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1503. Last Moment Before All Ants Fall Out of a Plank

Description

We have a wooden plank of the length n units. Some ants are walking on the plank, each ant moves with a speed of 1 unit per second. Some of the ants move to the left, the other move to the right.

When two ants moving in two different directions meet at some point, they change their directions and continue moving again. Assume changing directions does not take any additional time.

When an ant reaches one end of the plank at a time t, it falls out of the plank immediately.

Given an integer n and two integer arrays left and right, the positions of the ants moving to the left and the right, return the moment when the last ant(s) fall out of the plank.

 

Example 1:

Input: n = 4, left = [4,3], right = [0,1]
Output: 4
Explanation: In the image above:
-The ant at index 0 is named A and going to the right.
-The ant at index 1 is named B and going to the right.
-The ant at index 3 is named C and going to the left.
-The ant at index 4 is named D and going to the left.
The last moment when an ant was on the plank is t = 4 seconds. After that, it falls immediately out of the plank. (i.e., We can say that at t = 4.0000000001, there are no ants on the plank).

Example 2:

Input: n = 7, left = [], right = [0,1,2,3,4,5,6,7]
Output: 7
Explanation: All ants are going to the right, the ant at index 0 needs 7 seconds to fall.

Example 3:

Input: n = 7, left = [0,1,2,3,4,5,6,7], right = []
Output: 7
Explanation: All ants are going to the left, the ant at index 7 needs 7 seconds to fall.

 

Constraints:

  • 1 <= n <= 104
  • 0 <= left.length <= n + 1
  • 0 <= left[i] <= n
  • 0 <= right.length <= n + 1
  • 0 <= right[i] <= n
  • 1 <= left.length + right.length <= n + 1
  • All values of left and right are unique, and each value can appear only in one of the two arrays.

Solutions

Solution 1: Brain Teaser

The key point of the problem is that when two ants meet and then turn around, it is equivalent to the two ants continuing to move in their original directions. Therefore, we only need to find the maximum distance moved by any ant.

Note that the lengths of the $\textit{left}$ and $\textit{right}$ arrays may be $0$.

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

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class Solution:
    def getLastMoment(self, n: int, left: List[int], right: List[int]) -> int:
        ans = 0
        for x in left:
            ans = max(ans, x)
        for x in right:
            ans = max(ans, n - x)
        return ans

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class Solution {
    public int getLastMoment(int n, int[] left, int[] right) {
        int ans = 0;
        for (int x : left) {
            ans = Math.max(ans, x);
        }
        for (int x : right) {
            ans = Math.max(ans, n - x);
        }
        return ans;
    }
}

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class Solution {
public:
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int ans = 0;
        for (int& x : left) {
            ans = max(ans, x);
        }
        for (int& x : right) {
            ans = max(ans, n - x);
        }
        return ans;
    }
};

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func getLastMoment(n int, left []int, right []int) (ans int) {
    for _, x := range left {
        ans = max(ans, x)
    }
    for _, x := range right {
        ans = max(ans, n-x)
    }
    return
}

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function getLastMoment(n: number, left: number[], right: number[]): number {
    let ans = 0;
    for (const x of left) {
        ans = Math.max(ans, x);
    }
    for (const x of right) {
        ans = Math.max(ans, n - x);
    }
    return ans;
}

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