Dynamic Programming

Back to DSA & Algorithms/00 - Roadmap Overview

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What is Dynamic Programming?

Dynamic Programming (DP) is an optimization technique for problems with:

1. Optimal substructure: The optimal solution contains optimal solutions to subproblems
2. Overlapping subproblems: The same subproblems are solved multiple times

Instead of recomputing, DP stores results and reuses them.

The Mental Model

Imagine computing Fibonacci naively:

fib(5) = fib(4) + fib(3)
fib(4) = fib(3) + fib(2)    ← fib(3) computed AGAIN
fib(3) = fib(2) + fib(1)    ← fib(2) computed AGAIN

Without DP: ~15 function calls (exponential)
With DP: 5 lookups (linear)

Two Approaches

java
// Top-Down: Add cache to recursion
import java.util.HashMap;

public class Fibonacci {

    private static HashMap<Integer, Integer> memo = new HashMap<>();

    public static int fibTopDown(int n) {
        if (n <= 1) return n;
        if (memo.containsKey(n)) return memo.get(n);
        int result = fibTopDown(n - 1) + fibTopDown(n - 2);
        memo.put(n, result);
        return result;
    }

    // Bottom-Up: Fill table iteratively
    public static int fibBottomUp(int n) {
        if (n <= 1) return n;
        int[] dp = new int[n + 1];
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }

    // Space-Optimized: Only keep what you need
    public static int fibOptimized(int n) {
        if (n <= 1) return n;
        int prev2 = 0, prev1 = 1;
        for (int i = 2; i <= n; i++) {
            int curr = prev1 + prev2;
            prev2 = prev1;
            prev1 = curr;
        }
        return prev1;
    }
}

The DP Framework (for EVERY problem)

1. Define the state: What variables describe a subproblem? → dp[i], dp[i][j]
2. Find the transition: How does dp[i] relate to smaller subproblems?
3. Determine base cases: What are the smallest subproblems?
4. Identify the answer: Which cell in the table is the final answer?
5. Optimize space (optional): Can you reduce the table to fewer rows/variables?

How to Identify DP Problems

Signal phrases:

• "Find the minimum cost to..."
• "Find the maximum profit from..."
• "How many ways to..."
• "Is it possible to..."
• "Find the longest/shortest subsequence/path..."

Ask: "Does making a choice now affect future choices?" If yes → likely DP.

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Pattern 1: 1D DP (Linear Sequence)

What it is

State depends on previous elements in a sequence. dp[i] = answer considering the first i elements.

Climbing Stairs

java
// How many ways to reach step n? Can climb 1 or 2 steps at a time.

// State: dp[i] = number of ways to reach step i
// Transition: dp[i] = dp[i-1] + dp[i-2]
//   (you can reach step i from step i-1 or step i-2)
// Base: dp[0] = 1, dp[1] = 1

public static int climbStairs(int n) {
    if (n <= 2) return n;
    int prev2 = 1, prev1 = 2;
    for (int i = 3; i <= n; i++) {
        int curr = prev1 + prev2;
        prev2 = prev1;
        prev1 = curr;
    }
    return prev1;
}

House Robber

java
// Can't rob adjacent houses. Maximize stolen amount.
// [2, 7, 9, 3, 1] → Rob houses 0, 2, 4 = 2+9+1=12

// State: dp[i] = max money from first i houses
// Transition: dp[i] = max(dp[i-1], dp[i-2] + nums[i])
//   Either skip house i (take dp[i-1])
//   Or rob house i (take dp[i-2] + nums[i])

public static int rob(int[] nums) {
    if (nums.length == 0) return 0;
    if (nums.length == 1) return nums[0];

    int prev2 = 0, prev1 = 0;
    for (int num : nums) {
        int curr = Math.max(prev1, prev2 + num);
        prev2 = prev1;
        prev1 = curr;
    }
    return prev1;
}

// Walkthrough: [2, 7, 9, 3, 1]
// num=2: curr = max(0, 0+2) = 2.  prev2=0, prev1=2
// num=7: curr = max(2, 0+7) = 7.  prev2=2, prev1=7
// num=9: curr = max(7, 2+9) = 11. prev2=7, prev1=11
// num=3: curr = max(11, 7+3) = 11. prev2=11, prev1=11
// num=1: curr = max(11, 11+1) = 12. Answer: 12

Longest Increasing Subsequence (LIS)

java
// Find length of longest strictly increasing subsequence.
// [10, 9, 2, 5, 3, 7, 101, 18] → [2, 3, 7, 101] length 4

// O(n²) approach:
// State: dp[i] = length of LIS ending at index i
// Transition: dp[i] = max(dp[j] + 1) for all j < i where nums[j] < nums[i]

import java.util.Arrays;
import java.util.ArrayList;

public static int lengthOfLIS(int[] nums) {
    int n = nums.length;
    int[] dp = new int[n];
    Arrays.fill(dp, 1);  // Every element is an LIS of length 1

    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (nums[j] < nums[i]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
    }

    int max = 0;
    for (int val : dp) max = Math.max(max, val);
    return max;
}

// O(n log n) approach using binary search:
// Maintain a "tails" array where tails[i] = smallest tail element
// for an increasing subsequence of length i+1

public static int lengthOfLISFast(int[] nums) {
    ArrayList<Integer> tails = new ArrayList<>();
    for (int num : nums) {
        int pos = Collections.binarySearch(tails, num);
        if (pos < 0) pos = -(pos + 1);  // insertion point
        if (pos == tails.size()) {
            tails.add(num);
        } else {
            tails.set(pos, num);
        }
    }
    return tails.size();
}

Problems

• 70. Climbing Stairs - Easy
• 198. House Robber - Medium
• 213. House Robber II - Medium
• 91. Decode Ways - Medium
• 139. Word Break - Medium
• 300. Longest Increasing Subsequence - Medium
• 152. Maximum Product Subarray - Medium

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Pattern 2: 2D DP (Grid / Two Sequences)

What it is

State depends on two variables. Either navigating a grid or comparing two sequences.

Grid: Unique Paths

java
// Robot moves right or down. Count paths from top-left to bottom-right.

// State: dp[i][j] = number of paths to reach cell (i, j)
// Transition: dp[i][j] = dp[i-1][j] + dp[i][j-1]
// Base: dp[0][j] = 1, dp[i][0] = 1 (only one way along edges)

public static int uniquePaths(int m, int n) {
    int[][] dp = new int[m][n];
    for (int i = 0; i < m; i++) dp[i][0] = 1;
    for (int j = 0; j < n; j++) dp[0][j] = 1;

    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m - 1][n - 1];
}

Two Sequences: Longest Common Subsequence

java
// Find longest common subsequence of two strings.
// "abcde" and "ace" → "ace" (length 3)

// State: dp[i][j] = LCS length of text1[0..i-1] and text2[0..j-1]
// Transition:
//   If text1.charAt(i-1) == text2.charAt(j-1): dp[i][j] = dp[i-1][j-1] + 1 (extend match)
//   Else: dp[i][j] = max(dp[i-1][j], dp[i][j-1]) (skip one character)

public static int longestCommonSubsequence(String text1, String text2) {
    int m = text1.length(), n = text2.length();
    int[][] dp = new int[m + 1][n + 1];

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }

    return dp[m][n];
}

Edit Distance (Levenshtein)

java
// Minimum operations (insert, delete, replace) to convert word1 → word2
// "horse" → "ros" = 3

// State: dp[i][j] = edit distance between word1[0..i-1] and word2[0..j-1]
// Transition:
//   Match:    dp[i][j] = dp[i-1][j-1]  (no operation needed)
//   Replace:  dp[i][j] = dp[i-1][j-1] + 1
//   Insert:   dp[i][j] = dp[i][j-1] + 1   (insert char from word2)
//   Delete:   dp[i][j] = dp[i-1][j] + 1   (delete char from word1)

public static int minDistance(String word1, String word2) {
    int m = word1.length(), n = word2.length();
    int[][] dp = new int[m + 1][n + 1];

    // Base cases
    for (int i = 0; i <= m; i++) dp[i][0] = i;  // Delete all from word1
    for (int j = 0; j <= n; j++) dp[0][j] = j;  // Insert all from word2

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                dp[i][j] = dp[i - 1][j - 1];
            } else {
                dp[i][j] = 1 + Math.min(
                    dp[i - 1][j - 1],  // Replace
                    Math.min(
                        dp[i - 1][j],   // Delete
                        dp[i][j - 1]    // Insert
                    )
                );
            }
        }
    }

    return dp[m][n];
}

Problems

• 62. Unique Paths - Medium
• 64. Minimum Path Sum - Medium
• 1143. Longest Common Subsequence - Medium
• 72. Edit Distance - Medium
• 97. Interleaving String - Medium

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Pattern 3: Knapsack

What it is

Choose items with weights/values to maximize value within a capacity constraint.

0/1 Knapsack (Each Item Used At Most Once)

java
// State: dp[i][w] = max value using first i items with capacity w
// Transition: dp[i][w] = max(
//     dp[i-1][w],                          // Skip item i
//     dp[i-1][w - weight[i]] + value[i]    // Take item i (if it fits)
// )

public static int knapsack(int[] weights, int[] values, int capacity) {
    int n = weights.length;
    int[][] dp = new int[n + 1][capacity + 1];

    for (int i = 1; i <= n; i++) {
        for (int w = 0; w <= capacity; w++) {
            dp[i][w] = dp[i - 1][w];  // Skip
            if (weights[i - 1] <= w) {
                dp[i][w] = Math.max(dp[i][w],
                    dp[i - 1][w - weights[i - 1]] + values[i - 1]);
            }
        }
    }

    return dp[n][capacity];
}

// Space optimization: only need previous row
public static int knapsackOptimized(int[] weights, int[] values, int capacity) {
    int[] dp = new int[capacity + 1];
    for (int i = 0; i < weights.length; i++) {
        for (int w = capacity; w >= weights[i]; w--) {  // RIGHT TO LEFT!
            dp[w] = Math.max(dp[w], dp[w - weights[i]] + values[i]);
        }
    }
    return dp[capacity];
}

// Why right to left? To avoid using the same item twice.
// If left to right: dp[w] might use updated dp[w - weight[i]] from THIS row
// Right to left ensures we use values from the PREVIOUS row.

Unbounded Knapsack (Items Can Be Reused)

java
// Coin Change: minimum coins to make amount
// State: dp[a] = min coins to make amount a
// Transition: dp[a] = min(dp[a - coin] + 1) for each coin

public static int coinChange(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, Integer.MAX_VALUE);
    dp[0] = 0;

    for (int a = 1; a <= amount; a++) {
        for (int coin : coins) {
            if (coin <= a && dp[a - coin] != Integer.MAX_VALUE) {
                dp[a] = Math.min(dp[a], dp[a - coin] + 1);
            }
        }
    }

    return dp[amount] != Integer.MAX_VALUE ? dp[amount] : -1;
}

Subset Sum / Partition

java
// Can we partition array into two subsets with equal sum?
// [1, 5, 11, 5] → {1, 5, 5} and {11}, both sum to 11

public static boolean canPartition(int[] nums) {
    int total = 0;
    for (int num : nums) total += num;
    if (total % 2 != 0) return false;  // Odd total can't be split equally

    int target = total / 2;
    boolean[] dp = new boolean[target + 1];
    dp[0] = true;

    for (int num : nums) {
        for (int j = target; j >= num; j--) {  // Right to left (0/1)
            dp[j] = dp[j] || dp[j - num];
        }
    }

    return dp[target];
}

Problems

• 416. Partition Equal Subset Sum - Medium
• 494. Target Sum - Medium
• 322. Coin Change - Medium
• 518. Coin Change II - Medium

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Pattern 4: State Machine DP

What it is

Define states (like "holding stock", "in cooldown") and transitions between them.

Stock Problems

java
// Buy and Sell Stock with Cooldown
// States: hold (have stock), sold (just sold, must cooldown), rest (no stock, can buy)

public static int maxProfit(int[] prices) {
    int hold = -prices[0];  // Bought on day 0
    int sold = 0;           // Just sold (in cooldown)
    int rest = 0;           // No stock, free to buy

    for (int i = 1; i < prices.length; i++) {
        int prevHold = hold;
        hold = Math.max(hold, rest - prices[i]);    // Keep holding OR buy
        rest = Math.max(rest, sold);                 // Keep resting OR cooldown finished
        sold = prevHold + prices[i];                 // Sell what we're holding
    }

    return Math.max(sold, rest);
}

// Each day we transition between states:
// rest → hold (buy)
// hold → sold (sell)
// sold → rest (cooldown ends)
// hold → hold (keep holding)
// rest → rest (keep resting)

Problems

• 121. Best Time to Buy and Sell Stock - Easy
• 122. Best Time to Buy and Sell Stock II - Medium
• 309. Best Time to Buy and Sell Stock with Cooldown - Medium
• 188. Best Time to Buy and Sell Stock IV - Hard

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Pattern 5: Interval DP

What it is

Solve for all intervals [i, j], building from small intervals to large ones.

java
// Longest Palindromic Subsequence
public static int longestPalindromeSubseq(String s) {
    int n = s.length();
    int[][] dp = new int[n][n];

    for (int i = 0; i < n; i++) {
        dp[i][i] = 1;  // Single characters are palindromes
    }

    for (int length = 2; length <= n; length++) {
        for (int i = 0; i <= n - length; i++) {
            int j = i + length - 1;
            if (s.charAt(i) == s.charAt(j)) {
                dp[i][j] = dp[i + 1][j - 1] + 2;
            } else {
                dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
            }
        }
    }

    return dp[0][n - 1];
}

Problems

• 516. Longest Palindromic Subsequence - Medium
• 5. Longest Palindromic Substring - Medium
• 312. Burst Balloons - Hard

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Things You MUST Be Aware Of

1. Start with Brute Force Recursion

java
// ALWAYS start by writing the recursive solution.
// Then add memoization (HashMap<String, Integer> or int[][] memo with sentinel values).
// Then convert to bottom-up if needed.
// Don't try to jump straight to bottom-up — it's error-prone.

2. Space Optimization

java
// If dp[i] only depends on dp[i-1] → use 2 rows or 2 variables
// If dp[i][j] only depends on dp[i-1][j] and dp[i][j-1] → use 1 row

// 0/1 knapsack: iterate RIGHT TO LEFT to avoid reusing items
// Unbounded knapsack: iterate LEFT TO RIGHT (reuse is allowed)

3. Coin Change: Min vs Count

java
// Minimum coins (optimization): dp[a] = Math.min(dp[a], dp[a - c] + 1)
// Number of ways (counting):    for each coin c: dp[a] += dp[a - c]
// Different transitions for different questions!

4. The Hardest Part is Defining the State

java
// If your state doesn't capture enough info, you'll get wrong answers.
// If your state has too many dimensions, you'll be too slow.
// Ask: "What do I need to know to make the next decision?"

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DP Identification Cheat Sheet