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author | Jack Sangdahl <[email protected]> | 2024-12-21 17:32:46 -0700 |
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committer | Jack Sangdahl <[email protected]> | 2024-12-21 17:32:46 -0700 |
commit | 27f5c468d6732af671cebc250ad5079e98f2b7e2 (patch) | |
tree | f039954848db59f47fe524eaa823bd6619b43bf2 /find_building_alice_bob_meet.cpp | |
parent | c4bf635f931df151edbbbf47be3b37c730d9beed (diff) | |
download | leetcode-27f5c468d6732af671cebc250ad5079e98f2b7e2.tar.gz leetcode-27f5c468d6732af671cebc250ad5079e98f2b7e2.zip |
2940. find building where alice and bob can meet
Diffstat (limited to 'find_building_alice_bob_meet.cpp')
-rw-r--r-- | find_building_alice_bob_meet.cpp | 59 |
1 files changed, 59 insertions, 0 deletions
diff --git a/find_building_alice_bob_meet.cpp b/find_building_alice_bob_meet.cpp new file mode 100644 index 0000000..9067bcf --- /dev/null +++ b/find_building_alice_bob_meet.cpp @@ -0,0 +1,59 @@ +// 2940. Find Building Where Alice and Bob Can Meet +#include "leetcode.h" + +/* + * you are given a 0-indexed array 'heights' of positive integers where + * 'heights[i]' represents the height of the i'th building. if a person is in + * building 'i', they can move to any other building 'j' if and only if 'i < j' + * and 'heights[i] < heights[j]'. you are also given an array 'queries' where + * 'queries[i] = [a_i, b_i]'. on the i'th query, alice is in building 'a_i', and + * bob is in building 'b_i'. return array 'ans' where 'ans[i]' is the index of + * the leftmost building where alice and bob can meet on i'th query. if alice + * and bob cannot move to a common building on query 'i', set 'ans[i]' to -1 + */ + +class Solution { +public: + vector<int> leftmostBuildingQueries(vector<int> &heights, + vector<vector<int>> &queries) { + int hn = heights.size(), qn = queries.size(); + vector<vector<vector<int>>> q(hn); + priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq; + vector<int> ans(qn, -1); + for (int qi = 0; qi < qn; qi++) { + int i = queries[qi][0], j = queries[qi][1]; + if (i < j && heights[i] < heights[j]) + ans[qi] = j; + else if (i > j && heights[i] > heights[j]) + ans[qi] = i; + else if (i == j) + ans[qi] = i; + else + q[max(i, j)].push_back({max(heights[i], heights[j]), qi}); + } + for (int i = 0; i < hn; i++) { + while (!pq.empty() && pq.top()[0] < heights[i]) { + ans[pq.top()[1]] = i; + pq.pop(); + } + for (auto &j : q[i]) + pq.push(j); + } + return ans; + } +}; + +int main() { + Solution obj; + vector<int> h1 = {6, 4, 8, 5, 2, 7}, h2 = {5, 3, 8, 2, 6, 1, 4, 6}; + vvd(int) q1 = {{0, 1}, {0, 3}, {2, 4}, {3, 4}, {2, 2}}, + q2 = {{0, 7}, {3, 5}, {5, 2}, {3, 0}, {1, 6}}; + vector<int> lmbq1 = obj.leftmostBuildingQueries(h1, q1); + vector<int> lmbq2 = obj.leftmostBuildingQueries(h2, q2); + for (auto i : lmbq1) + printf("%d ", i); // expect: 2 5 -1 5 2 + printf("\n"); + for (auto i : lmbq2) + printf("%d ", i); // expect: 7 6 -1 4 6 + printf("\n"); +} |