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Reorder an array such that sum of left half is not equal to sum of right half
  • Difficulty Level : Easy
  • Last Updated : 22 Apr, 2021

Given an array arr[] of even length, the task is to check whether is it possible to reorder the array element such that the sum of the left half is not equal to the sum of the right half of the array. If it is possible then print “Yes” with the reordered sequence else print “No”.
Examples: 
 

Input: arr[] = {1, 2, 2, 1, 3, 1} 
Output: 
Yes 
1 1 1 2 2 3 
Explanation: 
sum of left half = 1 + 2 + 2 = 5 
sum of right half = 2 + 2 + 3 = 7 
both the sums are not equal.
Input: arr[] = {1, 1} 
Output: No 
Explanation: 
There is no such arrangement possible. 
 

Approach: If all the elements of the array are equal then we can’t reorder the array elements to fulfill the given conditions. Else in the sorted sequence of the given array, the sum of the left half and the right half of the array will always be unequal.
Below is the implementation of the above approach:
 

C




// C program for the above approach
#include <stdio.h>
#include <stdlib.h>
 
// A comparator function used by qsort
int compare(const void * a, const void * b)
{
    return (*(int*)a - *(int*)b);
}
 
// Function to print the required
// reordering of array if possible
void printArr(int arr[], int n)
{
     
    // Sort the array in increasing order
    qsort(arr, n, sizeof(int), compare);
 
    // If all elements are equal, then
    // it is not possible
    if (arr[0] == arr[n - 1])
    {
        printf("No\n");
    }
 
    // Else print the sorted array arr[]
    else
    {
        printf("Yes\n");
        for(int i = 0; i < n; i++)
        {
            printf("%d ", arr[i]);
        }
    }
}
 
// Driver Code
int main()
{
     
    // Given array
    int arr[] = { 1, 2, 2, 1, 3, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    printArr(arr, N);
    return 0;
}
 
// This code is contributed by equbalzeeshan

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required
// reordering of array if possible
void printArr(int arr[], int n)
{
    // Sort the array in increasing order
    sort(arr, arr + n);
 
    // If all elements are equal, then
    // it is not possible
    if (arr[0] == arr[n - 1]) {
        cout << "No" << endl;
    }
 
    // Else print the sorted array arr[]
    else {
 
        cout << "Yes" << endl;
        for (int i = 0; i < n; i++) {
            cout << arr[i] << " ";
        }
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 2, 1, 3, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    printArr(arr, N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to print the required
// reordering of array if possible
public static void printArr(int[] arr, int n)
{
     
    // Sort the array in increasing order
    Arrays.sort(arr);
 
    // If all elements are equal, then
    // it is not possible
    if (arr[0] == arr[n - 1])
    {
        System.out.println("No");
    }
 
    // Else print the sorted array arr[]
    else
    {
        System.out.println("Yes");
        for(int i = 0; i < n; i++)
        {
            System.out.print(arr[i] + " ");
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 1, 2, 2, 1, 3, 1 };
 
    int N = arr.length;
 
    // Function call
    printArr(arr, N);
}
}
 
// This code is contributed by equbalzeeshan

Python3




# Python3 program for the above approach
 
# Function to print the required
# reordering of the array if possible
def printArr(arr, n):
 
    # Sort the array in increasing
    # order
    arr.sort()
 
    # If all elements are equal,
    # then it is not possible
    if(arr[0] == arr[n - 1]):
        print("No")
 
    # Else print the sorted
    # array arr[]
    else:
        print("Yes")
        for i in range(n):
            print(arr[i], end = " ")
        print()
 
# Driver Code
if __name__ == '__main__':
 
    # Given array
    arr = [ 1, 2, 2, 1, 3, 1 ]
    N = len(arr)
 
    # Function Call
    printArr(arr, N)
 
# This code is contributed by Shivam Singh

C#




// C# code for the above approach
using System;
 
class GFG{
 
// Function to print the required
// reordering of array if possible
public static void printArr(int[] arr, int n)
{
     
    // Sort the array in increasing order
    Array.Sort(arr);
 
    // If all elements are equal, then
    // it is not possible
    if (arr[0] == arr[n - 1])
    {
        Console.Write("No\n");
    }
 
    // Else print the sorted array arr[]
    else
    {
        Console.Write("Yes\n");
        for(int i = 0; i < n; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }
}
 
// Driver code
public static void Main()
{
     
    // Given array
    int[] arr = new int [6]{ 1, 2, 2, 1, 3, 1 };
 
    int N = arr.Length;
 
    // Function call
    printArr(arr, N);
}
}
 
// This code is contributed by equbalzeeshan

Javascript




<script>
 
// Javascript program for the above approach
 
    // Function to print the required
    // reordering of array if possible
    function printArr(arr , n)
    {
 
        // Sort the array in increasing order
        arr.sort();
 
        // If all elements are equal, then
        // it is not possible
        if (arr[0] == arr[n - 1]) {
            document.write("No<br/>");
        }
 
        // Else print the sorted array arr
        else {
            document.write("Yes<br/>");
            for (i = 0; i < n; i++) {
                document.write(arr[i] + " ");
            }
        }
    }
 
    // Driver code
     
 
        // Given array
        var arr = [ 1, 2, 2, 1, 3, 1 ];
 
        var N = arr.length;
 
        // Function call
        printArr(arr, N);
 
// This code contributed by umadevi9616
 
</script>
Output: 
Yes
1 1 1 2 2 3

 

Time Complexity: O(N*log(N))
 

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