Lecture 64 :- Aggresive Cows

The Aggressive Cows problem is a classic optimization problem related to placing cows in stalls. The problem is stated as follows:

You are given N stalls in a row and C cows. Each stall can have at most one cow. You have to place the C cows in the stalls such that the minimum distance between any two cows is maximized. Your task is to find this maximum minimum distance.

To solve this problem, we can use binary search to find the maximum minimum distance between cows. We will set the search space to be the range from 1 (minimum distance possible) to the maximum distance between the first and last stall. Then, we will check if it is possible to place the cows in such a way that the minimum distance between them is greater than or equal to the mid value (the value we are currently searching for). If it is possible, we try to maximize this minimum distance by moving to the right half of the search space. Otherwise, we move to the left half.

Here's the C++ implementation of the Aggressive Cows problem:

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#include <iostream> #include <vector> #include <algorithm> bool isPossible(std::vector<int>& stalls, int cows, int distance) {    int count = 1;    int prev = stalls[0];    for (int i = 1; i < stalls.size(); i++) {        if (stalls[i] - prev >= distance) {            count++;            prev = stalls[i];        }    }    return count >= cows; } int maxMinDistance(std::vector<int>& stalls, int cows) {    std::sort(stalls.begin(), stalls.end());    int left = 1;    int right = stalls[stalls.size() - 1] - stalls[0];    int result = -1;    while (left <= right) {        int mid = left + (right - left) / 2;        if (isPossible(stalls, cows, mid)) {            result = mid;            left = mid + 1;        } else {            right = mid - 1;        }    }    return result; } int main() {    std::vector<int> stalls = {1, 2, 8, 4, 9};    int cows = 3;    int maxMinDist = maxMinDistance(stalls, cows);    std::cout << "Maximum minimum distance between cows: " << maxMinDist << std::endl;    return 0; }

Example Output:

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Maximum minimum distance between cows: 3

In this code, the isPossible function checks if it is possible to place the cows in such a way that the minimum distance between them is greater than or equal to distance. The maxMinDistance function uses binary search to find the maximum minimum distance between cows. The time complexity of this algorithm is O(N * log(maxDist)), where N is the number of stalls and maxDist is the maximum distance between the first and last stall.