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Diffstat (limited to 'max_tranfer_requests.cpp')
-rw-r--r-- | max_tranfer_requests.cpp | 46 |
1 files changed, 46 insertions, 0 deletions
diff --git a/max_tranfer_requests.cpp b/max_tranfer_requests.cpp new file mode 100644 index 0000000..c23f66b --- /dev/null +++ b/max_tranfer_requests.cpp @@ -0,0 +1,46 @@ +// 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(int) & 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)); + 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(int) r1 = {{0, 1}, {1, 0}, {0, 1}, {1, 2}, {2, 0}, {3, 4}}; + vvd(int) 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 +} |