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-rw-r--r--other/make_sum_divisible_p.cpp36
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diff --git a/other/make_sum_divisible_p.cpp b/other/make_sum_divisible_p.cpp
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--- a/other/make_sum_divisible_p.cpp
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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
-}