Jump Game II LeetCode Solution

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

Jump Game II 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 C++ Solution ->

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 Java Solution ->

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 JavaScript Solution ->

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 Python Solution ->

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)