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In this post, you will find the solution for the **Jump Game II** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Jump Game II– LeetCode Problem

Jump Game II– LeetCode Problem

**Problem:**

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:2Explanation: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 <= 10`

^{4}`0 <= nums[i] <= 1000`

Jump Game II– LeetCode Solutions

Jump Game II in C++

class Solution { public: int jump(vector<int>& nums) { int ans = 0; int end = 0; int farthest = 0; for (int i = 0; i < nums.size() - 1; ++i) { farthest = max(farthest, i + nums[i]); if (farthest >= nums.size() - 1) { ++ans; break; } if (i == end) { ++ans; end = farthest; } } return ans; } };

Jump Game II in Java

class Solution { public int jump(int[] nums) { int ans = 0; int end = 0; int farthest = 0; for (int i = 0; i < nums.length - 1; ++i) { farthest = Math.max(farthest, i + nums[i]); if (farthest >= nums.length - 1) { ++ans; break; } if (i == end) { ++ans; end = farthest; } } return ans; } }

Jump Game II in Python

class Solution: def jump(self, nums: List[int]) -> int: ans = 0 end = 0 farthest = 0 for i in range(len(nums) - 1): farthest = max(farthest, i + nums[i]) if farthest >= len(nums) - 1: ans += 1 break if i == end: ans += 1 end = farthest return ans