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] = 1000Subsets 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]} = 1011Therefore, 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 <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; } |
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<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 |
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 |
C#
// C# Program to implement // the above approach using System;using System.Collections.Generic; class GFG { // Function to rearrange the array such // that no same-indexed subset have sum // equal to that in the original array static void printNewArray(List<int> a, int n) { // Initialize a vector List<Tuple<int, int>> v = new List<Tuple<int, int>>(); // Iterate the array for (int i = 0; i < n; i++) { v.Add(new Tuple<int, int>(a[i],i)); } // Sort the vector v.Sort(); 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[(i + 1) % n].Item2] = v[i].Item1; } // Print the answer for (int i = 0; i < n; i++) { Console.Write(ans[i] + " "); } } // Driver code static void Main() { List<int> a = new List<int>(new int[]{4, 1, 2, 5, 3}); int n = a.Count; printNewArray(a, n); }}// This code is contributed by divyesh072019 |
Javascript
<script> // Javascript 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 function printNewArray(a, n) { // Initialize a vector var v = []; // Iterate the array for (var i = 0; i < n; i++) { v.push([ a[i], i]); } // Sort the vector v.sort(); var ans = Array(n); // Shift of elements to the // index of its next cyclic element for (var i = 0; i < n; i++) { ans[v[(i + 1) % n][1]] = v[i][0]; } // Print the answer for (var i = 0; i < n; i++) { document.write( ans[i] + " "); } } // Driver Code var a = [4, 1, 2, 5, 3]; var n = a.length; printNewArray(a, n); </script> |
Output:
3 5 1 4 2
Time Complexity: O(N log N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
