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// 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 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");
}
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