Leetcode 55. Jump Game

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

Analysis:

Typical greedy algorithm.

We need to have  a global maxReach. Then go through the whole array. Update the maxReach value between itself and nums[i] + i. Then check if it’s greater than nums.length – 1

Time complexity: O(n)

Space complexity: O(1)

Code is below:

public class Solution {
    public boolean canJump(int[] nums) {
        // O(n)
        
        if (nums == null || nums.length == 0) {
            return false;
        }
        
        int maxReach = 0;
        for (int i = 0; i < nums.length && i <= maxReach; i++) {             maxReach = Math.max(maxReach, nums[i] + i);             if (maxReach >= nums.length - 1) {
                return true;
            }
        }
        
        return false;
    }
}