Leetcode 45. Jump Game II

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]. 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

It can be solved by using simple BFS like (or you can say it’s a kind of two pointers)

class Solution:
    def jump(self, nums: List[int]) -> int:
        jump = 0

        l = r = 0

        while r < len(nums) - 1:
            max_range = l
            for index in range(l, r + 1):
                max_range = max(max_range, index + nums[index])
            
            l = r + 1
            r = max_range
            jump += 1
        
        return jump