diff options
Diffstat (limited to 'other/max_tranfer_requests.cpp')
-rw-r--r-- | other/max_tranfer_requests.cpp | 46 |
1 files changed, 0 insertions, 46 deletions
diff --git a/other/max_tranfer_requests.cpp b/other/max_tranfer_requests.cpp deleted file mode 100644 index 2f0e499..0000000 --- a/other/max_tranfer_requests.cpp +++ /dev/null @@ -1,46 +0,0 @@ -// 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 -} |