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Dynamic Programming
Last Updated : 26 May, 2026
title: Dynamic Programming in Java
DSA in Java49 words6 headingsExamples included
title: Dynamic Programming in Java description: DP concepts aur classic problems
Complex problem ko overlapping subproblems mein todna. Memoization ya Tabulation.
1. Fibonacci — Tabulation (Bottom-up)
long fib(int n) {
if (n <= 1) return n;
long[] dp = new long[n + 1];
dp[0] = 0; dp[1] = 1;
for (int i = 2; i <= n; i++) dp[i] = dp[i-1] + dp[i-2];
return dp[n];
}
2. 0/1 Knapsack
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]; // Don't take item i
if (weights[i-1] <= w) // Take item i
dp[i][w] = Math.max(dp[i][w], values[i-1] + dp[i-1][w - weights[i-1]]);
}
}
return dp[n][capacity];
}
3. Longest Common Subsequence
int lcs(String s1, String s2) {
int m = s1.length(), n = s2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
if (s1.charAt(i-1) == s2.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];
}
4. Coin Change (Min coins)
int minCoins(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int i = 1; i <= amount; i++)
for (int coin : coins)
if (coin <= i) dp[i] = Math.min(dp[i], dp[i - coin] + 1);
return dp[amount] > amount ? -1 : dp[amount];
}
5. Longest Increasing Subsequence
int lis(int[] arr) {
int n = arr.length;
int[] dp = new int[n];
Arrays.fill(dp, 1);
int maxLen = 1;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++)
if (arr[j] < arr[i]) dp[i] = Math.max(dp[i], dp[j] + 1);
maxLen = Math.max(maxLen, dp[i]);
}
return maxLen;
}
6. Edit Distance
int editDistance(String s1, String s2) {
int m = s1.length(), n = s2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) dp[i][0] = i;
for (int j = 0; j <= n; j++) dp[0][j] = j;
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
if (s1.charAt(i-1) == s2.charAt(j-1)) dp[i][j] = dp[i-1][j-1];
else dp[i][j] = 1 + Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1]));
return dp[m][n];
}
Exam Focus
Revise definitions, diagrams, examples, and short-answer points for Dynamic Programming.
Interview Use
Prepare one clear explanation, one practical example, and one common mistake for this Java topic.
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