Smallest subarray whose sum is multiple of array size - GeeksforGeeks

Smallest subarray whose sum is multiple of array size - GeeksforGeeks
Given an array of size N, we need to find smallest sized subarray whose sum is divisible by array size N.

We can solve this problem considering below facts,

Let S[i] denotes sum of first i elements i.e.  
   S[i] = a[1] + a[2] .. + a[i]
Now subarray arr(i, i + x) has sum multiple of N then,
  (arr(i] + arr[i+1] + .... + arr[i + x])) % N = 0
  (S[i+x] – S[i] ) % N  = 0
  S[i] % N = S[i + x] % N 

We need to find the minimum value of x for which the above condition holds. This can be implemented in single iteration with O(N) time-complexity using another array modIdx of size N. Array modIdx is initialized with all elements as -1. modIdx[k] is to be updated with i in each iteration, where k = sum % N.
Now in each iteration we need to update modIdx[k] according to the value of sum % N.

We need to check two things,
If at any instant k = 0 and it is the first time we are updating modIdx[0] (i.e. modIdx[0] was -1)
Then we assign x to i + 1, because (i + 1) will be the length of subarray whose sum is multiple of N
In other case whenever we get a mod value, if this index is not -1, that means it is updated by some other sum value, whose index is stored at that index, we update x with this difference value, i.e. by i – modIdx[k].
After each above operation we update the minimum value of length and corresponding starting index and end index for the subarray. Finally, this gives the solution to our problem.

void printSubarrayMultipleOfN(int arr[], int N)

{

    // A direct index table to see if sum % N

    // has appeared before or not.  

    int modIdx[N];

 

    //  initialize all mod index with -1

    for (int i = 0; i < N; i++)

        modIdx[i] = -1;

 

    // initializing minLen and curLen with larger

    // values

    int minLen = N + 1;

    int curLen = N + 1;

 

    // To store sum of array elements

    int sum = 0;

 

    //  looping for each value of array

    int l, r;

    for (int i = 0; i < N; i++)

    {

        sum += arr[i];

        sum %= N;

 

        // If this is the first time we have

        // got mod value as 0, then S(0, i) % N

        // == 0

        if (modIdx[sum] == -1 && sum == 0)

            curLen = i + 1;

 

        // If we have reached this mod before then

        // length of subarray will be i - previous_position

        if (modIdx[sum] != -1)

            curLen = i - modIdx[sum];

 

        //  choose minimum length os subarray till now

        if (curLen < minLen)

        {

            minLen = curLen;

 

            //  update left and right indices of subarray

            l = modIdx[sum] + 1;

            r = i;

        }

        modIdx[sum] = i;

    }

 

    //  print subarray

    for (int i = l; i <= r; i++)

        cout << arr[i] << " ";

    cout << endl;

}

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