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Cut all the rods with some length such that the sum of cut-off length is maximized

Last Updated : 11 Jul, 2025
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Given N rods of different lengths. The task is to cut all the rods with some maximum integer height 'h' such that the sum of cut-off lengths of the rod is maximized and must be greater than M. Print -1 if no such cut is possible. 
Note: A rod cannot be cut also. 
Examples:
 

Input: N = 7, M = 8, a[] = {1, 2, 3, 5, 4, 7, 6} 
Output: 3 
Rod 1 and 2 are untouched, and rod 3, 4, 5, 6, 7 are cut with the cut-off lengths being (3-3) + (4-3) + (5-3) + (7-3) + (6-3) which is equal to 10 which is greater than M = 8. 
Input: N = 4, M = 2, a[] = {1, 2, 3, 3} 
Output: 2


 


Approach: 
 

  • Sort the array in ascending order
  • Run a binary search with values low=0 and high=length[n-1], such that mid=(low+high)/2.
  • Run a loop from n-1 till 0 adding the height of the rod cut-off to the sum.
  • If the sum is greater than or equal to m, assign low with mid+1 otherwise high will be updated with mid.
  • After Binary search is completed the answer will be low-1. 
     


Below is the implementation of the above approach: 
 

C++ 
// C++ program to find the maximum possible
// length of rod which will be cut such that
// sum of cut off lengths will be maximum
#include <bits/stdc++.h>
using namespace std;

// Function to run Binary Search to
// find maximum cut off length
int binarySearch(int adj[], int target, int length)
{

    int low = 0;
    int high = adj[length - 1];
    while (low < high) {

        // f is the flag variable
        // sum is for the total length cutoff
        int f = 0, sum = 0;

        int mid = low + (high - low) / 2;

        // Loop from higher to lower
        // for optimization
        for (int i = length - 1; i >= 0; i--) {

            // Only if length is greater
            // than cut-off length
            if (adj[i] > mid) {
                sum = sum + adj[i] - mid;
            }

            // When total cut off length becomes greater
            // than desired cut off length
            if (sum >= target) {
                f = 1;
                low = mid + 1;
                break;
            }
        }

        // If flag variable is not set
        // Change high
        if (f == 0)
            high = mid;
    }

    // returning the maximum cut off length
    return low - 1;
}

// Driver Function
int main()
{
    int n1 = 7;
    int n2 = 8;

    int adj[] = { 1, 2, 3, 4, 5, 7, 6 };

    // Sorting the array in ascending order
    sort(adj, adj + n1);

    // Calling the binarySearch Function
    cout << binarySearch(adj, n2, n1);
}
Java 
// Java program to find the 
// maximum possible length 
// of rod which will be cut 
// such that sum of cut off 
// lengths will be maximum
import java.util.*;

class GFG
{
// Function to run Binary 
// Search to find maximum 
// cut off length
static int binarySearch(int adj[], 
                        int target,
                        int length)
{
int low = 0;
int high = adj[length - 1];
while (low < high) 
{

    // f is the flag variable
    // sum is for the total
    // length cutoff
    int f = 0, sum = 0;

    int mid = low + (high - low) / 2;

    // Loop from higher to lower
    // for optimization
    for (int i = length - 1; 
            i >= 0; i--) 
    {

        // Only if length is greater
        // than cut-off length
        if (adj[i] > mid) 
        {
            sum = sum + adj[i] - mid;
        }

        // When total cut off length 
        // becomes greater than
        // desired cut off length
        if (sum >= target) 
        {
            f = 1;
            low = mid + 1;
            break;
        }
    }

    // If flag variable is 
    // not set Change high
    if (f == 0)
        high = mid;
}

// returning the maximum 
// cut off length
return low - 1;
}

// Driver Code
public static void main(String args[])
{
    int n1 = 7;
    int n2 = 8;

    int adj[] = { 1, 2, 3, 4, 5, 7, 6 };

    // Sorting the array 
    // in ascending order
    Arrays.sort(adj);

    // Calling the binarySearch Function
    System.out.println(binarySearch(adj, n2, n1));
}
}

// This code is contributed
// by Arnab Kundu
Python3 
# Python 3 program to find the 
# maximum possible length of 
# rod which will be cut such 
# that sum of cut off lengths 
# will be maximum

# Function to run Binary Search 
# to find maximum cut off length 
def binarySearch(adj, target, length) :
    low = 0
    high = adj[length - 1]
    
    while (low < high) :

        # f is the flag variable 
        # sum is for the total 
        # length cutoff

        # multiple assignments
        f, sum = 0, 0

        # take integer value 
        mid = low + (high - low) // 2;

        # Loop from higher to lower 
        # for optimization 
        for i in range(length - 1, -1 , -1) :
            
            # Only if length is greater 
            # than cut-off length 
            if adj[i] > mid :
                sum = sum + adj[i] - mid
                
            # When total cut off length 
            # becomes greater than 
            # desired cut off length
            if sum >= target :
                f = 1
                low = mid + 1
                break

        # If flag variable is 
        # not set. Change high 
        if f == 0 :
            high = mid

    # returning the maximum 
    # cut off length 
    return low - 1

# Driver code
if __name__ == "__main__" :

    n1 = 7
    n2 = 8

    # adj = [1,2,3,3]
    adj = [ 1, 2, 3, 4, 5, 7, 6]

    # Sorting the array 
    # in ascending order
    adj.sort()

    # Calling the binarySearch Function 
    print(binarySearch(adj, n2, n1))

# This code is contributed
# by ANKITRAI1
C# 
// C# program to find the 
// maximum possible length 
// of rod which will be cut 
// such that sum of cut off 
// lengths will be maximum
using System;

class GFG
{
// Function to run Binary 
// Search to find maximum 
// cut off length
static int binarySearch(int []adj, 
                        int target,
                        int length)
{
int low = 0;
int high = adj[length - 1];
while (low < high) 
{

    // f is the flag variable
    // sum is for the total
    // length cutoff
    int f = 0, sum = 0;

    int mid = low + (high - low) / 2;

    // Loop from higher to lower
    // for optimization
    for (int i = length - 1; 
            i >= 0; i--) 
    {

        // Only if length is greater
        // than cut-off length
        if (adj[i] > mid) 
        {
            sum = sum + adj[i] - mid;
        }

        // When total cut off length 
        // becomes greater than
        // desired cut off length
        if (sum >= target) 
        {
            f = 1;
            low = mid + 1;
            break;
        }
    }

    // If flag variable is 
    // not set Change high
    if (f == 0)
        high = mid;
}

// returning the maximum 
// cut off length
return low - 1;
}

// Driver Code
public static void Main()
{
    int n1 = 7;
    int n2 = 8;

    int []adj = {1, 2, 3, 4, 5, 7, 6};

    // Sorting the array 
    // in ascending order
    Array.Sort(adj);

    // Calling the binarySearch Function
    Console.WriteLine(binarySearch(adj, n2, n1));
}
}

// This code is contributed
// by Subhadeep Gupta
PHP 
<?php 
// PHP program to find the maximum 
// possible length of rod which will 
// be cut such that sum of cut off 
// lengths will be maximum

// Function to run Binary Search 
// to find maximum cut off length
function binarySearch(&$adj, $target, 
                            $length)
{
    $low = 0;
    $high = $adj[$length - 1];
    while ($low < $high)
    {

        // f is the flag variable
        // sum is for the total 
        // length cutoff
        $f = 0;
        $sum = 0;

        $mid = $low + ($high - $low) / 2;

        // Loop from higher to lower
        // for optimization
        for ($i = $length - 1; $i >= 0; $i--) 
        {

            // Only if length is greater
            // than cut-off length
            if ($adj[$i] > $mid) 
            {
                $sum = $sum + $adj[$i] - $mid;
            }

            // When total cut off length becomes 
            // greater than desired cut off length
            if ($sum >= $target) 
            {
                $f = 1;
                $low = $mid + 1;
                break;
            }
        }

        // If flag variable is not
        // set Change high
        if ($f == 0)
            $high = $mid;
    }

    // returning the maximum cut off length
    return $low - 1;
}

// Driver Code
$n1 = 7;
$n2 = 8;

$adj = array( 1, 2, 3, 4, 5, 7, 6 );

// Sorting the array in ascending order
sort($adj);

// Calling the binarySearch Function
echo (int)binarySearch($adj, $n2, $n1);

// This code is contributed by ChitraNayal
?>
JavaScript 
<script>

// Javascript program to find the 
// maximum possible length 
// of rod which will be cut 
// such that sum of cut off 
// lengths will be maximum

// Function to run Binary 
// Search to find maximum 
// cut off length
function binarySearch(adj,target,length)
{
let low = 0;
let high = adj[length - 1];
while (low < high) 
{
  
    // f is the flag variable
    // sum is for the total
    // length cutoff
    let f = 0, sum = 0;
  
    let mid = low + Math.floor((high - low) / 2);
  
    // Loop from higher to lower
    // for optimization
    for (let i = length - 1; 
            i >= 0; i--) 
    {
  
        // Only if length is greater
        // than cut-off length
        if (adj[i] > mid) 
        {
            sum = sum + adj[i] - mid;
        }
  
        // When total cut off length 
        // becomes greater than
        // desired cut off length
        if (sum >= target) 
        {
            f = 1;
            low = mid + 1;
            break;
        }
    }
  
    // If flag variable is 
    // not set Change high
    if (f == 0)
        high = mid;
}
  
// returning the maximum 
// cut off length
return low - 1;
}
    
// Driver Code    
let n1 = 7;
let n2 = 8;
let adj=[1, 2, 3, 4, 5, 7, 6 ];

// Sorting the array 
// in ascending order
adj.sort(function(a,b){return a-b;});

// Calling the binarySearch Function
document.write(binarySearch(adj, n2, n1));
    
    
// This code is contributed by avanitrachhadiya2155

</script>

Output: 
3

 

Time Complexity: O(N * log N) 
Auxiliary Space: O(1)
 


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