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# Find a non empty subset in an array of N integers such that sum of elements of subset is divisible by N

• Difficulty Level : Hard
• Last Updated : 06 Mar, 2023

Given an array of N integers, the task is to find a non-empty subset such that the sum of elements of the subset is divisible by N. Output any such subset with its size and the indices(1-based indexing) of elements in the original array if it exists.
Prerequisites: Pigeonhole Principle
Examples:

```Input: arr[] = { 2, 3, 7, 1, 9 }
Output: 2
1 2
The required subset is { 2, 3 } whose indices are 1 and 2.

Input: arr[] = {2, 11, 4}
Output: 2
2 3 ```

A naive approach will be to generate all possible subsets by using the Power Set of the given array, calculate their respective sums and check if the sum is divisible by N.
Time Complexity: O(2N * N), O(2N) for generating all subsets, and O(N) for calculating the sum of every subset.
Efficient Approach: On considering Prefix Sums, we obtain:

prefixSum0 = arr0
prefixSum1 = arr0 + arr1
prefixSum2 = arr0 + arr1 + arr2

prefixSumN = arr0 + arr1 + arr2 + … + arrN
It can be seen easily that arrL + arrL+1 + … + arrR (L ≤ R) equals to prefixSumR –
prefixSumL-1. If the sum of any contiguous subsegment is divisible by N, then it
means the residue upon taking modulo N of prefixSumR – prefixSumL-1
is zero, i.e.
(prefixSumR – prefixSumL-1) % N = 0;
Splitting the modulo,
prefixSumR % N – prefixSumL-1 % N = 0
prefixSumR % N = prefixSumL-1 % N.

Since there are (N) values of prefixSums and N possible residues for N (0, 1, 2 … N-2, N-1). Hence, according to the pigeonhole principle there always exists a contiguous subsegment whose prefixSum extremities are equal. If at any instance, prefixL, then the first L indexes will give the subset.

Steps were to solve the above approach:

• Initialize an empty unordered_map mp that will store the residue of the prefix sum and its index.
• Initialize a variable sum to 0.
• Iterate over the array arr with index i from 0 to N-1:
• Add the value of arr[i] to sum.
• Take the residue of sum with N, i.e., sum = sum % N.
• If sum is equal to 0:
• Print the size of the subsegment which is (i+1).
• Print the indices of the subsegment which are from 0 to i.
• Return from the function.
• If sum is already in the map:
• Print the size of the subsegment which is (i – mp[sum]).
• Print the indices of the subsegment which are from (mp[sum]+1) to i.
• Return from the function.
• Else mp[sum]=i.

Below is the code to implement the above approach:

## C++

 `// CPP Program to find Non empty``// subset such that its elements' sum``// is divisible by N``#include ``using` `namespace` `std;` `// Function to print the subset index and size``void` `findNonEmptySubset(``int` `arr[], ``int` `N)``{` `    ``// Hash Map to store the indices of residue upon``    ``// taking modulo N of prefixSum``    ``unordered_map<``int``, ``int``> mp;` `    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < N; i++) {``        ``// Calculating the residue of prefixSum``        ``sum = (sum + arr[i]) % N;` `        ``// If the pre[i]%n==0``        ``if` `(sum == 0) {``            ``// print size``            ``cout << i + 1 << endl;` `            ``// Print the first i indices``            ``for` `(``int` `j = 0; j <= i; j++)``                ``cout << j + 1 << ``" "``;``            ``return``;``        ``}` `        ``// If this sum was seen earlier, then``        ``// the contiguous subsegment has been found``        ``if` `(mp.find(sum) != mp.end()) {``            ``// Print the size of subset``            ``cout << (i - mp[sum]) << endl;` `            ``// Print the indices of contiguous subset``            ``for` `(``int` `j = mp[sum] + 1; j <= i; j++)``                ``cout << j + 1 << ``" "``;` `            ``return``;``        ``}``        ``else``            ``mp[sum] = i;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 7, 1, 9 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findNonEmptySubset(arr, N);` `    ``return` `0;``}`

## Java

 `// Java Program to find Non``// empty subset such that``// its elements' sum is``// divisible by N``import` `java.io.*;``import` `java.util.HashMap;``import` `java.util.Map.Entry;``import` `java.util.Map;``import` `java.lang.*;` `class` `GFG``{``    ` `// Function to print the``// subset index and size``static` `void` `findNonEmptySubset(``int` `arr[],``                               ``int` `N)``{` `    ``// Hash Map to store the``    ``// indices of residue upon``    ``// taking modulo N of prefixSum``    ``HashMap mp =``                ``new` `HashMap();` `    ``int` `sum = ``0``;``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``// Calculating the``        ``// residue of prefixSum``        ``sum = (sum + arr[i]) % N;` `        ``// If the pre[i]%n==0``        ``if` `(sum == ``0``)``        ``{``            ``// print size``            ``System.out.print(i + ``1` `+ ``"\n"``);` `            ``// Print the first i indices``            ``for` `(``int` `j = ``0``; j <= i; j++)``                ``System.out.print(j + ``1` `+ ``" "``);``            ``return``;``        ``}` `        ``// If this sum was seen``        ``// earlier, then the``        ``// contiguous subsegment``        ``// has been found``        ``if` `(mp.containsKey(sum) == ``true``)``        ``{``            ``// Print the size of subset``            ``System.out.println((i -``                              ``mp.get(sum)));` `            ``// Print the indices of``            ``// contiguous subset``            ``for` `(``int` `j = mp.get(sum) + ``1``;``                     ``j <= i; j++)``                ``System.out.print(j + ``1` `+ ``" "``);` `            ``return``;``        ``}``        ``else``            ``mp.put(sum,i);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``2``, ``3``, ``7``, ``1``, ``9` `};``    ``int` `N = arr.length;` `    ``findNonEmptySubset(arr, N);``}``}`

## Python3

 `# Python3 Program to find Non``# empty subset such that its``# elements' sum is divisible``# by N` `# Function to print the subset``# index and size``def` `findNonEmptySubset(arr, N):` `    ``# Hash Map to store the indices``    ``# of residue upon taking modulo``    ``# N of prefixSum``    ``mp ``=` `{}` `    ``Sum` `=` `0``    ``for` `i ``in` `range``(N):``        ``# Calculating the residue of``        ``# prefixSum``        ``Sum` `=` `(``Sum` `+` `arr[i]) ``%` `N` `        ``# If the pre[i]%n==0``        ``if` `(``Sum` `=``=` `0``) :``            ``# print size``            ``print``(i ``+` `1``)` `            ``# Print the first i indices``            ``for` `j ``in` `range``(i ``+` `1``):``                ``print``(j ``+` `1``, end ``=` `" "``)``            ``return` `        ``# If this sum was seen earlier,``        ``# then the contiguous subsegment``        ``# has been found``        ``if` `Sum` `in` `mp :``          ` `            ``# Print the size of subset``            ``print``((i ``-` `mp[``Sum``]))` `            ``# Print the indices of contiguous``            ``# subset``            ``for` `j ``in` `range``(mp[``Sum``] ``+` `1``,``                           ``i ``+` `1``):``                ``print``(j ``+` `1``, end ``=` `" "``)` `            ``return``        ``else``:``            ``mp[``Sum``] ``=` `i``            ` `# Driver code``arr ``=` `[``2``, ``3``, ``7``, ``1``, ``9``]``N ``=` `len``(arr)``findNonEmptySubset(arr, N)` `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# Program to find Non``// empty subset such that``// its elements' sum is``// divisible by N` `using` `System;``using` `System.Collections ;` `class` `GFG``{    ``    ``// Function to print the``    ``// subset index and size``    ``static` `void` `findNonEmptySubset(``int` `[]arr, ``int` `N)``    ``{``    ` `        ``// Hash Map to store the``        ``// indices of residue upon``        ``// taking modulo N of prefixSum``        ``Hashtable mp = ``new` `Hashtable();``    ` `        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``            ``// Calculating the``            ``// residue of prefixSum``            ``sum = (sum + arr[i]) % N;``    ` `            ``// If the pre[i]%n==0``            ``if` `(sum == 0)``            ``{``                ``// print size``                ``Console.Write(i + 1 + ``"\n"``);``    ` `                ``// Print the first i indices``                ``for` `(``int` `j = 0; j <= i; j++)``                    ``Console.Write(j + 1 + ``" "``);``                ``return``;``            ``}``    ` `            ``// If this sum was seen``            ``// earlier, then the``            ``// contiguous subsegment``            ``// has been found``            ``if` `(mp.ContainsKey(sum) == ``true``)``            ``{``                ``// Print the size of subset``                    ``Console.WriteLine(i - Convert.ToInt32(mp[sum]));``    ` `                ``// Print the indices of``                ``// contiguous subset``                ``for` `(``int` `j = Convert.ToInt32(mp[sum]) + 1;``                        ``j <= i; j++)``                        ``Console.Write(j + 1 + ``" "``);``    ` `                ``return``;``            ``}``            ``else``                ``mp.Add(sum,i);``        ``}``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 2, 3, 7, 1, 9 };``        ``int` `N = arr.Length;``    ` `        ``findNonEmptySubset(arr, N);``    ``}``    ``// This code is contributed by Ryuga``}`

## Javascript

 ``

Output:

```2
1 2```

Time Complexity: O(N), where N is the size of the array.
Auxiliary Space: O(N)

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