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Maximum of smallest possible area that can get with exactly k cut of given rectangular

  • Difficulty Level : Hard
  • Last Updated : 30 May, 2021

Given n×m big rectangular area with unit squares and k times cut is allowed, the cut should be straight (horizontal or vertical) and should go along the edges of the unit square. What is the maximum possible area of the smallest piece he can get with exactly k cuts.

Examples : 

Input : 3 4 1
Output : 6

Input : 6 4 2
Output : 8

 

Image for 2nd input 
 



Image for 1st input 
 

As this is n×m rectangular area so there are (n-1) rows and (m-1) columns . So if k > (n + m – 2) then, then cut are not possible . Then, if k is less than that . There will be two cases 

  1. When k is less than max( n, m ) – 1 : In the 1st case, if k is less than max( n, m ) – 1, then either m * ( n / ( k+1 ) ) or n * ( m / ( k+1 ) ) is maximum, here we have divided by ( k + 1) because horizontally or vertically i.e ( m * n = total blocks ) is divided into ( k + 1 ) parts.
  2. When k is greater than or equal to max( n, m) – 1 : In the 2nd case, if k >= max( n, m ) – 1, then there will be cut on both rows as well as columns so, the maximum possible smallest area will be either m / ( k – n + 2) or n / ( k – m + 2 ) . In this case suppose if n > m then, firstly n-1 (row or column) cut is possible . After that (k – n) cut will be done on m – 1 . So, here we have adjusted this ( k – n ) cut such that smallest possible division should be maximum.

Code – Below is the implementation of the following approach  

C++




// C++ code for Maximum of smallest
// possible area that can get with
// exactly k cut of given rectangular
#include <bits/stdc++.h>
using namespace std;
 
void max_area(int n, int m, int k)
{
    if (k > (n + m - 2))
 
        cout << "Not possible" << endl;
 
    else {
 
        int result;
 
        // for the 1st case
        if (k < max(m, n) - 1) {
 
            result = max(m * (n / (k + 1)), n * (m / (k + 1)));
        }
 
        // for the second case
        else {
 
            result = max(m / (k - n + 2), n / (k - m + 2));
        }
 
        // print final result
        cout << result << endl;
    }
}
 
// driver code
int main()
{
 
    int n = 3, m = 4, k = 1;
    max_area(n, m, k);
}

Java




// Java code for Maximum of smallest
// possible area that can get with
// exactly k cut of given rectangular
 
class GFG {
     
    //Utility Function
    static void max_area(int n, int m, int k)
    {
        if (k > (n + m - 2))
     
            System.out.println("Not possible");
     
        else {
     
            int result;
     
            // for the 1st case
            if (k < Math.max(m, n) - 1)
            {
                result = Math.max(m * (n / (k + 1)),
                         n * (m / (k + 1)));
            }
     
            // for the second case
            else {
     
                result = Math.max(m / (k - n + 2),
                         n / (k - m + 2));
            }
     
            // print final result
            System.out.println(result);
        }
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 3, m = 4, k = 1;
        max_area(n, m, k);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python3 code for Maximum of smallest
# possible area that can get with
# exactly k cut of given rectangular
 
def max_area(n,m,k):
    if (k > (n + m - 2)):
        print("Not possible")
    else:
        # for the 1st case
        if (k < max(m,n) - 1):
            result = max(m * (n / (k + 1)), n * (m / (k + 1)));
 
        # for the second case
        else:
            result = max(m / (k - n + 2), n / (k - m + 2));
 
        # print final result
        print(result)
 
# driver code
n = 3
m = 4
k = 1
 
max_area(n, m, k)
 
# This code is contributed
# by Azkia Anam.

C#




// C# code for Maximum of smallest
// possible area that can get with
// exactly k cut of given rectangular
using System;
 
class GFG {
     
    //Utility Function
    static void max_area(int n, int m, int k)
    {
        if (k > (n + m - 2))
     
            Console.WriteLine("Not possible");
     
        else {
     
            int result;
     
            // for the 1st case
            if (k < Math.Max(m, n) - 1)
            {
                result = Math.Max(m * (n / (k + 1)),
                        n * (m / (k + 1)));
            }
     
            // for the second case
            else {
     
                result = Math.Max(m / (k - n + 2),
                        n / (k - m + 2));
            }
     
            // print final result
            Console.WriteLine(result);
        }
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 3, m = 4, k = 1;
         
        max_area(n, m, k);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP code for Maximum of smallest
// possible area that can get with
// exactly k cut of given rectangular
function max_area($n, $m, $k)
{
    if ($k > ($n + $m - 2))
 
        echo "Not possible" ,"\n";
 
    else
    {
        $result;
 
        // for the 1st case
        if ($k < max($m, $n) - 1)
        {
 
            $result = max($m * ($n / ($k + 1)),
                          $n * ($m / ($k + 1)));
        }
 
        // for the second case
        else
        {
 
            $result = max($m / ($k - $n + 2),
                          $n / ($k - $m + 2));
        }
 
        // print final result
    echo $result ,"\n";
    }
}
 
// Driver Code
$n = 3; $m = 4; $k = 1;
max_area($n, $m, $k);
 
// This code is contributed by ajit
?>

Javascript




<script>
 
// JavaScript code for Maximum of smallest
// possible area that can get with
// exactly k cut of given rectangular
 
// Utility Function
function max_area(n, m, k)
{
    if (k > (n + m - 2))
   
        document.write("Not possible");
    else
    {
        let result;
   
        // For the 1st case
        if (k < Math.max(m, n) - 1)
        {
            result = Math.max(m * (n / (k + 1)),
                              n * (m / (k + 1)));
        }
   
        // For the second case
        else
        {
            result = Math.max(m / (k - n + 2),
                              n / (k - m + 2));
        }
   
        // Print final result
        document.write(result);
    }
}
 
// Driver Code
let n = 3, m = 4, k = 1;
 
max_area(n, m, k);
 
// This code is contributed by susmitakundugoaldanga
 
</script>

Output : 
 

6

Time Complexity: O(1)
Auxiliary Space: O(1)
 

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