Given an array of size N, we need to find smallest sized subarray whose sum is divisible by array size N.
Input : arr = [1, 1, 2, 2, 4, 2] Output : [2 4] Size of array, N = 6 Following subarrays have sum as multiple of N [1, 1, 2, 2], [2, 4], [1, 1, 2, 2, 4, 2] The smallest among all is [2 4]
We can solve this problem considering below facts,
Let S[i] denotes sum of first i elements i.e. S[i] = a + a .. + 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 (i.e. modIdx 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.
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