45. Jump Game II

Medium LeetCode

You are given a 0-indexed array of integers nums of length n. You are initially positioned at index 0. Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at index i, you can jump to any index (i + j) where: 0 <= j <= nums[i] and i + j < n. Return the minimum number of jumps to reach index n - 1. The test cases are generated such that you can reach index 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
Explanation: The minimum number of jumps to reach the last index is 2.

Constraints:

• 1 <= nums.length <= 10^4
• 0 <= nums[i] <= 1000
• It's guaranteed that you can reach nums[n - 1].

How to solve the problem

• Greedy

class Solution:
    def jump(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return 0
        end = 0  # current jump boundary
        far = 0  # farthest index we can reach
        result = 0
        for i in range(len(nums)-1):
            far = max(far, i+nums[i])
            if i == end:  # end of current jump, need one more jump
                result += 1
                end = far
                if end >= len(nums) - 1:
                    return result
        return result

Complexity

• Greedy: Time complexity: O(n), Space complexity: O(1) - Where n is the length of the array. We traverse the array once and use constant extra space.