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+// 2300. Successful Pairs of Spells and Potions
+#include "leetcode.h"
+
+/*
+ * given two positive integer arrays 'spells' and 'potions', of length n and m
+ * respectively where 'spells[i]' represents the strength of the i'th spell and
+ * 'potions[j] represents the strength of the j'th potion. also given an integer
+ * success. a spell and potion pair is considered successful if the product of
+ * their strengths is at least 'success'. return an integer array 'pairs' of
+ * lengthn where 'pairs[i]' is the number of potions that will form a successful
+ * pair with the i'th spell.
+ */
+
+class Solution {
+public:
+ vector<int> successfulPairs(vector<int> &spells, vector<int> &potions,
+ long long success) {
+ int m = potions.size();
+ vector<int> ans;
+ sort(potions.begin(), potions.end());
+ for (auto i : spells) {
+ int left = 0, right = m - 1, mid;
+ while (left <= right) {
+ mid = (left + right) >> 1;
+ if ((long long)i * potions[mid] >= success)
+ right = mid - 1;
+ else
+ left = mid + 1;
+ }
+ ans.push_back(m - left);
+ }
+ return ans;
+ }
+};
+
+int main() {
+ Solution obj;
+ vector<int> s1 = {5, 1, 3}, s2 = {3, 1, 2};
+ vector<int> p1 = {1, 2, 4, 4, 5}, p2 = {8, 5, 8};
+ for (auto i : obj.successfulPairs(s1, p1, 7))
+ printf("%d ", i); // expec: 4 0 3
+ printf("\n");
+ for (auto i : obj.successfulPairs(s2, p2, 16))
+ printf("%d ", i); // expec: 2 0 2
+}