You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

• 0 <= j <= nums[i] and
• i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

Example

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

/**
 * @param {number[]} nums
 * @return {number}
 */
var jump = function(N) {
    //Create a variable to store the length of the array
    let len = N.length - 1, curr = -1, next = 0, ans = 0
    //Loop through the array
    for (let i = 0; next < len; i++) {
        //Check if the current index is greater than the current
        if (i > curr) ans++, curr = next
        //Set the next index
        next = Math.max(next, N[i] + i)
    }
    //Return ans
    return ans
};