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-rw-r--r--other/max_tranfer_requests.cpp46
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diff --git a/other/max_tranfer_requests.cpp b/other/max_tranfer_requests.cpp
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--- a/other/max_tranfer_requests.cpp
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-// 1601. Maximum Number of Achievable Transfer Requests
-#include "leetcode.h"
-
-/*
- * we have 'n' buildings numbered from 0 to 'n - 1'. each building has a number
- * of employees. it's transfer season and some employees want to change the
- * building they reside in. you are an an array 'requests' where 'requests[i] =
- * [fromi, toi]' represents an employee's request to transfer from building
- * 'fromi' to building 'toi'. all buildings are full. so a list of requests is
- * achievable only if for each building, the net change in emplyee transfers is
- * zero. this means the number of emplyees leaving is equal to the number of
- * employees moving in. return the maximum number of achievable requests
- */
-
-class Solution {
-public:
- int maximumRequests(int n, vvd & requests) {
- int ans = 0, k = requests.size(), r = 1 << k, x;
- for (int i = 1; i < r; i++) {
- int *a = (int *)calloc(n, sizeof);
- int curr = 0;
- for (int j = 0; j < k; j++) {
- if (int(i >> j) & 1) {
- a[requests[j][0]]++;
- a[requests[j][1]]--;
- curr++;
- }
- }
- for (x = 0; x < n; x++)
- if (a[x])
- break;
- if (x == n)
- ans = max(ans, curr);
- }
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vvd r1 = {{0, 1}, {1, 0}, {0, 1}, {1, 2}, {2, 0}, {3, 4}};
- vvd r2 = {{0, 0}, {1, 2}, {2, 1}}, r3 = {{0, 3}, {3, 1}, {1, 2}, {2, 0}};
- printf("%d\n", obj.maximumRequests(5, r1)); // expect: 5
- printf("%d\n", obj.maximumRequests(3, r2)); // expect: 3
- printf("%d\n", obj.maximumRequests(4, r3)); // expect: 4
-}