Last updated on February 22nd, 2024 at 03:58 am

Here, We see Jump Game II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

List of all LeetCode Solution

Jump Game II LeetCode Solution

Problem Statement

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].

Example 1:
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.

Example 2:
Input: nums = [2,3,0,1,4]
Output: 2

Jump Game II LeetCode Solution C++

class Solution {
public:
    int jump(vector<int>& nums) {
        int n = nums.size(), step = 0, start = 0, end = 0;
        while (end < n - 1) {
            step++; 
			int maxend = end + 1;
			for (int i = start; i <= end; i++) {
                if (i + nums[i] >= n - 1) return step;
				maxend = max(maxend, i + nums[i]);
			}
            start = end + 1;
            end = maxend;
        }
		return step;        
    }
};
Code language: C++ (cpp)

Jump Game II LeetCode Solution Java

class Solution {
    public int jump(int[] nums) {
        final int size = nums.length;
        int destination = size-1;
        int curCoverage = 0, lastJumpIdx = 0;
        int timesOfJump = 0;
        if( size == 1 ){
            return 0;
        }
        for( int i = 0 ; i < size ; i++){
            curCoverage = Math.max(curCoverage, i + nums[i] );
            if( i == lastJumpIdx ){
                lastJumpIdx = curCoverage;
                timesOfJump++;
                if( curCoverage >= destination){
                    return timesOfJump;
                }
            }
        }
        return timesOfJump;        
    }
}
Code language: Java (java)

Jump Game II LeetCode Solution Javascript

var jump = function(nums) {
    const size = nums.length;
    let destination = size-1;
    let curCoverage = 0, lastJumpIdx = 0;
    let timesOfJump = 0;
    if( size == 1 ){
        return 0;
    }
    for( let i = 0 ; i < size ; i++){
        curCoverage = Math.max(curCoverage, i + nums[i] );
        if( i == lastJumpIdx ){
            lastJumpIdx = curCoverage;
            timesOfJump++;
            if( curCoverage >= destination){
                return timesOfJump;
            }
        }
    }
    return timesOfJump;    
};
Code language: JavaScript (javascript)

Jump Game II LeetCode Solution Python

class Solution(object):
    def jump(self, nums):
        n, start, end, step = len(nums), 0, 0, 0
        while end < n - 1:
            step += 1
            maxend = end + 1
            for i in range(start, end + 1):
                if i + nums[i] >= n - 1:
                    return step
                maxend = max(maxend, i + nums[i])
            start, end = end + 1, maxend
        return step
Code language: Python (python)