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

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.

Yes

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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. 

PHP

$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

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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