Divide array into two parts with equal sum according to the given constraints

Last Updated : 07 Aug, 2023

Given an array arr[] of N integers, the task is to select an integer x (which may or may not be present in the array) and remove all of its occurrences from the array and divide the remaining array into two non-empty sub-sets such that:

The elements of the first set are strictly smaller than x.
The elements of the second set are strictly greater than x.
The sum of the elements of both the sets is equal.

If such an integer exists then print Yes otherwise print No.

Examples:

Input: arr[] = {1, 2, 2, 5}
Output: Yes
Choose x = 3, after removing all of its occurrences the array becomes arr[] = {1, 2, 2, 5} {1, 2, 2} and {5} are the required sub-sets.

Input: arr[] = {2, 1}
Output: No

Approach:

The idea is to first sort the array and for all the numbers lying between 1 to maximum number present in the array, apply binary search and check if on removing all its occurrences from the array, sum of elements present on its left side (which are smaller than it) and sum of elements present on the right side (which are greater than it) is equal.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that checks if the given
// conditions are satisfied
void IfExists(int arr[], int n)
{
    // To store the prefix sum
    // of the array elements
    int sum[n];
 
    // Sort the array
    sort(arr, arr + n);
 
    sum[0] = arr[0];
 
    // Compute the prefix sum array
    for (int i = 1; i < n; i++)
        sum[i] = sum[i - 1] + arr[i];
 
    // Maximum element in the array
    int max = arr[n - 1];
 
    // Variable to check if there exists any number
    bool flag = false;
 
    for (int i = 1; i <= max; i++) {
 
        // Stores the index of the largest
        // number present in the array
        // smaller than i
        int findex = 0;
 
        // Stores the index of the smallest
        // number present in the array
        // greater than i
        int lindex = 0;
 
        int l = 0;
        int r = n - 1;
 
        // Find index of smallest number
        // greater than i
        while (l <= r) {
            int m = (l + r) / 2;
 
            if (arr[m] < i) {
                findex = m;
                l = m + 1;
            }
            else
                r = m - 1;
        }
 
        l = 1;
        r = n;
        flag = false;
 
        // Find index of smallest number
        // greater than i
        while (l <= r) {
            int m = (r + l) / 2;
 
            if (arr[m] > i) {
                lindex = m;
                r = m - 1;
            }
            else
                l = m + 1;
        }
 
        // If there exists a number
        if (sum[findex] == sum[n - 1] - sum[lindex - 1]) {
            flag = true;
            break;
        }
    }
 
    // If no such number exists
    // print no
    if (flag)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 5 };
    int n = sizeof(arr) / sizeof(int);
    IfExists(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach 
import java.util.*;
 
class GFG
{
 
// Function that checks if the given 
// conditions are satisfied 
static void IfExists(int arr[], int n) 
{ 
    // To store the prefix sum 
    // of the array elements 
    int sum[] = new int[n]; 
 
    // Sort the array 
    Arrays.sort(arr); 
 
    sum[0] = arr[0]; 
 
    // Compute the prefix sum array 
    for (int i = 1; i < n; i++) 
        sum[i] = sum[i - 1] + arr[i]; 
 
    // Maximum element in the array 
    int max = arr[n - 1]; 
 
    // Variable to check if there exists any number 
    boolean flag = false; 
 
    for (int i = 1; i <= max; i++) 
    { 
 
        // Stores the index of the largest 
        // number present in the array 
        // smaller than i 
        int findex = 0; 
 
        // Stores the index of the smallest 
        // number present in the array 
        // greater than i 
        int lindex = 0; 
 
        int l = 0; 
        int r = n - 1; 
 
        // Find index of smallest number 
        // greater than i 
        while (l <= r) 
        { 
            int m = (l + r) / 2; 
 
            if (arr[m] < i) 
            { 
                findex = m; 
                l = m + 1; 
            } 
            else
                r = m - 1; 
        } 
 
        l = 1; 
        r = n; 
        flag = false; 
 
        // Find index of smallest number 
        // greater than i 
        while (l <= r) 
        { 
            int m = (r + l) / 2; 
 
            if (arr[m] > i) 
            { 
                lindex = m; 
                r = m - 1; 
            } 
            else
                l = m + 1; 
        } 
 
        // If there exists a number 
        if (sum[findex] == sum[n - 1] - sum[lindex - 1]) 
        { 
            flag = true; 
            break; 
        } 
    } 
 
    // If no such number exists 
    // print no 
    if (flag) 
        System.out.println("Yes"); 
    else
        System.out.println("No"); 
} 
 
// Driver code 
public static void main(String args[])
{ 
    int arr[] = { 1, 2, 2, 5 }; 
    int n = arr.length; 
    IfExists(arr, n); 
}
} 
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach 
 
# Function that checks if the given 
# conditions are satisfied 
def IfExists(arr, n) :
 
    # To store the prefix sum 
    # of the array elements 
    sum = [0] * n; 
 
    # Sort the array 
    arr.sort(); 
 
    sum[0] = arr[0]; 
 
    # Compute the prefix sum array 
    for i in range(1, n) : 
        sum[i] = sum[i - 1] + arr[i]; 
 
    # Maximum element in the array 
    max = arr[n - 1]; 
 
    # Variable to check if there 
    # exists any number 
    flag = False; 
 
    for i in range(1, max + 1) :
 
        # Stores the index of the largest 
        # number present in the array 
        # smaller than i 
        findex = 0; 
 
        # Stores the index of the smallest 
        # number present in the array 
        # greater than i 
        lindex = 0; 
 
        l = 0; 
        r = n - 1; 
 
        # Find index of smallest number 
        # greater than i 
        while (l <= r) :
            m = (l + r) // 2; 
 
            if (arr[m] < i) :
                findex = m; 
                l = m + 1; 
             
            else :
                r = m - 1; 
         
        l = 1; 
        r = n; 
        flag = False; 
 
        # Find index of smallest number 
        # greater than i 
        while (l <= r) : 
            m = (r + l) // 2; 
 
            if (arr[m] > i) : 
                lindex = m; 
                r = m - 1; 
             
            else :
                l = m + 1; 
         
        # If there exists a number 
        if (sum[findex] == sum[n - 1] -
                           sum[lindex - 1]) : 
            flag = True; 
            break; 
         
    # If no such number exists 
    # print no 
    if (flag) : 
        print("Yes"); 
    else :
        print("No"); 
 
# Driver code 
if __name__ == "__main__" : 
 
    arr = [ 1, 2, 2, 5 ]; 
     
    n = len(arr) ;
    IfExists(arr, n); 
     
# This code is contributed by Ryuga

C#

// C# implementation of the approach 
using System;
 
class GFG
{
     
// Function that checks if the given 
// conditions are satisfied 
static void IfExists(int[] arr, int n) 
{ 
    // To store the prefix sum 
    // of the array elements 
    int[] sum = new int[n]; 
 
    // Sort the array 
    Array.Sort(arr); 
 
    sum[0] = arr[0]; 
 
    // Compute the prefix sum array 
    for (int i = 1; i < n; i++) 
        sum[i] = sum[i - 1] + arr[i]; 
 
    // Maximum element in the array 
    int max = arr[n - 1]; 
 
    // Variable to check if there exists any number 
    bool flag = false; 
 
    for (int i = 1; i <= max; i++) 
    { 
 
        // Stores the index of the largest 
        // number present in the array 
        // smaller than i 
        int findex = 0; 
 
        // Stores the index of the smallest 
        // number present in the array 
        // greater than i 
        int lindex = 0; 
 
        int l = 0; 
        int r = n - 1; 
 
        // Find index of smallest number 
        // greater than i 
        while (l <= r) 
        { 
            int m = (l + r) / 2; 
 
            if (arr[m] < i) 
            { 
                findex = m; 
                l = m + 1; 
            } 
            else
                r = m - 1; 
        } 
 
        l = 1; 
        r = n; 
        flag = false; 
 
        // Find index of smallest number 
        // greater than i 
        while (l <= r) 
        { 
            int m = (r + l) / 2; 
 
            if (arr[m] > i) 
            { 
                lindex = m; 
                r = m - 1; 
            } 
            else
                l = m + 1; 
        } 
 
        // If there exists a number 
        if (sum[findex] == sum[n - 1] - sum[lindex - 1]) 
        { 
            flag = true; 
            break; 
        } 
    } 
 
    // If no such number exists 
    // print no 
    if (flag) 
        Console.WriteLine("Yes"); 
    else
        Console.WriteLine("No"); 
} 
 
// Driver code 
public static void Main()
{ 
    int[] arr = { 1, 2, 2, 5 }; 
    int n = arr.Length; 
    IfExists(arr, n); 
}
} 
 
// This code is contributed by Code_Mech.

Javascript

// Java implementation of the approach 
 
// Function that checks if the given 
// conditions are satisfied
function ifExists(arr, n) {
    // To store the prefix sum 
    // of the array elements
    const sum = new Array(n).fill(0);
    // Sort the array 
    arr.sort((a, b) => a - b);
   
    sum[0] = arr[0];
    // Compute the prefix sum array 
    for (let i = 1; i < n; i++) {
      sum[i] = sum[i - 1] + arr[i];
    }
  // Maximum element in the array 
    const max = arr[n - 1];
     
    // Variable to check if there exists any number 
    let flag = false;
   
    for (let i = 1; i <= max; i++) {
         // Stores the index of the largest 
        // number present in the array 
       // smaller than i 
      let findex = 0;
       
       // Stores the index of the smallest 
      // number present in the array 
      // greater than i
      let lindex = 0;
       
      let l = 0;
      let r = n - 1;
   
          // Find index of smallest number 
        // greater than i 
      while (l <= r) {
        const m = Math.floor((l + r) / 2);
   
        if (arr[m] < i) {
          findex = m;
          l = m + 1;
        } else {
          r = m - 1;
        }
      }
   
      l = 1;
      r = n;
      flag = false;
   
      while (l <= r) {
        const m = Math.floor((r + l) / 2);
   
        if (arr[m] > i) {
          lindex = m;
          r = m - 1;
        } else {
          l = m + 1;
        }
      }
   
      if (sum[findex] === sum[n - 1] - sum[lindex - 1]) {
        flag = true;
        break;
      }
    }
   
    if (flag) {
      console.log("Yes");
    } else {
      console.log("No");
    }
}
   
// Driver code 
const arr = [1, 2, 2, 5];
const n = arr.length;
ifExists(arr, n);
 
// This code is contributed by Dwaipayan Bandyopadhyay

PHP

<?php
// PHP implementation of the approach
 
// Function that checks if the given
// conditions are satisfied
function IfExists($arr, $n)
{
    // To store the prefix $sum
    // of the array elements
    $sum = array_fill(0, $n, 0);
 
    // Sort the array
    sort($arr);
 
    $sum[0] = $arr[0];
 
    // Compute the prefix sum array
    for ($i = 1; $i < $n; $i++)
        $sum[$i] = $sum[$i - 1] + $arr[$i];
 
    // Maximum element in the array
    $max = $arr[$n - 1];
 
    // Variable to check if there exists any number
    $flag = false;
 
    for ($i = 1; $i <= $max; $i++) 
    {
 
        // Stores the index of the largest
        // number present in the array
        // smaller than i
        $findex = 0;
 
        // Stores the index of the smallest
        // number present in the array
        // greater than i
        $lindex = 0;
 
        $l = 0;
        $r = $n - 1;
 
        // Find index of smallest number
        // greater than i
        while ($l <= $r) 
        {
            $m = ($l + $r) / 2; 
            if ($arr[$m] < $i) 
            {
                $findex = $m;
                $l = $m + 1;
            }
            else
                $r = $m - 1;
        }
 
        $l = 1;
        $r = $n;
        $flag = false;
 
        // Find index of smallest number
        // greater than i
        while ($l <= $r)
        {
            $m = ($r + $l) / 2;
 
            if ($arr[$m] > $i) 
            {
                $lindex = $m;
                $r = $m - 1;
            }
            else
                $l = $m + 1;
        }
 
        // If there exists a number
        if ($sum[$findex] == $sum[$n - 1] -
                             $sum[$lindex - 1])
        {
            $flag = true;
            break;
        }
    }
 
    // If no such number exists
    // print no
    if ($flag == true)
        echo "Yes";
    else
        echo "No";
}
 
// Driver code
$arr = array(1, 2, 2, 5 );
$n = sizeof($arr);
IfExists($arr, $n);
 
// This code is contributed by ihritik
?>

Output

Yes

Divide array into two sub-arrays such that their averages are equal

sakshi_srivastava

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Divide array into two parts with equal sum according to the given constraints

C++

Java

Python3

C#

Javascript

PHP

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