README_EN.md

45. Jump Game II

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Description

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

 

Constraints:

• 1 <= nums.length <= 104
• 0 <= nums[i] <= 1000

Solutions

Python3

class Solution:
    def jump(self, nums: List[int]) -> int:
        end = mx = steps = 0
        for i, num in enumerate(nums[:-1]):
            mx = max(mx, i + num)
            if i == end:
                end = mx
                steps += 1
        return steps

Java

class Solution {
    public int jump(int[] nums) {
        int end = 0;
        int mx = 0;
        int steps = 0;
        for (int i = 0; i < nums.length - 1; ++i) {
            mx = Math.max(mx, i + nums[i]);
            if (i == end) {
                end = mx;
                ++steps;
            }
        }
        return steps;
    }
}

C++

class Solution {
public:
    int jump(vector<int>& nums) {
        int mx = 0, steps = 0, end = 0;
        for (int i = 0; i < nums.size() - 1; ++i) {
            mx = max(mx, i + nums[i]);
            if (i == end) {
                end = mx;
                ++steps;
            }
        }
        return steps;
    }
};

Go

func jump(nums []int) int {
	mx, steps, end := 0, 0, 0
	for i := 0; i < len(nums)-1; i++ {
		mx = max(mx, i+nums[i])
		if i == end {
			end = mx
			steps++
		}
	}
	return steps
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

C#

public class Solution {
    public int Jump(int[] nums) {
        int end = 0;
        int mx = 0;
        int steps = 0;
        for (int i = 0; i < nums.Length - 1; ++i)
        {
            mx = Math.Max(mx, i + nums[i]);
            if (i == end)
            {
                end = mx;
                ++steps;
            }
        }
        return steps;
    }
}

C

#define min(a, b) a < b ? a : b
int jump(int* nums, int numsSize) {
    int dp[numsSize];
    for (int i = 0; i < numsSize; i++) {
        dp[i] = numsSize;
    }
    dp[0] = 0;
    for (int i = 0; i < numsSize - 1; i++) {
        for (int j = i + 1; j < (min(i + nums[i] + 1, numsSize)); j++) {
            dp[j] = min(dp[j], dp[i] + 1);
        }
    }
    return dp[numsSize - 1];
}

Rust

impl Solution {
    pub fn jump(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut dp = vec![i32::MAX; n];
        dp[0] = 0;
        for i in 0..n - 1 {
            for j in 1..=nums[i] as usize {
                if i + j >= n {
                    break;
                }
                dp[i + j] = dp[i + j].min(dp[i] + 1);
            }
        }
        dp[n - 1]
    }
}

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