Leetcode - 45. Jump Game II

The question says that we can be always able to reach the end , only that we need to find is the min steps we can take.

We need 0 jumps to be at index 0 since we are starting there

Now from 2 we can jump to either jump to 3 or 1

we can represent this window by l , r pointers

now the next jump window is from 1 to 4 ( since this is the point they can jump max from window [3,1]

CODE:

/**
 * @param {number[]} nums
 * @return {number}
 */
var jump = function(nums) {
    let result = 0 ;
    let l = 0 ;
    let r = 0;
    while(r<nums.length-1){

        let farthest = 0 ;
        for(let i=l;i<=r;i++){
            farthest = Math.max(farthest , i+ nums[i]);
        }
        l = r + 1 ;
        r = farthest;
        result++;
    }
    return result
};

Please go through the video by neetcode if you do not understand