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  1. 11-November

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

Previous03. Build an Array With Stack OperationsNext07. Eliminate Maximum Number of Monsters

Last updated 1 year ago

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The problem can be found at the following link:

My Approach

  1. Initialize an integer variable lastFallTime to 0, which will be used to keep track of the last moment when an ant falls off the plank.

  2. Iterate through the positions of ants moving to the left (from the left vector):

    • For each leftPosition in the left vector, update lastFallTime by taking the maximum of its current value and the value of leftPosition. This represents the farthest position reached by ants moving to the left.

  3. Iterate through the positions of ants moving to the right (from the right vector):

    • For each rightPosition in the right vector, calculate the farthest position reached by ants moving to the right from the right end of the plank. To do this, subtract rightPosition from n (the length of the plank) to get the distance from the right end to the ant's position. Update lastFallTime by taking the maximum of its current value and this calculated value.

  4. After iterating through both the left and right positions, the lastFallTime variable will hold the last moment when an ant falls off the plank.

  5. Return the value of lastFallTime, which represents the last moment at which an ant falls off the plank.

Time and Auxiliary Space Complexity

  • Time Complexity: O(n)

  • Auxiliary Space Complexity: O(1)

Code (C++)


class Solution {
public:
    
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int lastFallTime = 0;
        
        // Determine the farthest position reached by ants moving to the left.
        for (int leftPosition : left) {
            lastFallTime = max(lastFallTime, leftPosition);
        }
        
        // Determine the farthest position reached by ants moving to the right,
        // considering they start from the right end of the plank (n - position).
        for (int rightPosition : right) {
            lastFallTime = max(lastFallTime, n - rightPosition);
        }
        
        return lastFallTime;
    }
};

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rishabhv12/Daily-Leetcode-Solution