Place the prisoners into cells to maximize the minimum difference between any two Last Updated : 12 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given an array cell[] of N elements, which represent the positions of the cells in a prison. Also, given an integer P which is the number of prisoners, the task is to place all the prisoners in the cells in an ordered manner such that the minimum distance between any two prisoners is as large as possible. Finally, print the maximized distance.Examples: Input: cell[] = {1, 2, 8, 4, 9}, P = 3 Output: 3 The three prisoners will be placed at the cells numbered 1, 4 and 8 with the minimum distance 3 which is the maximum possible.Input: cell[] = {10, 12, 18}, P = 2 Output: 8 The three possible placements are {10, 12}, {10, 18} and {12, 18}. Approach: This problem can be solved using binary search. As the minimum distance between two cells in which prisoners will be kept has to be maximized, the search space will be of distance, starting from 0 (if two prisoners are kept in the same cell) and ending at cell[N - 1] - cell[0] (if one prisoner is kept in the first cell, and the other one is kept in the last cell). Initialize L = 0 and R = cell[N - 1] - cell[0] then apply the binary search. For every mid, check whether the prisoners can be placed such that the minimum distance between any two prisoners is at least mid If yes then try to increase this distance in order to maximize the answer and check again.If not then try to decrease the distance.Finally, print the maximized distance. Below is the implementation of the above approach: C++ // C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if the prisoners // can be placed such that the minimum distance // between any two prisoners is at least sep bool canPlace(int a[], int n, int p, int sep) { // Considering the first prisoner // is placed at 1st cell int prisoners_placed = 1; // If the first prisoner is placed at // the first cell then the last_prisoner_placed // will be the first prisoner placed // and that will be in cell[0] int last_prisoner_placed = a[0]; for (int i = 1; i < n; i++) { int current_cell = a[i]; // Checking if the prisoner can be // placed at ith cell or not if (current_cell - last_prisoner_placed >= sep) { prisoners_placed++; last_prisoner_placed = current_cell; // If all the prisoners got placed // then return true if (prisoners_placed == p) { return true; } } } return false; } // Function to return the maximized distance int maxDistance(int cell[], int n, int p) { // Sort the array so that binary // search can be applied on it sort(cell, cell + n); // Minimum possible distance // for the search space int start = 0; // Maximum possible distance // for the search space int end = cell[n - 1] - cell[0]; // To store the result int ans = 0; // Binary search while (start <= end) { int mid = start + ((end - start) / 2); // If the prisoners can be placed such that // the minimum distance between any two // prisoners is at least mid if (canPlace(cell, n, p, mid)) { // Update the answer ans = mid; start = mid + 1; } else { end = mid - 1; } } return ans; } // Driver code int main() { int cell[] = { 1, 2, 8, 4, 9 }; int n = sizeof(cell) / sizeof(int); int p = 3; cout << maxDistance(cell, n, p); return 0; } Java // Java implementation of the approach import java.util.*; class GFG { // Function that returns true if the prisoners // can be placed such that the minimum distance // between any two prisoners is at least sep static boolean canPlace(int a[], int n, int p, int sep) { // Considering the first prisoner // is placed at 1st cell int prisoners_placed = 1; // If the first prisoner is placed at // the first cell then the last_prisoner_placed // will be the first prisoner placed // and that will be in cell[0] int last_prisoner_placed = a[0]; for (int i = 1; i < n; i++) { int current_cell = a[i]; // Checking if the prisoner can be // placed at ith cell or not if (current_cell - last_prisoner_placed >= sep) { prisoners_placed++; last_prisoner_placed = current_cell; // If all the prisoners got placed // then return true if (prisoners_placed == p) { return true; } } } return false; } // Function to return the maximized distance static int maxDistance(int cell[], int n, int p) { // Sort the array so that binary // search can be applied on it Arrays.sort(cell); // Minimum possible distance // for the search space int start = 0; // Maximum possible distance // for the search space int end = cell[n - 1] - cell[0]; // To store the result int ans = 0; // Binary search while (start <= end) { int mid = start + ((end - start) / 2); // If the prisoners can be placed such that // the minimum distance between any two // prisoners is at least mid if (canPlace(cell, n, p, mid)) { // Update the answer ans = mid; start = mid + 1; } else { end = mid - 1; } } return ans; } // Driver code public static void main (String[] args) { int cell[] = { 1, 2, 8, 4, 9 }; int n = cell.length; int p = 3; System.out.println(maxDistance(cell, n, p)); } } // This code is contributed by AnkitRai01 Python3 # Python3 implementation of the approach # Function that returns true if the prisoners # can be placed such that the minimum distance # between any two prisoners is at least sep def canPlace(a, n, p, sep): # Considering the first prisoner # is placed at 1st cell prisoners_placed = 1 # If the first prisoner is placed at # the first cell then the last_prisoner_placed # will be the first prisoner placed # and that will be in cell[0] last_prisoner_placed = a[0] for i in range(1, n): current_cell = a[i] # Checking if the prisoner can be # placed at ith cell or not if (current_cell - last_prisoner_placed >= sep): prisoners_placed += 1 last_prisoner_placed = current_cell # If all the prisoners got placed # then return true if (prisoners_placed == p): return True return False # Function to return the maximized distance def maxDistance(cell, n, p): # Sort the array so that binary # search can be applied on it cell = sorted(cell) # Minimum possible distance # for the search space start = 0 # Maximum possible distance # for the search space end = cell[n - 1] - cell[0] # To store the result ans = 0 # Binary search while (start <= end): mid = start + ((end - start) // 2) # If the prisoners can be placed such that # the minimum distance between any two # prisoners is at least mid if (canPlace(cell, n, p, mid)): # Update the answer ans = mid start = mid + 1 else : end = mid - 1 return ans # Driver code cell= [1, 2, 8, 4, 9] n = len(cell) p = 3 print(maxDistance(cell, n, p)) # This code is contributed by mohit kumar 29 C# // C# implementation of the approach using System; using System.Collections; class GFG { // Function that returns true if the prisoners // can be placed such that the minimum distance // between any two prisoners is at least sep static bool canPlace(int []a, int n, int p, int sep) { // Considering the first prisoner // is placed at 1st cell int prisoners_placed = 1; // If the first prisoner is placed at // the first cell then the last_prisoner_placed // will be the first prisoner placed // and that will be in cell[0] int last_prisoner_placed = a[0]; for (int i = 1; i < n; i++) { int current_cell = a[i]; // Checking if the prisoner can be // placed at ith cell or not if (current_cell - last_prisoner_placed >= sep) { prisoners_placed++; last_prisoner_placed = current_cell; // If all the prisoners got placed // then return true if (prisoners_placed == p) { return true; } } } return false; } // Function to return the maximized distance static int maxDistance(int []cell, int n, int p) { // Sort the array so that binary // search can be applied on it Array.Sort(cell); // Minimum possible distance // for the search space int start = 0; // Maximum possible distance // for the search space int end = cell[n - 1] - cell[0]; // To store the result int ans = 0; // Binary search while (start <= end) { int mid = start + ((end - start) / 2); // If the prisoners can be placed such that // the minimum distance between any two // prisoners is at least mid if (canPlace(cell, n, p, mid)) { // Update the answer ans = mid; start = mid + 1; } else { end = mid - 1; } } return ans; } // Driver code public static void Main() { int []cell = { 1, 2, 8, 4, 9 }; int n = cell.Length; int p = 3; Console.WriteLine(maxDistance(cell, n, p)); } } // This code is contributed by AnkitRai01 JavaScript <script> // javascript implementation of the approach // Function that returns true if the prisoners // can be placed such that the minimum distance // between any two prisoners is at least sep function canPlace(a , n , p , sep) { // Considering the first prisoner // is placed at 1st cell var prisoners_placed = 1; // If the first prisoner is placed at // the first cell then the last_prisoner_placed // will be the first prisoner placed // and that will be in cell[0] var last_prisoner_placed = a[0]; for (i = 1; i < n; i++) { var current_cell = a[i]; // Checking if the prisoner can be // placed at ith cell or not if (current_cell - last_prisoner_placed >= sep) { prisoners_placed++; last_prisoner_placed = current_cell; // If all the prisoners got placed // then return true if (prisoners_placed == p) { return true; } } } return false; } // Function to return the maximized distance function maxDistance(cell , n , p) { // Sort the array so that binary // search can be applied on it cell.sort(); // Minimum possible distance // for the search space var start = 0; // Maximum possible distance // for the search space var end = cell[n - 1] - cell[0]; // To store the result var ans = 0; // Binary search while (start <= end) { var mid = start + parseInt(((end - start) / 2)); // If the prisoners can be placed such that // the minimum distance between any two // prisoners is at least mid if (canPlace(cell, n, p, mid)) { // Update the answer ans = mid; start = mid + 1; } else { end = mid - 1; } } return ans; } // Driver code var cell = [ 1, 2, 8, 4, 9 ]; var n = cell.length; var p = 3; document.write(maxDistance(cell, n, p)); // This code is contributed by Rajput-Ji </script> Output: 3 Comment More infoAdvertise with us M mishrakishan1 Follow Improve Article Tags : Divide and Conquer DSA Arrays Practice Tags : ArraysDivide and Conquer Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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