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

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(2^N * N), O(2^N) 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
…
prefixSum_N = arr₀ + arr₁ + arr₂ + … + arr_N
It can be seen easily that arr_L + arr_L+1 + … + arr_R (L ? R) equals to prefixSum_R – 
prefixSum_L-1. If the sum of any contiguous subsegment is divisible by N, then it 
means the residue upon taking modulo N of prefixSum_R – prefixSum_L-1 
is zero, i.e.
(prefixSum_R – prefixSum_L-1) % N = 0;
Splitting the modulo, 
prefixSum_R % N – prefixSum_L-1 % N = 0 
prefixSum_R % N = prefixSum_L-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, prefix_L, 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: 
 

2
1 2

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