Subarray Sums Divisible by K
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 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
分析
sum(0,i)%k = sum(0,j)%k, i+1->j之间可以被K整除。 这里注意先把当前cnt[cur] 加入res, 再increase cnt[cur] 因为每次当前Index只需要之前的数字,不是更新后数字
class Solution:
def subarraysDivByK(self, A: List[int], K: int) -> int:
n = len(A)
mm = {}
mm[0] = 1
res = prefix = 0
for a in A:
prefix += a
rem = prefix%K
if rem in mm:
res += mm[rem]
mm[rem] +=1
else:
mm[rem] = 1
return res
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