Maximum Number of Jumps to Reach the Last Index - LeetCode Medium | Dynamic Programming | O(n^2) #Sh

Añadido: 2026-05-16

[00:00:00]What if I told you the secret to maximum jumps is dynamic programming?

[00:00:04]We're given an array of integers and a target value, and we need to find the maximum number of jumps to reach the last index.

[00:00:10]A brute force approach would be to try all possible jumps, but that's not efficient.

[00:00:15]Instead, we can use dynamic programming to build up a solution.

[00:00:19]We define a dynamic programming array DP, where DP I represents the maximum number of jumps from index zero to index I.

[00:00:26]We initialize DP zero to zero, since we're already at index zero, and then we iterate through the array.

[00:00:32]For each index I, we check all previous indices J that we can jump from and update DP I if we find a longer jump sequence.

[00:00:40]Finally, we return DP N1, which represents the maximum number of jumps to reach the last index.

[00:00:46]The time complexity is O N squared, and the space complexity is O N, where N is the length of the array.

[00:00:52]Let's take a look at the code to see how it works.

[00:00:55]Don't forget to like and subscribe for more LeetCode solutions and explanations.