In this post, we are going to solve the Jump Game II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Problem
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]
andi + j < n
Return the minimum number of jumps to reach nums[n - 1]
. The test cases are generated such that you can 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
Constraints:
1 <= nums.length <= 104
0 <= nums[i] <= 1000
Now, let’s see the leetcode solution of Jump Game II Leetcode Solution.
Jump Game II Leetcode Solution in Python
class Solution: def jump(self, nums: List[int]) -> int: ans = 0 end = 0 farthest = 0 # Implicit BFS for i in range(len(nums) - 1): farthest = max(farthest, i + nums[i]) if farthest >= len(nums) - 1: ans += 1 break if i == end: # Visited all the items on the current level ans += 1 # Increment the level end = farthest # Make the queue size for the next level return ans
Jump Game II Leetcode Solution in CPP
class Solution { public: int jump(vector<int>& nums) { int ans = 0; int end = 0; int farthest = 0; // Implicit BFS 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) { // Visited all the items on the current level ++ans; // Increment the level end = farthest; // Make the queue size for the next level } } return ans; } };
Jump Game II Leetcode Solution in Java
class Solution { public int jump(int[] nums) { int ans = 0; int end = 0; int farthest = 0; // Implicit BFS 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) { // Visited all the items on the current level ++ans; // Increment the level end = farthest; // Make the queue size for the next level } } return ans; } }
Note: This problem Jump Game II is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.