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++

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// 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; 

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Java

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// 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

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Output: 

3 5 1 4 2

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

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Improved By : offbeat