45 lines
1.9 KiB
C++
45 lines
1.9 KiB
C++
class cmp {
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public:
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// overloading the function call operator
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bool operator()(pair<int,pair<int,int>> p1, pair<int,pair<int,int>> p2){
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return p1.first < p2.first;
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}
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};
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class Solution {
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public:
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vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
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int size1 = nums1.size();
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int size2 = nums2.size();
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vector<vector<int> > ans;
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priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,cmp> pq;
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for(int i=0; i<size1; i++){
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for(int j=0; j<size2; j++){
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int sum=nums1[i]+nums2[j];
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if(pq.size()<k){
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pq.push( {sum , {nums1[i] , nums2[j]} } );
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}
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// sum is smaller than the sum stored at top(), which is the max in the heap, then let that max go;
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else if(sum < pq.top().first){
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pq.pop();
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pq.push( {sum , {nums1[i] , nums2[j]} } );
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}else{
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// else means, we have k elements in the heap, and sum is larger than the max; then all remaining indices (in this current row) would just generate larger sums.
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// so this break is not breaking out of the entire nested loops, but rather, just break out of the inner loop, because we now know matrix[i][j] is already excluded, then of course, matrix[i][j+1], matrix[i][j+2], etc., should all be excluded.
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break;
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}
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}
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}
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while(!pq.empty()){
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// this is a max heap, so every time we get is the largest; therefore data in ans is sorted (based on the sum) in descending order
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ans.push_back({pq.top().second.first , pq.top().second.second});
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pq.pop();
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}
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// reverse it so we get it in ascending order.
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reverse(ans.begin(), ans.end());
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return ans;
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}
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};
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