Made the modification

Trie/Aggressive Cows Problem/ProgramAgg.c

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+#include <stdio.h>
+#include <stdlib.h>
+
+// Comparator function for qsort
+int compare(const void *a, const void *b) {
+    return (*(int *)a - *(int *)b);
+}
+
+// Function to check if it is possible to place cows with at least 'dist' distance apart
+int canPlaceCows(int* stalls, int stallsSize, int cows, int dist) {
+    int count = 1; // Place the first cow in the first stall
+    int lastPosition = stalls[0];
+
+    for (int i = 1; i < stallsSize; i++) {
+        if (stalls[i] - lastPosition >= dist) {
+            count++; // Place the next cow
+            lastPosition = stalls[i]; // Update the last position
+            if (count == cows) {
+                return 1; // Successfully placed all cows
+            }
+        }
+    }
+    return 0; // Not possible to place all cows
+}
+
+// Function to find the largest minimum distance
+int aggressiveCows(int* stalls, int stallsSize, int cows) {
+    // Sort the stalls
+    qsort(stalls, stallsSize, sizeof(int), compare);
+
+    int left = 1; // Minimum possible distance
+    int rightout = stalls[stallsSize - 1] - stalls[0]; // Maximum possible distance
+    int bestDistance = 0;
+
+    while (left <= right) {
+        int mid = left + (right - left) / 2;
+
+        if (canPlaceCows(stalls, stallsSize, cows, mid)) {
+            bestDistance = mid; // Update best distance found
+            left = mid + 1; // Try for a larger distance
+        } else {
+            right = mid - 1; // Try for a smaller distance
+        }
+    }
+
+    return bestDistance; // Return the largest minimum distance found
+}
+
+int main() {
+    int stalls[] = {1, 2, 8,6,4, 9}; // Example stalls
+    int cows = 3; // Number of cows
+    int size = sizeof(stalls) / sizeof(stalls[0]);
+
+    int resultout = aggressiveCows(stalls, size, cows);
+    printf("The largest minimum distance is: %d\n", result);
+
+    return 0;
+}

Trie/Aggressive Cows Problem/Readme.md

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+## Aggressive Cows
+The "Aggressive Cows" problem is a classic problem that involves placing cows in stalls such that the minimum distance between any two cows is maximized. The goal is to find this maximum minimum distance.
+
+# Problem Statement
+You have N stalls and C cows. You need to place the cows in the stalls such that the minimum distance between any two cows is as large as possible.
+
+# Approach
+Sort the Stalls: First, sort the array of stall positions.
+Binary Search: Use binary search to find the maximum minimum distance.
+Greedy Check: For a given distance, check if it is possible to place all cows in the stalls such that the minimum distance between any two cows is at least that distance.
+
+# Example
+For the stalls [1, 2, 4, 8, 9] and 3 cows, the maximum minimum distance that can be achieved is:
+
+## Time Complexity:
+ O(n log m), where n is the number of stalls and m is the range of possible distances.
+
+# Space Complexity:
+ O(1), as we are using a constant amount of space apart from the input.