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// 1590. Make Sum Divisible by P
#include "leetcode.h"
/*
* given an array of positive integers 'nums', remove the smallest subarray
* (possibly empty) such that the sum of the remaining elements is divisible by
* 'p'. it is not allowed to remove the whole array. return the length of the
* smallest subarray that you need to remove, or -1 if it is impossible. a
* subarray is defined as a contiguous block of elements in the array.
*/
class Solution {
public:
int minSubarray(vector<int> &nums, int p) {
int n = nums.size(), ans = n, req = 0, curr = 0;
for (auto i : nums)
req = (req + i) % p;
unordered_map<int, int> dp = {{0, -1}};
for (int i = 0; i < n; i++) {
curr = (curr + nums[i]) % p;
dp[curr] = i;
int need = (curr - req + p) % p;
if (dp.count(need))
ans = min(ans, i - dp[need]);
}
return ans < n ? ans : -1;
}
};
int main() {
Solution obj;
vector<int> n1 = {3, 1, 4, 2}, n2 = {6, 3, 5, 2}, n3 = {1, 2, 3};
printf("%d\n", obj.minSubarray(n1, 6)); // expect: 1
printf("%d\n", obj.minSubarray(n2, 9)); // expect: 2
printf("%d\n", obj.minSubarray(n3, 3)); // expect: 0
}
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