# Reorder an array such that sum of left half is not equal to sum of right half

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 ` `#include ` ` `  `// 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 == 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); ` ` `  `    ``// Function call ` `    ``printArr(arr, N); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by equbalzeeshan `

## C++

 `// C++ program for the above approach ` `#include ` `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 == 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); ` ` `  `    ``// 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 == 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` `{ 1, 2, 2, 1, 3, 1 };  ` ` `  `    ``int` `N = arr.Length;  ` ` `  `    ``// Function call  ` `    ``printArr(arr, N);  ` `} ` `} ` ` `  `// This code is contributed by equbalzeeshan `

Output:

```Yes
1 1 1 2 2 3
```

Time Complexity: O(N*log(N))

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