0900 - 0999
0974 - Subarray Sums Divisible by K (Medium)
Problem Statement
Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible byk.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [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]Example 2:
Input: nums = [5], k = 9
Output: 0Constraints:
1 <= nums.length <= 3 * 10^4-10^4 <= nums[i] <= 10^42 <= k <= 10^4
Approach 1: Hash Map + Prefix Sum
class Solution {
public:
vector<int> generatePrefixSum(vector<int>& a) {
int n = a.size();
vector<int> pref(n);
pref[0] = a[0];
for (int i = 1; i < n; i++) pref[i] = pref[i - 1] + a[i];
return pref;
}
int subarraysDivByK(vector<int>& nums, int k) {
// generate prefix sum
vector<int> pref = generatePrefixSum(nums);
// store the modular cnt
vector<int> m(k);
// base value
m[0] = 1;
// init ans
int ans = 0;
// iterate each sum
for (auto& x : pref) {
// if x is negative, turn it to positive modular equivalent
if (x < 0) x = (x % k + k) % k;
// we need to find pairs such that (pref[j] - pref[i]) % k == 0
// hence, we look for pref[j] % k == pref[i] % k
// why ..?
// pref[j] = a * k + x
// pref[i] = b * k + y
// pref[j] - pref[i] = (a * k + x) - (b * k + y)
// pref[j] - pref[i] = k * (a - b) + (x - y)
// -> (pref[j] - pref[i]) % k == 0 if & only if (x - y) == 0
m[x % k]++;
}
// n choose 2
for (auto& x : m) ans += x * (x - 1) / 2;
return ans;
}
};