Find numbers of subarray of an array whose sum is divided by given number [duplicate]

Answer 1

I may have an easier solution. in O(n) time and O(n+k) space. where n is the size of array & k is the number we are checking the divisibility with.

Consider the array as A[n] and the number is K

1. create another array SUM_TILL_NOW[n].
2. for each A[i] fill SUM_TILL_NOW [i]= SUM_TILL_NOW[i-1]+A[i] %K; (SUM_TILL_NOW[0]= A[0])
3. find two numbers which are equal in this new Array.

To do that create a new array CHECK[] of size K.

Iterate over the SUM_TILL_NOW array and check if CHECK[SUM_TILL_NOW[i]] set.

If not set it to i.

else CHECK[SUM_TILL_NOW[i]], i is the subset where the sum is divisible by K.

Below is a c++ function of the same.

#include 
#include 

using namespace std;

void printrange(int* A, int N, int K)
{
    int STN[N], C[K];
    memset(C, -1, K*sizeof(int));
    int i;
    int sum=A[0];
    STN[0]= (A[0]%K);
    for (i= 1; i< N; i++)
    {
        sum+= A[i];
        STN[i]= sum%K;
    }
    for(i=0; i< N; i++)
    {
        if(C[STN[i]] == -1)
            C[STN[i]] =i;
        else
        {
            cout<< C[STN[i]]+1 <<" "<< i;
            break;
        }
    }
}

int main()
{
    int A[]= {6, 9, 2, 1, 8, 6, 2, 5};
    printrange(A, sizeof(A)/sizeof(A[0]), 7);
    return 0;
}