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# Jump Game II LeetCode Solution

## Problem – Jump Game II LeetCode Solution

Given an array of non-negative integers `nums`, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

You can assume that you can always reach the last index.

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``````

Constraints:

• `1 <= nums.length <= 104`
• `0 <= nums[i] <= 1000`

## Jump Game II LeetCode Solution in Java

``````public int jump(int[] A) {
int jumps = 0, curEnd = 0, curFarthest = 0;
for (int i = 0; i < A.length - 1; i++) {
curFarthest = Math.max(curFarthest, i + A[i]);
if (i == curEnd) {
jumps++;
curEnd = curFarthest;
}
}
return jumps;
}
``````

## Jump Game II LeetCode Solution in Python

``````    def jump(self, nums):
if len(nums) <= 1: return 0
l, r = 0, nums
times = 1
while r < len(nums) - 1:
times += 1
nxt = max(i + nums[i] for i in range(l, r + 1))
l, r = r, nxt
return times
``````

## Jump Game II LeetCode Solution in 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;
}
};
``````
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