# Rearrange an Array such that Sum of same-indexed subsets differ from their Sum in the original Array

Given an array A[] consisting of N distinct integers, the task is to rearrange the given array such that the sum of every same-indexed non-empty subsets of size less than N, is not equal to their sum in the original array.
Examples:

Input: A[] = {1000, 100, 10, 1}
Output: 100 10 1 1000
Explanation:
Original Array A[] = {1000, 100, 10, 1}
Final Array B[] = {100, 10, 1, 1000}
Subsets of size 1:

```A = 1000  B = 100
A = 100   B = 10
A = 10    B = 1
A = 1     B = 1000

```

Subsets of size 2:

```{A, A} = 1100  {B, B} = 110
{A, A} = 1010  {B, B} = 101
{A, A} = 110   {B, B} = 11
.....
Similarly, all same-indexed subsets of size 2 have a different sum.

```

Subsets of size 3:

```{A, A, A} = 1110  {B, B, B} = 111
{A, A, A} = 1011 {B, B, B} = 1101
{A, A, A} = 111  {B, B, B} = 1011

```

Therefore, no same-indexed subsets have equal sum.
Input: A[] = {1, 2, 3, 4, 5}
Output: 5 1 2 3 4

Approach:
The idea is to simply replace every array element except one, by a smaller element. Follow the steps below to solve the problem:

• Store the array elements in pairs of {A[i], i}.
• Sort the pairs in ascending order Of the array elements
• Now, traverse the sorted order, and insert each element at the original index of its next greater element(i.e. at the index v[(i + 1) % n].second). This ensures that every index except one now has a smaller element than the previous value stored in it.

Proof:
Let S = { arr1, arr2, …, arrk } be a subset.
If u does not belong to S initially, upon insertion of u into S, the sum of the subset changes.
Similarly, if u belongs to S, let S’ contains all the elements not present in S. This means that u do not belong to S’. Then, by the same reasoning above, the sum of the subset S’ differs from its original sum.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; `   `// Function to rearrange the array such ` `// that no same-indexed subset have sum ` `// equal to that in the original array ` `void` `printNewArray(vector<``int``> a, ``int` `n) ` `{ ` `    ``// Initialize a vector ` `    ``vector > v; `   `    ``// Iterate the array ` `    ``for` `(``int` `i = 0; i < n; i++) { `   `        ``v.push_back({ a[i], i }); ` `    ``} `   `    ``// Sort the vector ` `    ``sort(v.begin(), v.end()); `   `    ``int` `ans[n]; `   `    ``// Shift of elements to the ` `    ``// index of its next cyclic element ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``ans[v[(i + 1) % n].second] ` `            ``= v[i].first; ` `    ``} `   `    ``// Print the answer ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cout << ans[i] << ``" "``; ` `    ``} ` `} `   `// Driver Code ` `int` `main() ` `{ ` `    ``vector<``int``> a = { 4, 1, 2, 5, 3 }; `   `    ``int` `n = a.size(); `   `    ``printNewArray(a, n); `   `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.io.*;` `import` `java.util.*; `   `class` `GFG{` `    `  `static` `class` `pair` `{` `    ``int` `first, second;` `    `  `    ``pair(``int` `first, ``int` `second)` `    ``{` `        ``this``.first = first;` `        ``this``.second = second;` `    ``}` `}`   `// Function to rearrange the array such ` `// that no same-indexed subset have sum ` `// equal to that in the original array ` `static` `void` `printNewArray(List a, ``int` `n) ` `{ ` `    `  `    ``// Initialize a vector ` `    ``List v = ``new` `ArrayList<>(); ` `    `  `    ``// Iterate the array ` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{ ` `        ``v.add(``new` `pair(a.get(i), i)); ` `    ``} ` `    `  `    ``// Sort the vector ` `    ``Collections.sort(v, (pair s1, pair s2) ->` `    ``{` `        ``return` `s1.first - s2.first;` `    ``});` `    `  `    ``int` `ans[] = ``new` `int``[n]; `   `    ``// Shift of elements to the ` `    ``// index of its next cyclic element ` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{ ` `        ``ans[v.get((i + ``1``) % n).second] = v.get(i).first; ` `    ``} ` `    `  `    ``// Print the answer ` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{ ` `        ``System.out.print(ans[i] + ``" "``); ` `    ``} ` `}`   `// Driver Code ` `public` `static` `void` `main(String args[])` `{ ` `    ``List a = Arrays.asList(``4``, ``1``, ``2``, ``5``, ``3``); `   `    ``int` `n = a.size(); ` `    ``printNewArray(a, n); ` `} ` `} `   `// This code is contributed by offbeat`

## Python3

 `# Python3 Program to implement ` `# the above approach `   `# Function to rearrange the array such ` `# that no same-indexed subset have sum ` `# equal to that in the original array ` `def` `printNewArray(a, n):`   `    ``# Initialize a vector ` `    ``v ``=` `[]`   `    ``# Iterate the array ` `    ``for` `i ``in` `range` `(n):` `        ``v.append((a[i], i ))` `    `  `    ``# Sort the vector ` `    ``v.sort()`   `    ``ans ``=` `[``0``] ``*` `n`   `    ``# Shift of elements to the ` `    ``# index of its next cyclic element ` `    ``for` `i ``in` `range` `(n):` `        ``ans[v[(i ``+` `1``) ``%` `n][``1``]] ``=` `v[i][``0``]` `   `  `    ``# Print the answer ` `    ``for` `i ``in` `range` `(n):` `        ``print` `(ans[i], end ``=` `" "``)`   `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``:  ` `    ``a ``=` `[``4``, ``1``, ``2``, ``5``, ``3``]` `    ``n ``=` `len``(a)` `    ``printNewArray(a, n)`   `# This code is contributed by Chitranayal`

Output:

```3 5 1 4 2

```

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

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