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[0] = 1000  B[0] = 100
A[1] = 100   B[1] = 10
A[2] = 10    B[2] = 1
A[3] = 1     B[3] = 1000




Subsets of size 2: 

{A[0], A[1]} = 1100  {B[0], B[1]} = 110
{A[0], A[2]} = 1010  {B[0], B[2]} = 101
{A[1], A[2]} = 110   {B[1], B[2]} = 11
.....
Similarly, all same-indexed subsets of size 2 have a different sum.




Subsets of size 3: 
 

{A[0], A[1], A[2]} = 1110  {B[0], B[1], B[2]} = 111
{A[0], A[2], A[3]} = 1011 {B[0], B[2], B[3]} = 1101
{A[1], A[2], A[3]} = 111  {B[1], B[2], B[3]} = 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
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<pair<int, int> > 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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<Integer> a, int n)
{
     
    // Initialize a vector
    List<pair> 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<Integer> a = Arrays.asList(4, 1, 2, 5, 3);
 
    int n = a.size();
    printNewArray(a, n);
}
}
 
// This code is contributed by offbeat

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


Output: 

3 5 1 4 2



 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : offbeat, chitranayal