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Place the prisoners into cells to maximize the minimum difference between any two
  • Difficulty Level : Hard
  • Last Updated : 07 Apr, 2021

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:
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:
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 whther 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

 




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