README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0045.Jump%20Game%20II/README_EN.md	Greedy Array Dynamic Programming

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
• It's guaranteed that you can reach nums[n - 1].

Solutions

Solution 1: Greedy Algorithm

We can use a variable $mx$ to record the farthest position that can be reached from the current position, a variable $last$ to record the position of the last jump, and a variable $ans$ to record the number of jumps.

Next, we traverse each position $i$ in $[0,..n - 2]$. For each position $i$, we can calculate the farthest position that can be reached from the current position through $i + nums[i]$. We use $mx$ to record this farthest position, that is, $mx = max(mx, i + nums[i])$. Then, we check whether the current position has reached the boundary of the last jump, that is, $i = last$. If it has reached, then we need to make a jump, update $last$ to $mx$, and increase the number of jumps $ans$ by $1$.

Finally, we return the number of jumps $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Similar problems:

• 55. Jump Game
• 1024. Video Stitching
• 1326. Minimum Number of Taps to Open to Water a Garden

Python3

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

Java

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

C++

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

Go

func jump(nums []int) (ans int) {
	mx, last := 0, 0
	for i, x := range nums[:len(nums)-1] {
		mx = max(mx, i+x)
		if last == i {
			ans++
			last = mx
		}
	}
	return
}

TypeScript

function jump(nums: number[]): number {
    let [ans, mx, last] = [0, 0, 0];
    for (let i = 0; i < nums.length - 1; ++i) {
        mx = Math.max(mx, i + nums[i]);
        if (last === i) {
            ++ans;
            last = mx;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn jump(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut mx = 0;
        let mut last = 0;
        for i in 0..(nums.len() - 1) {
            mx = mx.max(i as i32 + nums[i]);
            if last == i as i32 {
                ans += 1;
                last = mx;
            }
        }
        ans
    }
}

C#

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

C

int jump(int* nums, int numsSize) {
    int ans = 0;
    int mx = 0;
    int last = 0;
    for (int i = 0; i < numsSize - 1; ++i) {
        mx = (mx > i + nums[i]) ? mx : (i + nums[i]);
        if (last == i) {
            ++ans;
            last = mx;
        }
    }
    return ans;
}

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function jump($nums) {
        $ans = 0;
        $mx = 0;
        $last = 0;

        for ($i = 0; $i < count($nums) - 1; $i++) {
            $mx = max($mx, $i + $nums[$i]);
            if ($last == $i) {
                $ans++;
                $last = $mx;
            }
        }

        return $ans;
    }
}