# Rearrange array to make product of prefix sum array non zero

• Difficulty Level : Medium
• Last Updated : 11 May, 2021

Given an array, arr[ ] of size N, the task is to rearrange the given array such that the product of all the elements of its prefix sum array is not equal to 0. If it is not possible to rearrange the array that satisfies the given condition, then print -1.

Examples:

Input: arr[] = {1, -1, -2, 3}
Output: 3 1 -1 -2
Explanation:
Prefix sum after rearranging the given array to {3, 1, -1, -2} are {3, 4, 3, 1} and product all the elements of its prefix sum array = (3 * 4 * 3 * 1) = 36.
Therefore, the required array is {3, 1, -1, -2}

Input: arr = {1, 1, -1, -1}
Output: -1

Approach: The idea is to sort the given array either in ascending order or descending order so that any element of its prefix sum not equal to 0. Follow the steps below to solve the problem:

• Calculate the sum of elements of the given array, say totalSum.
• If totalSum = 0 then print -1.
• If totalSum > 0 then print the given array in decreasing order so that any elements of its prefix sum not equal to 0.
• If totalSum < 0 then print the given array in ascending order so that any elements of its prefix sum not equal to 0.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to print array elements``void` `printArr(``int` `arr[], ``int` `N)``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``cout << arr[i] << ``" "``;``    ``}``}` `// Function to rearrange array``// that satisfies the given condition``void` `rearrangeArr(``int` `arr[], ``int` `N)``{``    ``// Stores sum of elements``    ``// of the given array``    ``int` `totalSum = 0;` `    ``// Calculate totalSum``    ``for` `(``int` `i = 0; i < N; i++) {``        ``totalSum += arr[i];``    ``}` `    ``// If the totalSum is equal to 0``    ``if` `(totalSum == 0) {` `        ``// No possible way to``        ``// rearrange array``        ``cout << ``"-1"` `<< endl;``    ``}` `    ``// If totalSum exceeds 0``    ``else` `if` `(totalSum > 0) {` `        ``// Rearrange the array``        ``// in descending order``        ``sort(arr, arr + N,``             ``greater<``int``>());``        ``printArr(arr, N);``    ``}` `    ``// Otherwise``    ``else` `{` `        ``// Rearrange the array``        ``// in ascending order``        ``sort(arr, arr + N);``        ``printArr(arr, N);``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, -1, -2, 3 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``rearrangeArr(arr, N);``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{` `// Function to print array elements``static` `void` `printArr(``int` `arr[], ``int` `N)``{``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``System.out.print(arr[i] + ``" "``);``    ``}``}` `// Function to rearrange array``// that satisfies the given condition``static` `void` `rearrangeArr(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores sum of elements``    ``// of the given array``    ``int` `totalSum = ``0``;` `    ``// Calculate totalSum``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``totalSum += arr[i];``    ``}` `    ``// If the totalSum is equal to 0``    ``if` `(totalSum == ``0``)``    ``{``        ` `        ``// No possible way to``        ``// rearrange array``        ``System.out.print(``"-1"` `+ ``"\n"``);``    ``}` `    ``// If totalSum exceeds 0``    ``else` `if` `(totalSum > ``0``)``    ``{``        ` `        ``// Rearrange the array``        ``// in descending order``        ``Arrays.sort(arr);``        ``arr = reverse(arr);``        ``printArr(arr, N);``    ``}``    ` `    ``// Otherwise``    ``else``    ``{``        ` `        ``// Rearrange the array``        ``// in ascending order``        ``Arrays.sort(arr);``        ``printArr(arr, N);``    ``}``}` `static` `int``[] reverse(``int` `a[])``{``    ``int` `i, n = a.length, t;``    ``for``(i = ``0``; i < n / ``2``; i++)``    ``{``        ``t = a[i];``        ``a[i] = a[n - i - ``1``];``        ``a[n - i - ``1``] = t;``    ``}``    ``return` `a;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, -``1``, -``2``, ``3` `};``    ``int` `N = arr.length;``    ` `    ``rearrangeArr(arr, N);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to rearrange array``# that satisfies the given condition``def` `rearrangeArr(arr, N):``    ` `    ``# Stores sum of elements``    ``# of the given array``    ``totalSum ``=` `0``    ` `    ``# Calculate totalSum``    ``for` `i ``in` `range``(N):``        ``totalSum ``+``=` `arr[i]` `    ``# If the totalSum is equal to 0``    ``if` `(totalSum ``=``=` `0``):``        ` `        ``# No possible way to``        ``# rearrange array``        ``print``(``-``1``)` `    ``# If totalSum exceeds 0``    ``elif` `(totalSum > ``0``):``        ` `        ``# Rearrange the array``        ``# in descending order``        ``arr.sort(reverse ``=` `True``)``        ``print``(``*``arr, sep ``=` `' '``)``        ` `    ``# Otherwise``    ``else``:` `        ``# Rearrange the array``        ``# in ascending order``        ``arr.sort()``        ``print``(``*``arr, sep ``=` `' '``)` `# Driver Code``arr ``=` `[ ``1``, ``-``1``, ``-``2``, ``3` `]``N ``=` `len``(arr)` `rearrangeArr(arr, N);` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to print array elements``static` `void` `printArr(``int` `[]arr, ``int` `N)``{``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``Console.Write(arr[i] + ``" "``);``    ``}``}` `// Function to rearrange array``// that satisfies the given condition``static` `void` `rearrangeArr(``int` `[]arr, ``int` `N)``{``    ` `    ``// Stores sum of elements``    ``// of the given array``    ``int` `totalSum = 0;` `    ``// Calculate totalSum``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``totalSum += arr[i];``    ``}` `    ``// If the totalSum is equal to 0``    ``if` `(totalSum == 0)``    ``{``        ` `        ``// No possible way to``        ``// rearrange array``        ``Console.Write(``"-1"` `+ ``"\n"``);``    ``}` `    ``// If totalSum exceeds 0``    ``else` `if` `(totalSum > 0)``    ``{``        ` `        ``// Rearrange the array``        ``// in descending order``        ``Array.Sort(arr);``        ``arr = reverse(arr);``        ``printArr(arr, N);``    ``}``    ` `    ``// Otherwise``    ``else``    ``{``        ` `        ``// Rearrange the array``        ``// in ascending order``        ``Array.Sort(arr);``        ``printArr(arr, N);``    ``}``}` `static` `int``[] reverse(``int` `[]a)``{``    ``int` `i, n = a.Length, t;``    ``for``(i = 0; i < n / 2; i++)``    ``{``        ``t = a[i];``        ``a[i] = a[n - i - 1];``        ``a[n - i - 1] = t;``    ``}``    ``return` `a;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, -1, -2, 3 };``    ``int` `N = arr.Length;``    ` `    ``rearrangeArr(arr, N);``}``}` `// This code is contributed by Princi Singh`

## Javascript

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Output:

`3 1 -1 -2`

Time Complexity: O(N logN)
Space Complexity: O(1)

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