Leetcode 1503: Last Moment Before All Ants Fall Out of a Plank

Input: You are given two integer arrays: 'left' and 'right'. The array 'left' represents the positions of ants moving to the left, and the array 'right' represents the positions of ants moving to the right.

Example: n = 5, left = [3, 4], right = [1, 2]

Constraints:

• 1 <= n <= 10^4

• 0 <= left.length <= n + 1

• 0 <= right.length <= n + 1

• 1 <= left.length + right.length <= n + 1

• All values in 'left' and 'right' are unique, and no position is repeated in both arrays.

Output: The output should be an integer representing the time when the last ant falls off the plank.

Example: 6

Constraints:

• The time is a non-negative integer.

Goal: The goal is to compute the moment when the last ant falls off the plank.

Steps:

• For each ant moving to the right, calculate how much time it takes to fall off the right end of the plank.

• For each ant moving to the left, calculate how much time it takes to fall off the left end of the plank.

• The time when the last ant falls off is the maximum of all these times.

Goal: The array 'left' and 'right' represent positions on the plank, and the total number of ants must be between 1 and n + 1.

Steps:

• The plank's length, n, is between 1 and 10^4.

• Each position in 'left' and 'right' is distinct and falls within the range [0, n].

Assumptions:

• The input arrays 'left' and 'right' contain unique values.

• The time for ants to change directions when they meet is negligible.

• Input: n = 5, left = [3, 4], right = [1, 2]

• Explanation: The last ant to fall off is the one at position 4, which will take 5 seconds to reach the left end.

• Input: n = 6, left = [], right = [0, 1, 2, 3, 4, 5]

• Explanation: The ant at position 0 will take 6 seconds to fall off the plank, and all other ants fall off before it.

Approach: To solve the problem, calculate the time for each ant to fall off the plank. The solution is simply the maximum time of all ants.

Observations:

• The ants' movements are independent, so we just need to compute how long it will take for each ant to fall off the plank.

• The time for ants moving to the left is just their position, while the time for ants moving to the right is the plank's length minus their position.

Steps:

• Step 1: For ants in the 'right' array, compute the time as n - position.

• Step 2: For ants in the 'left' array, compute the time as position.

• Step 3: Return the maximum of these times as the answer.

Empty Inputs:

• If either 'left' or 'right' is empty, we only consider the times for the non-empty array.

Large Inputs:

• Ensure that the solution handles the maximum value of n, which is 10^4.

Special Values:

• If all ants are moving in one direction, the time will be the time it takes the farthest ant to fall off.

Constraints:

• The input arrays 'left' and 'right' should respect the given constraints.

int getLastMoment(int n, vector<int>& left, vector<int>& right) {
    
    int mx = 0;
    for(int i = 0; i < right.size(); i++)
        mx = max(mx, n - right[i]);
    
    for(int i = 0; i < left.size(); i++)
        mx = max(mx, left[i]);        
    
    return mx;
}

1 : Function Definition

int getLastMoment(int n, vector<int>& left, vector<int>& right) {

Defines the function that calculates the last moment any object in the `left` or `right` array reaches the boundary of size `n`.

2 : Variable Initialization

    int mx = 0;

Initializes a variable `mx` to store the maximum moment of time as the loop progresses.

3 : Loop

    for(int i = 0; i < right.size(); i++)

Loops through each element in the `right` array.

4 : Mathematical Operations

        mx = max(mx, n - right[i]);

For each element in the `right` array, calculates the moment when it reaches the right boundary. The time is updated in the `mx` variable.

5 : Loop

    for(int i = 0; i < left.size(); i++)

Loops through each element in the `left` array.

6 : Mathematical Operations

        mx = max(mx, left[i]);

For each element in the `left` array, it calculates the moment when it reaches the left boundary. The time is updated in the `mx` variable.

7 : Return

    return mx;

Returns the maximum time calculated from both the left and right arrays.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n), as we iterate through both the 'left' and 'right' arrays once.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant, as we only need a few variables to store the maximum time.

Leetcode 1503: Last Moment Before All Ants Fall Out of a Plank

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