Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Note:

1. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

Solution: Count prefix sums

let c[i] denotes the counts of prefix_sum % K init: c[0] = 1
Whenever we end up with the same prefix sum (after modulo), which means there are subarrys end with current element that is divisible by K (0 modulo).

e.g. A = [4,5,0,-2,-3,1], K = 5
[4,5] has prefix sum of 4, which happens at index 0 [4], and index 1, [4,5]
[4,5,0] also has a prefix sum of 4, which means [4, {5,0}], [4,5, {0}] are divisible by K.

ans += (c[prefix_sum] – 1)
i = 0, prefix_sum = 0, c[(0+4)%5] = c[4] = 1, ans = 0
i = 1, prefix_sum = 4+5, c[(4+5)%5] = c[4] = 2, ans = 0+2-1=0 => [5]
i = 2, prefix_sum = 4+0, c[(4+0)%5] = c[4] = 3, ans = 1+3-1=3 => [5], [5,0], [0]
i = 3, prefix_sum = 4-2, c[(4-2)%5] = c[2] = 1, ans = 3
i = 4, prefix_sum = 2-3, c[(2-3+5)%5] = c[4] = 4, ans = 3+4-1=6 => [5],[5,0],[0],[5,0,-2,-3], [0,-2,-3],[-2,-3]
i = 5, prefix_sum = 4+1, c[(4+1)%5] = c[0] = 2, ans = 6 + 2 – 1 =>
[5],[5,0],[0],[5,0,-2,-3], [0,-2,-3],[-2,-3], [4,5,0,-2,-3,1]

Time complexity: O(n)
Space complexity: O(n)

## Python3

If you like my articles / videos, donations are welcome.

Buy anything from Amazon to support our website

Paypal
Venmo
huahualeetcode

## Be First to Comment

Mission News Theme by Compete Themes.