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# Find the Array Permutation having sum of elements at odd indices greater than sum of elements at even indices

• Difficulty Level : Easy
• Last Updated : 22 Jul, 2021

Given an array arr[] consisting of N integers, the task is to find the permutation of array elements such that the sum of odd indices elements is greater than or equal to the sum of even indices elements.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 1 4 2 3
Explanation:
Consider the permutation of the given array as {1, 4, 2, 3}.
Now, the sum of elements at odd indices = (4 + 3) = 7 and the sum of elements at even indices = (1 + 2) = 3.
As the sum at odd indices elements is greater than the sum of even indices element. Therefore, print the current permutation.

Input: arr[] = {123, 45, 67, 89, 60, 33}
Output: 33 123 45 89 60 67

Naive Approach: The simplest approach to solve the given problem is to generate all possible permutations of the given array and print that permutation of the array whose sum of odd indices elements is greater than or equal to the sum of even indices elements.

Time Complexity: O(N*N!)
Auxiliary Space: O(N)

Efficient Approach: The above approach can also be optimized by sorting the given array and using Two Pointer Approach. Follow the steps below to solve the problem:

Below is the implementation of the above approach :

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the permutation of``// array elements such that the sum of``// elements at odd āindices is greater``// than sum of elements at even indices``void` `rearrangeArray(``int` `arr[], ``int` `n)``{``    ``// Sort the given array``    ``sort(arr, arr + n);` `    ``// Initialize the two pointers``    ``int` `j = n - 1;``    ``int` `i = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `k = 0; k < n; k++) {` `        ``// Check if k is even``        ``if` `(k % 2 == 0) {``            ``cout << arr[i] << ``" "``;` `            ``// Increment the value``            ``// of i``            ``i++;``        ``}``        ``else` `{``            ``cout << arr[j] << ``" "``;` `            ``// Decrement the value``            ``// of j``            ``j--;``        ``}``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 123, 45, 67, 89, 60, 33 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``rearrangeArray(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to find the permutation of``// array elements such that the sum of``// elements at odd āindices is greater``// than sum of elements at even indices``static` `void` `rearrangeArray(``int` `arr[], ``int` `n)``{``    ` `    ``// Sort the given array``    ``Arrays.sort(arr);` `    ``// Initialize the two pointers``    ``int` `j = n - ``1``;``    ``int` `i = ``0``;` `    ``// Traverse the array arr[]``    ``for``(``int` `k = ``0``; k < n; k++)``    ``{``        ` `        ``// Check if k is even``        ``if` `(k % ``2` `== ``0``)``        ``{``            ``System.out.print(arr[i] + ``" "``);` `            ``// Increment the value``            ``// of i``            ``i++;``        ``}``        ``else``        ``{``            ``System.out.print(arr[j] + ``" "``);` `            ``// Decrement the value``            ``// of j``            ``j--;``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``123``, ``45``, ``67``, ``89``, ``60``, ``33` `};``    ``int` `N = arr.length;``    ` `    ``rearrangeArray(arr, N);``}``}` `// This code is contributed by avijitmondal1998`

## Python3

 `# Python3 program for the above approach` `# Function to find the permutation of``# array elements such that the sum of``# elements at odd āindices is greater``# than sum of elements at even indices``def` `rearrangeArray(arr, n):``    ` `    ``# Sort the given array``    ``arr ``=` `sorted``(arr)` `    ``# Initialize the two pointers``    ``j ``=` `n ``-` `1``    ``i ``=` `0` `    ``# Traverse the array arr[]``    ``for` `k ``in` `range``(n):``        ` `        ``# Check if k is even``        ``if` `(k ``%` `2` `=``=` `0``):``            ``print``(arr[i], end ``=` `" "``)``            ` `            ``# Increment the value``            ``# of i``            ``i ``+``=` `1``        ``else``:``            ``print``(arr[j], end ``=` `" "``)` `            ``# Decrement the value``            ``# of j``            ``j ``-``=` `1` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[ ``123``, ``45``, ``67``, ``89``, ``60``, ``33` `]``    ``N ``=` `len``(arr)``    ` `    ``rearrangeArray(arr, N)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find the permutation of``// array elements such that the sum of``// elements at odd āindices is greater``// than sum of elements at even indices``static` `void` `rearrangeArray(``int``[] arr, ``int` `n)``{``    ` `    ``// Sort the given array``    ``Array.Sort(arr);` `    ``// Initialize the two pointers``    ``int` `j = n - 1;``    ``int` `i = 0;` `    ``// Traverse the array arr[]``    ``for``(``int` `k = 0; k < n; k++)``    ``{``        ` `        ``// Check if k is even``        ``if` `(k % 2 == 0)``        ``{``            ``Console.Write(arr[i] + ``" "``);` `            ``// Increment the value``            ``// of i``            ``i++;``        ``}``        ``else``        ``{``            ``Console.Write(arr[j] + ``" "``);` `            ``// Decrement the value``            ``// of j``            ``j--;``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = { 123, 45, 67, 89, 60, 33 };``    ``int` `N = arr.Length;` `    ``rearrangeArray(arr, N);``}``}` `// This code is contributed by subham348`

## Javascript

 ``
Output:
`33 123 45 89 60 67`

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

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