Jump Game II LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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 Solutions

Time: O(n)
Space: O(1)

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;
  }
};

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;
  }
}

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

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