# Jump Game II LeetCode Solution

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

## 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)```
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