><strong>Recurrence:</strong> dp[i] = dp[i-1] + dp[i-2] (arrive from one step back OR two steps back)</li>
<li><strong>Base cases:</strong> dp[0] = 1 (one way to be at the start), dp[1] = 1 (one way to reach step 1)</li>
<li><strong>Order:</strong> Left to right (i = 2, 3, ..., n)</li>
<li><strong>Space:</strong> Only need last two values, so O(1) instead of O(n)</li>
</ol>
</div>
</section>
<!-- ======================= SECTION 4 ======================= -->
<section id="fibonacci">
<h2>4. Fibonacci (The Classic Intro)</h2>
<p>Every DP tutorial starts with Fibonacci for a reason: it perfectly illustrates why DP exists. The Fibonacci sequence is 0, 1, 1, 2, 3, 5, 8, 13, 21, ... where each number is the sum of the two before it.</p>
<div class="formula-box">fib(n) = fib(n-1) + fib(n-2), where fib(0) = 0, fib(1) = 1</div>
<h3>Approach 1: Naive Recursion -- O(2^n) time, O(n) space</h3>
<p>This is what NOT to do. Each call branches into two more, creating exponential work.</p>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">fib</span>(n):
<span class="keyword">if</span> n <= <span class="number">1</span>:
<span class="keyword">return</span> n
<span class="keyword">return</span> <span class="function">fib</span>(n - <span class="number">1</span>) + <span class="function">fib</span>(n - <span class="number">2</span>)
<span class="comment"># fib(40) takes SECONDS. fib(50) may never finish.</span></code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">fib</span>(n) {
<span class="keyword">if</span> (n <= <span class="number">1</span>) <span class="keyword">return</span> n;
<span class="keyword">return</span> <span class="function">fib</span>(n - <span class="number">1</span>) + <span class="function">fib</span>(n - <span class="number">2</span>);
}</code></pre>
<h3>Approach 2: Top-Down with Memoization -- O(n) time, O(n) space</h3>
<p>Same recursion, but we remember every result. Each fib(k) is computed exactly once.</p>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">fib</span>(n, memo={}):
<span class="keyword">if</span> n <= <span class="number">1</span>:
<span class="keyword">return</span> n
<span class="keyword">if</span> n <span class="keyword">in</span> memo:
<span class="keyword">return</span> memo[n]
memo[n] = <span class="function">fib</span>(n - <span class="number">1</span>, memo) + <span class="function">fib</span>(n - <span class="number">2</span>, memo)
<span class="keyword">return</span> memo[n]
<span class="comment"># fib(1000) returns instantly</span></code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">fib</span>(n, memo = {}) {
<span class="keyword">if</span> (n <= <span class="number">1</span>) <span class="keyword">return</span> n;
<span class="keyword">if</span> (memo[n] !== <span class="keyword">undefined</span>) <span class="keyword">return</span> memo[n];
memo[n] = <span class="function">fib</span>(n - <span class="number">1</span>, memo) + <span class="function">fib</span>(n - <span class="number">2</span>, memo);
<span class="keyword">return</span> memo[n];
}</code></pre>
<h3>Approach 3: Bottom-Up Tabulation -- O(n) time, O(n) space</h3>
<p>Build the table iteratively from the base cases up. No recursion at all.</p>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">fib</span>(n):
<span class="keyword">if</span> n <= <span class="number">1</span>:
<span class="keyword">return</span> n
dp = [<span class="number">0</span>] * (n + <span class="number">1</span>)
dp[<span class="number">1</span>] = <span class="number">1</span>
<span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, n + <span class="number">1</span>):
dp[i] = dp[i - <span class="number">1</span>] + dp[i - <span class="number">2</span>]
<span class="keyword">return</span> dp[n]</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">fib</span>(n) {
<span class="keyword">if</span> (n <= <span class="number">1</span>) <span class="keyword">return</span> n;
<span class="keyword">const</span> dp = <span class="keyword">new</span> <span class="builtin">Array</span>(n + <span class="number">1</span>).fill(<span class="number">0</span>);
dp[<span class="number">1</span>] = <span class="number">1</span>;
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">2</span>; i <= n; i++) {
dp[i] = dp[i - <span class="number">1</span>] + dp[i - <span class="number">2</span>];
}
<span class="keyword">return</span> dp[n];
}</code></pre>
<div class="example-box">
<div class="label">State Table Walkthrough</div>
<p>Computing fib(6) bottom-up:</p>
<table>
<thead><tr><th>i</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th></tr></thead>
<tbody><tr><td>dp[i]</td><td>0</td><td>1</td><td>1</td><td>2</td><td>3</td><td>5</td><td>8</td></tr></tbody>
</table>
<p>Each cell is just the sum of the two cells to its left. That is it.</p>
</div>
<h3>Approach 4: Space-Optimized -- O(n) time, O(1) space</h3>
<p>We only ever look at the last two values, so why store the whole array?</p>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">fib</span>(n):
<span class="keyword">if</span> n <= <span class="number">1</span>:
<span class="keyword">return</span> n
prev2, prev1 = <span class="number">0</span>, <span class="number">1</span>
<span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, n + <span class="number">1</span>):
curr = prev1 + prev2
prev2 = prev1
prev1 = curr
<span class="keyword">return</span> prev1</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">fib</span>(n) {
<span class="keyword">if</span> (n <= <span class="number">1</span>) <span class="keyword">return</span> n;
<span class="keyword">let</span> prev2 = <span class="number">0</span>, prev1 = <span class="number">1</span>;
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">2</span>; i <= n; i++) {
<span class="keyword">const</span> curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
<span class="keyword">return</span> prev1;
}</code></pre>
<table>
<thead><tr><th>Approach</th><th>Time</th><th>Space</th></tr></thead>
<tbody>
<tr><td>Naive recursion</td><td>O(2^n)</td><td>O(n)</td></tr>
<tr><td>Top-down memo</td><td>O(n)</td><td>O(n)</td></tr>
<tr><td>Bottom-up table</td><td>O(n)</td><td>O(n)</td></tr>
<tr><td>Space-optimized</td><td>O(n)</td><td>O(1)</td></tr>
</tbody>
</table>
</section>
<!-- ======================= SECTION 5 ======================= -->
<section id="climbing-stairs">
<h2>5. Climbing Stairs</h2>
<p><strong>Problem:</strong> You are climbing a staircase with n steps. Each time you can climb 1 or 2 steps. How many distinct ways can you reach the top?</p>
<div class="formula-box">
State: dp[i] = number of ways to reach step i<br>
Recurrence: dp[i] = dp[i-1] + dp[i-2]<br>
Base cases: dp[0] = 1, dp[1] = 1
</div>
<p>Wait -- this is just Fibonacci! To reach step i, you either came from step i-1 (one step) or step i-2 (two steps). The total ways is the sum of both.</p>
<div class="example-box">
<div class="label">State Table: n = 5</div>
<table>
<thead><tr><th>Step</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr></thead>
<tbody><tr><td>Ways</td><td>1</td><td>1</td><td>2</td><td>3</td><td>5</td><td>8</td></tr></tbody>
</table>
<p>There are 8 distinct ways to climb 5 stairs.</p>
</div>
<h3>Top-Down</h3>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">climbStairs</span>(n, memo={}):
<span class="keyword">if</span> n <= <span class="number">1</span>:
<span class="keyword">return</span> <span class="number">1</span>
<span class="keyword">if</span> n <span class="keyword">in</span> memo:
<span class="keyword">return</span> memo[n]
memo[n] = <span class="function">climbStairs</span>(n - <span class="number">1</span>, memo) + <span class="function">climbStairs</span>(n - <span class="number">2</span>, memo)
<span class="keyword">return</span> memo[n]</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">climbStairs</span>(n, memo = {}) {
<span class="keyword">if</span> (n <= <span class="number">1</span>) <span class="keyword">return</span> <span class="number">1</span>;
<span class="keyword">if</span> (memo[n] !== <span class="keyword">undefined</span>) <span class="keyword">return</span> memo[n];
memo[n] = <span class="function">climbStairs</span>(n - <span class="number">1</span>, memo) + <span class="function">climbStairs</span>(n - <span class="number">2</span>, memo);
<span class="keyword">return</span> memo[n];
}</code></pre>
<h3>Bottom-Up (Space-Optimized)</h3>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">climbStairs</span>(n):
<span class="keyword">if</span> n <= <span class="number">1</span>:
<span class="keyword">return</span> <span class="number">1</span>
prev2, prev1 = <span class="number">1</span>, <span class="number">1</span>
<span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">2</span>, n + <span class="number">1</span>):
curr = prev1 + prev2
prev2 = prev1
prev1 = curr
<span class="keyword">return</span> prev1</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">climbStairs</span>(n) {
<span class="keyword">if</span> (n <= <span class="number">1</span>) <span class="keyword">return</span> <span class="number">1</span>;
<span class="keyword">let</span> prev2 = <span class="number">1</span>, prev1 = <span class="number">1</span>;
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">2</span>; i <= n; i++) {
<span class="keyword">const</span> curr = prev1 + prev2;
prev2 = prev1;
prev1 = curr;
}
<span class="keyword">return</span> prev1;
}</code></pre>
<p><strong>Time:</strong> O(n) | <strong>Space:</strong> O(1)</p>
</section>
<!-- ======================= SECTION 6 ======================= -->
<section id="1d-dp">
<h2>6. 1D DP Problems</h2>
<!-- House Robber -->
<h3>House Robber</h3>
<p><strong>Problem:</strong> You are a robber planning to rob houses along a street. Each house has a certain amount of money. You cannot rob two adjacent houses (alarm triggers). Find the maximum amount you can rob.</p>
<div class="formula-box">
State: dp[i] = max money robbing houses 0..i<br>
Recurrence: dp[i] = max(dp[i-1], dp[i-2] + nums[i])<br>
Base cases: dp[0] = nums[0], dp[1] = max(nums[0], nums[1])
</div>
<p>At each house, you have two choices: <strong>skip it</strong> (take dp[i-1]) or <strong>rob it</strong> (take dp[i-2] + this house's money). Pick whichever is more.</p>
<div class="example-box">
<div class="label">State Table: nums = [2, 7, 9, 3, 1]</div>
<table>
<thead><tr><th>i</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th></tr></thead>
<tbody>
<tr><td>nums[i]</td><td>2</td><td>7</td><td>9</td><td>3</td><td>1</td></tr>
<tr><td>dp[i]</td><td>2</td><td>7</td><td>11</td><td>11</td><td>12</td></tr>
</tbody>
</table>
<p>dp[2] = max(7, 2+9) = 11. dp[3] = max(11, 7+3) = 11. dp[4] = max(11, 11+1) = 12. Rob houses 0, 2, 4 for $12.</p>
</div>
<h3>Top-Down</h3>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">rob</span>(nums):
memo = {}
<span class="keyword">def</span> <span class="function">dp</span>(i):
<span class="keyword">if</span> i < <span class="number">0</span>:
<span class="keyword">return</span> <span class="number">0</span>
<span class="keyword">if</span> i <span class="keyword">in</span> memo:
<span class="keyword">return</span> memo[i]
memo[i] = <span class="builtin">max</span>(<span class="function">dp</span>(i - <span class="number">1</span>), <span class="function">dp</span>(i - <span class="number">2</span>) + nums[i])
<span class="keyword">return</span> memo[i]
<span class="keyword">return</span> <span class="function">dp</span>(<span class="builtin">len</span>(nums) - <span class="number">1</span>)</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">rob</span>(nums) {
<span class="keyword">const</span> memo = {};
<span class="keyword">function</span> <span class="function">dp</span>(i) {
<span class="keyword">if</span> (i < <span class="number">0</span>) <span class="keyword">return</span> <span class="number">0</span>;
<span class="keyword">if</span> (memo[i] !== <span class="keyword">undefined</span>) <span class="keyword">return</span> memo[i];
memo[i] = Math.<span class="function">max</span>(<span class="function">dp</span>(i - <span class="number">1</span>), <span class="function">dp</span>(i - <span class="number">2</span>) + nums[i]);
<span class="keyword">return</span> memo[i];
}
<span class="keyword">return</span> <span class="function">dp</span>(nums.length - <span class="number">1</span>);
}</code></pre>
<h3>Bottom-Up (Space-Optimized)</h3>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">rob</span>(nums):
<span class="keyword">if</span> <span class="keyword">not</span> nums:
<span class="keyword">return</span> <span class="number">0</span>
<span class="keyword">if</span> <span class="builtin">len</span>(nums) == <span class="number">1</span>:
<span class="keyword">return</span> nums[<span class="number">0</span>]
prev2, prev1 = <span class="number">0</span>, <span class="number">0</span>
<span class="keyword">for</span> num <span class="keyword">in</span> nums:
curr = <span class="builtin">max</span>(prev1, prev2 + num)
prev2 = prev1
prev1 = curr
<span class="keyword">return</span> prev1</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">rob</span>(nums) {
<span class="keyword">let</span> prev2 = <span class="number">0</span>, prev1 = <span class="number">0</span>;
<span class="keyword">for</span> (<span class="keyword">const</span> num <span class="keyword">of</span> nums) {
<span class="keyword">const</span> curr = Math.<span class="function">max</span>(prev1, prev2 + num);
prev2 = prev1;
prev1 = curr;
}
<span class="keyword">return</span> prev1;
}</code></pre>
<p><strong>Time:</strong> O(n) | <strong>Space:</strong> O(1)</p>
<!-- Maximum Subarray -->
<h3>Maximum Subarray (Kadane's Algorithm)</h3>
<p><strong>Problem:</strong> Find the contiguous subarray with the largest sum.</p>
<div class="formula-box">
State: dp[i] = maximum subarray sum ending at index i<br>
Recurrence: dp[i] = max(nums[i], dp[i-1] + nums[i])<br>
Base case: dp[0] = nums[0]
</div>
<p>At each element, you decide: start a new subarray here, or extend the previous one. If the running sum is negative, start fresh.</p>
<div class="example-box">
<div class="label">State Table: nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]</div>
<table>
<thead><tr><th>i</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th></tr></thead>
<tbody>
<tr><td>nums[i]</td><td>-2</td><td>1</td><td>-3</td><td>4</td><td>-1</td><td>2</td><td>1</td><td>-5</td><td>4</td></tr>
<tr><td>dp[i]</td><td>-2</td><td>1</td><td>-2</td><td>4</td><td>3</td><td>5</td><td>6</td><td>1</td><td>5</td></tr>
</tbody>
</table>
<p>Maximum is 6 (subarray [4, -1, 2, 1]).</p>
</div>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">maxSubArray</span>(nums):
max_sum = curr_sum = nums[<span class="number">0</span>]
<span class="keyword">for</span> num <span class="keyword">in</span> nums[<span class="number">1</span>:]:
curr_sum = <span class="builtin">max</span>(num, curr_sum + num)
max_sum = <span class="builtin">max</span>(max_sum, curr_sum)
<span class="keyword">return</span> max_sum</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">maxSubArray</span>(nums) {
<span class="keyword">let</span> maxSum = nums[<span class="number">0</span>], currSum = nums[<span class="number">0</span>];
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i < nums.length; i++) {
currSum = Math.<span class="function">max</span>(nums[i], currSum + nums[i]);
maxSum = Math.<span class="function">max</span>(maxSum, currSum);
}
<span class="keyword">return</span> maxSum;
}</code></pre>
<p><strong>Time:</strong> O(n) | <strong>Space:</strong> O(1)</p>
<!-- Coin Change -->
<h3>Coin Change</h3>
<p><strong>Problem:</strong> Given coin denominations and a target amount, find the minimum number of coins needed. If impossible, return -1.</p>
<div class="formula-box">
State: dp[i] = minimum coins needed for amount i<br>
Recurrence: dp[i] = min(dp[i - coin] + 1) for each coin where coin <= i<br>
Base case: dp[0] = 0
</div>
<div class="example-box">
<div class="label">State Table: coins = [1, 3, 4], amount = 6</div>
<table>
<thead><tr><th>Amount</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th></tr></thead>
<tbody><tr><td>dp[i]</td><td>0</td><td>1</td><td>2</td><td>1</td><td>1</td><td>2</td><td>2</td></tr></tbody>
</table>
<p>dp[6] = 2 (use coins 3 + 3). Note: greedy (4+1+1 = 3 coins) fails here!</p>
</div>
<h3>Top-Down</h3>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">coinChange</span>(coins, amount):
memo = {}
<span class="keyword">def</span> <span class="function">dp</span>(remaining):
<span class="keyword">if</span> remaining == <span class="number">0</span>:
<span class="keyword">return</span> <span class="number">0</span>
<span class="keyword">if</span> remaining < <span class="number">0</span>:
<span class="keyword">return</span> <span class="builtin">float</span>(<span class="string">'inf'</span>)
<span class="keyword">if</span> remaining <span class="keyword">in</span> memo:
<span class="keyword">return</span> memo[remaining]
memo[remaining] = <span class="builtin">min</span>(<span class="function">dp</span>(remaining - c) + <span class="number">1</span> <span class="keyword">for</span> c <span class="keyword">in</span> coins)
<span class="keyword">return</span> memo[remaining]
result = <span class="function">dp</span>(amount)
<span class="keyword">return</span> result <span class="keyword">if</span> result != <span class="builtin">float</span>(<span class="string">'inf'</span>) <span class="keyword">else</span> -<span class="number">1</span></code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">coinChange</span>(coins, amount) {
<span class="keyword">const</span> memo = {};
<span class="keyword">function</span> <span class="function">dp</span>(remaining) {
<span class="keyword">if</span> (remaining === <span class="number">0</span>) <span class="keyword">return</span> <span class="number">0</span>;
<span class="keyword">if</span> (remaining < <span class="number">0</span>) <span class="keyword">return</span> <span class="number">Infinity</span>;
<span class="keyword">if</span> (memo[remaining] !== <span class="keyword">undefined</span>) <span class="keyword">return</span> memo[remaining];
<span class="keyword">let</span> min = <span class="number">Infinity</span>;
<span class="keyword">for</span> (<span class="keyword">const</span> coin <span class="keyword">of</span> coins) {
min = Math.<span class="function">min</span>(min, <span class="function">dp</span>(remaining - coin) + <span class="number">1</span>);
}
memo[remaining] = min;
<span class="keyword">return</span> min;
}
<span class="keyword">const</span> result = <span class="function">dp</span>(amount);
<span class="keyword">return</span> result === <span class="number">Infinity</span> ? -<span class="number">1</span> : result;
}</code></pre>
<h3>Bottom-Up</h3>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">coinChange</span>(coins, amount):
dp = [<span class="builtin">float</span>(<span class="string">'inf'</span>)] * (amount + <span class="number">1</span>)
dp[<span class="number">0</span>] = <span class="number">0</span>
<span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, amount + <span class="number">1</span>):
<span class="keyword">for</span> coin <span class="keyword">in</span> coins:
<span class="keyword">if</span> coin <= i:
dp[i] = <span class="builtin">min</span>(dp[i], dp[i - coin] + <span class="number">1</span>)
<span class="keyword">return</span> dp[amount] <span class="keyword">if</span> dp[amount] != <span class="builtin">float</span>(<span class="string">'inf'</span>) <span class="keyword">else</span> -<span class="number">1</span></code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">coinChange</span>(coins, amount) {
<span class="keyword">const</span> dp = <span class="keyword">new</span> <span class="builtin">Array</span>(amount + <span class="number">1</span>).fill(<span class="number">Infinity</span>);
dp[<span class="number">0</span>] = <span class="number">0</span>;
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i <= amount; i++) {
<span class="keyword">for</span> (<span class="keyword">const</span> coin <span class="keyword">of</span> coins) {
<span class="keyword">if</span> (coin <= i) {
dp[i] = Math.<span class="function">min</span>(dp[i], dp[i - coin] + <span class="number">1</span>);
}
}
}
<span class="keyword">return</span> dp[amount] === <span class="number">Infinity</span> ? -<span class="number">1</span> : dp[amount];
}</code></pre>
<p><strong>Time:</strong> O(amount * coins) | <strong>Space:</strong> O(amount)</p>
<div class="warning-box">
<div class="label">Why Greedy Fails</div>
<p>For coins [1, 3, 4] and amount 6, greedy picks 4+1+1 = 3 coins. DP correctly finds 3+3 = 2 coins. This is exactly why DP exists -- it explores all possibilities without brute-forcing every combination.</p>
</div>
<!-- Longest Increasing Subsequence -->
<h3>Longest Increasing Subsequence (LIS)</h3>
<p><strong>Problem:</strong> Find the length of the longest strictly increasing subsequence (not necessarily contiguous).</p>
<div class="formula-box">
State: dp[i] = length of LIS ending at index i<br>
Recurrence: dp[i] = max(dp[j] + 1) for all j < i where nums[j] < nums[i]<br>
Base case: dp[i] = 1 for all i (each element is a subsequence of length 1)
</div>
<div class="example-box">
<div class="label">State Table: nums = [10, 9, 2, 5, 3, 7, 101, 18]</div>
<table>
<thead><tr><th>i</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead>
<tbody>
<tr><td>nums[i]</td><td>10</td><td>9</td><td>2</td><td>5</td><td>3</td><td>7</td><td>101</td><td>18</td></tr>
<tr><td>dp[i]</td><td>1</td><td>1</td><td>1</td><td>2</td><td>2</td><td>3</td><td>4</td><td>4</td></tr>
</tbody>
</table>
<p>LIS length = 4. One possible LIS: [2, 3, 7, 101] or [2, 5, 7, 101] or [2, 5, 7, 18].</p>
</div>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">lengthOfLIS</span>(nums):
n = <span class="builtin">len</span>(nums)
dp = [<span class="number">1</span>] * n
<span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, n):
<span class="keyword">for</span> j <span class="keyword">in</span> <span class="builtin">range</span>(i):
<span class="keyword">if</span> nums[j] < nums[i]:
dp[i] = <span class="builtin">max</span>(dp[i], dp[j] + <span class="number">1</span>)
<span class="keyword">return</span> <span class="builtin">max</span>(dp)</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">lengthOfLIS</span>(nums) {
<span class="keyword">const</span> n = nums.length;
<span class="keyword">const</span> dp = <span class="keyword">new</span> <span class="builtin">Array</span>(n).fill(<span class="number">1</span>);
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i < n; i++) {
<span class="keyword">for</span> (<span class="keyword">let</span> j = <span class="number">0</span>; j < i; j++) {
<span class="keyword">if</span> (nums[j] < nums[i]) {
dp[i] = Math.<span class="function">max</span>(dp[i], dp[j] + <span class="number">1</span>);
}
}
}
<span class="keyword">return</span> Math.<span class="function">max</span>(...dp);
}</code></pre>
<p><strong>Time:</strong> O(n^2) | <strong>Space:</strong> O(n)</p>
<div class="tip-box">
<div class="label">For interviews</div>
<p>There is an O(n log n) solution using binary search + patience sorting. Know it exists and can explain the idea, but the O(n^2) DP solution is perfectly acceptable in most interviews. The O(n log n) version is rarely expected unless specifically asked.</p>
</div>
</section>
<!-- ======================= SECTION 7 ======================= -->
<section id="2d-dp">
<h2>7. 2D DP Problems</h2>
<p>When your state depends on two variables (two indices, a position in a grid, etc.), you need a 2D DP table.</p>
<!-- Unique Paths -->
<h3>Unique Paths</h3>
<p><strong>Problem:</strong> Given an m x n grid, starting at top-left, you can only move right or down. How many unique paths to the bottom-right?</p>
<div class="formula-box">
State: dp[i][j] = number of ways to reach cell (i, j)<br>
Recurrence: dp[i][j] = dp[i-1][j] + dp[i][j-1]<br>
Base cases: dp[0][j] = 1 (first row), dp[i][0] = 1 (first column)
</div>
<div class="example-box">
<div class="label">State Table: 3x4 grid</div>
<table>
<thead><tr><th></th><th>col 0</th><th>col 1</th><th>col 2</th><th>col 3</th></tr></thead>
<tbody>
<tr><td><strong>row 0</strong></td><td>1</td><td>1</td><td>1</td><td>1</td></tr>
<tr><td><strong>row 1</strong></td><td>1</td><td>2</td><td>3</td><td>4</td></tr>
<tr><td><strong>row 2</strong></td><td>1</td><td>3</td><td>6</td><td>10</td></tr>
</tbody>
</table>
<p>Answer: 10 unique paths. Each cell = sum of cell above + cell to the left.</p>
</div>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">uniquePaths</span>(m, n):
dp = [[<span class="number">1</span>] * n <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(m)]
<span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, m):
<span class="keyword">for</span> j <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, n):
dp[i][j] = dp[i - <span class="number">1</span>][j] + dp[i][j - <span class="number">1</span>]
<span class="keyword">return</span> dp[m - <span class="number">1</span>][n - <span class="number">1</span>]</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">uniquePaths</span>(m, n) {
<span class="keyword">const</span> dp = <span class="builtin">Array</span>.from({ length: m }, () =>
<span class="keyword">new</span> <span class="builtin">Array</span>(n).fill(<span class="number">1</span>)
);
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i < m; i++) {
<span class="keyword">for</span> (<span class="keyword">let</span> j = <span class="number">1</span>; j < n; j++) {
dp[i][j] = dp[i - <span class="number">1</span>][j] + dp[i][j - <span class="number">1</span>];
}
}
<span class="keyword">return</span> dp[m - <span class="number">1</span>][n - <span class="number">1</span>];
}</code></pre>
<p><strong>Time:</strong> O(m * n) | <strong>Space:</strong> O(m * n), can be optimized to O(n) using a single row</p>
<!-- Longest Common Subsequence -->
<h3>Longest Common Subsequence (LCS)</h3>
<p><strong>Problem:</strong> Given two strings, find the length of their longest common subsequence.</p>
<div class="formula-box">
State: dp[i][j] = LCS length of text1[0..i-1] and text2[0..j-1]<br>
Recurrence:<br>
If text1[i-1] == text2[j-1]: dp[i][j] = dp[i-1][j-1] + 1<br>
Else: dp[i][j] = max(dp[i-1][j], dp[i][j-1])<br>
Base cases: dp[0][j] = 0, dp[i][0] = 0
</div>
<div class="example-box">
<div class="label">State Table: text1 = "abcde", text2 = "ace"</div>
<table>
<thead><tr><th></th><th>""</th><th>a</th><th>c</th><th>e</th></tr></thead>
<tbody>
<tr><td><strong>""</strong></td><td>0</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td><strong>a</strong></td><td>0</td><td>1</td><td>1</td><td>1</td></tr>
<tr><td><strong>b</strong></td><td>0</td><td>1</td><td>1</td><td>1</td></tr>
<tr><td><strong>c</strong></td><td>0</td><td>1</td><td>2</td><td>2</td></tr>
<tr><td><strong>d</strong></td><td>0</td><td>1</td><td>2</td><td>2</td></tr>
<tr><td><strong>e</strong></td><td>0</td><td>1</td><td>2</td><td>3</td></tr>
</tbody>
</table>
<p>LCS length = 3. The LCS is "ace".</p>
</div>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">longestCommonSubsequence</span>(text1, text2):
m, n = <span class="builtin">len</span>(text1), <span class="builtin">len</span>(text2)
dp = [[<span class="number">0</span>] * (n + <span class="number">1</span>) <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(m + <span class="number">1</span>)]
<span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, m + <span class="number">1</span>):
<span class="keyword">for</span> j <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, n + <span class="number">1</span>):
<span class="keyword">if</span> text1[i - <span class="number">1</span>] == text2[j - <span class="number">1</span>]:
dp[i][j] = dp[i - <span class="number">1</span>][j - <span class="number">1</span>] + <span class="number">1</span>
<span class="keyword">else</span>:
dp[i][j] = <span class="builtin">max</span>(dp[i - <span class="number">1</span>][j], dp[i][j - <span class="number">1</span>])
<span class="keyword">return</span> dp[m][n]</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">longestCommonSubsequence</span>(text1, text2) {
<span class="keyword">const</span> m = text1.length, n = text2.length;
<span class="keyword">const</span> dp = <span class="builtin">Array</span>.from({ length: m + <span class="number">1</span> }, () =>
<span class="keyword">new</span> <span class="builtin">Array</span>(n + <span class="number">1</span>).fill(<span class="number">0</span>)
);
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i <= m; i++) {
<span class="keyword">for</span> (<span class="keyword">let</span> j = <span class="number">1</span>; j <= n; j++) {
<span class="keyword">if</span> (text1[i - <span class="number">1</span>] === text2[j - <span class="number">1</span>]) {
dp[i][j] = dp[i - <span class="number">1</span>][j - <span class="number">1</span>] + <span class="number">1</span>;
} <span class="keyword">else</span> {
dp[i][j] = Math.<span class="function">max</span>(dp[i - <span class="number">1</span>][j], dp[i][j - <span class="number">1</span>]);
}
}
}
<span class="keyword">return</span> dp[m][n];
}</code></pre>
<p><strong>Time:</strong> O(m * n) | <strong>Space:</strong> O(m * n)</p>
<!-- 0/1 Knapsack -->
<h3>0/1 Knapsack</h3>
<p><strong>Problem:</strong> Given items with weights and values, and a knapsack with weight capacity W, maximize the total value you can carry. Each item can only be used once.</p>
<div class="formula-box">
State: dp[i][w] = max value using items 0..i-1 with capacity w<br>
Recurrence:<br>
If weights[i-1] <= w: dp[i][w] = max(dp[i-1][w], dp[i-1][w - weights[i-1]] + values[i-1])<br>
Else: dp[i][w] = dp[i-1][w]<br>
Base cases: dp[0][w] = 0 for all w
</div>
<div class="example-box">
<div class="label">State Table: weights=[1,3,4,5], values=[1,4,5,7], capacity=7</div>
<table>
<thead><tr><th>item\cap</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead>
<tbody>
<tr><td><strong>0 items</strong></td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td><strong>item 1 (1,1)</strong></td><td>0</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr>
<tr><td><strong>item 2 (3,4)</strong></td><td>0</td><td>1</td><td>1</td><td>4</td><td>5</td><td>5</td><td>5</td><td>5</td></tr>
<tr><td><strong>item 3 (4,5)</strong></td><td>0</td><td>1</td><td>1</td><td>4</td><td>5</td><td>6</td><td>6</td><td>9</td></tr>
<tr><td><strong>item 4 (5,7)</strong></td><td>0</td><td>1</td><td>1</td><td>4</td><td>5</td><td>7</td><td>8</td><td>9</td></tr>
</tbody>
</table>
<p>Max value = 9 (items 2 and 3: weight 3+4=7, value 4+5=9).</p>
</div>
<pre><code><span class="lang-label">Python</span>
<span class="keyword">def</span> <span class="function">knapsack</span>(weights, values, capacity):
n = <span class="builtin">len</span>(weights)
dp = [[<span class="number">0</span>] * (capacity + <span class="number">1</span>) <span class="keyword">for</span> _ <span class="keyword">in</span> <span class="builtin">range</span>(n + <span class="number">1</span>)]
<span class="keyword">for</span> i <span class="keyword">in</span> <span class="builtin">range</span>(<span class="number">1</span>, n + <span class="number">1</span>):
<span class="keyword">for</span> w <span class="keyword">in</span> <span class="builtin">range</span>(capacity + <span class="number">1</span>):
<span class="keyword">if</span> weights[i - <span class="number">1</span>] <= w:
dp[i][w] = <span class="builtin">max</span>(
dp[i - <span class="number">1</span>][w],
dp[i - <span class="number">1</span>][w - weights[i - <span class="number">1</span>]] + values[i - <span class="number">1</span>]
)
<span class="keyword">else</span>:
dp[i][w] = dp[i - <span class="number">1</span>][w]
<span class="keyword">return</span> dp[n][capacity]</code></pre>
<pre><code><span class="lang-label">JavaScript</span>
<span class="keyword">function</span> <span class="function">knapsack</span>(weights, values, capacity) {
<span class="keyword">const</span> n = weights.length;
<span class="keyword">const</span> dp = <span class="builtin">Array</span>.from({ length: n + <span class="number">1</span> }, () =>
<span class="keyword">new</span> <span class="builtin">Array</span>(capacity + <span class="number">1</span>).fill(<span class="number">0</span>)
);
<span class="keyword">for</span> (<span class="keyword">let</span> i = <span class="number">1</span>; i <= n; i++) {
<span class="keyword">for</span> (<span class="keyword">let</span> w = <span class="number">0</span>; w <= capacity; w++) {
<span class="keyword">if</span> (weights[i - <span class="number">1</span>] <= w) {
dp[i][w] = Math.<span class="function">max</span>(
dp[i - <span class="number">1</span>][w],
dp[i - <span class="number">1</span>][w - weights[i - <span class="number">1</span>]] + values[i - <span class="number">1</span>]
);
} <span class="keyword">else</span> {
dp[i][w] = dp[i - <span class="number">1</span>][w];
}
}
}
<span class="keyword">return</span> dp[n][capacity];
}</code></pre>
<p><strong>Time:</strong> O(n * W) | <strong>Space:</strong> O(n * W), can be optimized to O(W) using a 1D array</p>
</section>
<!-- ======================= SECTION 8 ======================= -->
<section id="categories">
<h2>8. DP Categories Cheat Sheet</h2>
<p>Most DP problems fall into one of these categories. Recognizing the category is half the battle.</p>
<table>
<thead>
<tr>
<th>Category</th>
<th>Example Problems</th>
<th>Key Idea</th>
</tr>
</thead>
<tbody>
<tr>
<td><strong>Linear</strong></td>
<td>Fibonacci, Climbing Stairs, House Robber</td>
<td>dp[i] depends on previous elements</td>
</tr>
<tr>
<td><strong>Interval</strong></td>
<td>Palindromic Substrings, Matrix Chain Mult.</td>
<td>dp[i][j] for range i to j</td>
</tr>
<tr>
<td><strong>Grid</strong></td>
<td>Unique Paths, Min Path Sum</td>
<td>dp[i][j] for grid position</td>
</tr>
<tr>
<td><strong>String</strong></td>
<td>LCS, Edit Distance</td>
<td>dp[i][j] for indices in two strings</td>
</tr>
<tr>
<td><strong>Knapsack</strong></td>
<td>0/1 Knapsack, Coin Change, Subset Sum</td>
<td>Include or exclude item</td>
</tr>
<tr>
<td><strong>Tree</strong></td>
<td>House Robber III, Tree Diameter</td>
<td>DFS + memoization on tree nodes</td>
</tr>
</tbody>
</table>
<div class="tip-box">
<div class="label">Pattern Recognition</div>
<p>When you see a new DP problem, ask: "Which category is this?" If it involves a grid, try grid DP. If two strings, try string DP. If choosing to include/exclude items, try knapsack. This narrows down your approach immediately.</p>
</div>
</section>
<!-- ======================= SECTION 9 ======================= -->
<section id="leetcode">
<h2>9. LeetCode Problems by Category</h2>
<h3>1D DP</h3>
<table>
<thead><tr><th>Problem</th><th>Difficulty</th><th>Key Pattern</th></tr></thead>
<tbody>
<tr><td>70. Climbing Stairs</td><td><span class="tag green">Easy</span></td><td>Fibonacci variant</td></tr>
<tr><td>198. House Robber</td><td><span class="tag orange">Medium</span></td><td>Include/exclude adjacent</td></tr>
<tr><td>322. Coin Change</td><td><span class="tag orange">Medium</span></td><td>Unbounded knapsack</td></tr>
<tr><td>300. Longest Increasing Subsequence</td><td><span class="tag orange">Medium</span></td><td>LIS pattern</td></tr>
<tr><td>139. Word Break</td><td><span class="tag orange">Medium</span></td><td>String segmentation</td></tr>
<tr><td>91. Decode Ways</td><td><span class="tag orange">Medium</span></td><td>Fibonacci-like with conditions</td></tr>
</tbody>
</table>
<h3>2D DP</h3>
<table>
<thead><tr><th>Problem</th><th>Difficulty</th><th>Key Pattern</th></tr></thead>
<tbody>
<tr><td>62. Unique Paths</td><td><span class="tag orange">Medium</span></td><td>Grid traversal</td></tr>
<tr><td>1143. Longest Common Subsequence</td><td><span class="tag orange">Medium</span></td><td>Two-string comparison</td></tr>
<tr><td>72. Edit Distance</td><td><span class="tag orange">Medium</span></td><td>Two-string transformation</td></tr>
<tr><td>494. Target Sum</td><td><span class="tag orange">Medium</span></td><td>Knapsack variant</td></tr>
</tbody>
</table>
<h3>Classic Patterns</h3>
<table>
<thead><tr><th>Problem</th><th>Difficulty</th><th>Key Pattern</th></tr></thead>
<tbody>
<tr><td>0/1 Knapsack</td><td><span class="tag orange">Medium</span></td><td>Include/exclude with capacity</td></tr>
<tr><td>Unbounded Knapsack (Coin Change)</td><td><span class="tag orange">Medium</span></td><td>Reusable items with capacity</td></tr>
</tbody>
</table>
<div class="tip-box">
<div class="label">Study Order</div>
<p>Do them in this order: Climbing Stairs, House Robber, Coin Change, Unique Paths, LCS, then LIS. Each one builds on the concepts from the previous ones. After these six, the rest are variations.</p>
</div>
</section>
<!-- ======================= SECTION 10 ======================= -->
<section id="tips">
<h2>10. Tips for DP</h2>