// 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 leftmostBuildingQueries(vector &heights, vector> &queries) { int hn = heights.size(), qn = queries.size(); vector>> q(hn); priority_queue, vector>, greater>> pq; vector 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 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 lmbq1 = obj.leftmostBuildingQueries(h1, q1); vector 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"); }