# Jump Game LeetCode Solution

Here, We see Jump Game 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 an integer array `nums`. You are initially positioned at the array’s first index, and each element in the array represents your maximum jump length at that position.

Return `true` if you can reach the last index, or `false` otherwise.

```Example 1:

Input: nums = [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: nums = [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
```

## Jump Game Leetcode Solution C++

``````class Solution {
public:
bool canJump(vector<int>& nums) {
int i, minjump = 0;
for(i = nums.size()-2; i >= 0; i--){
minjump++;
if(nums[i] >= minjump)
minjump = 0;
}
if(minjump == 0)
return true;
else
return false;
}
};```Code language: PHP (php)```

## Jump Game Leetcode Solution Java

``````class Solution {
public boolean canJump(int[] nums) {
if(nums.length < 2) return true;

for(int curr = nums.length-2; curr>=0;curr--){
if(nums[curr] == 0){
int neededJumps = 1;
while(neededJumps > nums[curr]){
neededJumps++;
curr--;
if(curr < 0) return false;
}
}
}
return true;
}
}```Code language: PHP (php)```

## Jump Game Leetcode Solution JavaScript

``````var canJump = function(nums) {
let idx = 0;
let max = 0;
let target = nums.length - 1;
while(idx < nums.length) {
max = Math.max(max, idx + nums[idx]);
if (max >= target) {
return true;
}
if (max <= idx && nums[idx] === 0) {
return false;
}
idx++;
}
return false;
};```Code language: JavaScript (javascript)```

## Jump Game Leetcode Solution Python

``````class Solution(object):
def canJump(self, nums):
max_reach, n = 0, len(nums)
for i, x in enumerate(nums):
if max_reach < i: return False
if max_reach >= n - 1: return True
max_reach = max(max_reach, i + x)```Code language: HTML, XML (xml)```
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