
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find K Pairs with Smallest Sums in C++
Suppose we have two sorted arrays A1 and A2, and another value k. We have to define a pair (u, v) which is consists of one element from A1 and another element from A2. We have to find the k pairs like [(u1, v1), (u2, v2),…, (uk, vk)]. So if A1 = [1, 7, 11] and A2 = [2, 4, 6], and k = 3, then output will be [(1, 2), (1, 4), (1, 6)]
To solve this, we will follow these steps −
- Define one data type, that will take two values a and b, and index.
- create one priority queue, that will take key of type data and list of data as value.
- n := size of A1, m := size of A2
- if n is 0 or m is 0, then return
- create one matrix called ret
- for i in range 0 to n – 1
- insert (A1[i], A2[0], 0) as data into queue
- while queue is not empty, and k is not 0, then
- curr := top of queue, delete queue element
- insert first_val of curr, second_val of curr into ret
- if index of curr + 1 < m, then
- insert data with (first_val of curr, A2[index of curr + 1], index of curr + 1), into queue
- decrease k by 1
- return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h> #include <stack> using namespace std; struct Data{ int firstVal, secondVal, idx; Data(int a, int b, int c){ firstVal = a; secondVal = b; idx = c; } }; struct Comparator{ bool operator()(Data a, Data b){ return !(a.firstVal + a.secondVal < b.firstVal + b.secondVal); } }; class Solution { public: vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) { priority_queue <Data, vector <Data>, Comparator> pq; int n = nums1.size(); int m = nums2.size(); if(!n || !m)return {}; vector < vector <int> > ret; for(int i = 0; i < n; i++){ pq.push(Data(nums1[i], nums2[0], 0)); } while(!pq.empty() && k--){ Data curr = pq.top(); pq.pop(); ret.push_back({curr.firstVal, curr.secondVal}); if(curr.idx + 1 < m){ pq.push(Data(curr.firstVal, nums2[curr.idx + 1], curr.idx + 1)); } } return ret; } }; void print(vector <int> const &arr) { cout<<"["; for(int i=0; i < arr.size(); i++) std::cout << arr.at(i) <<","; cout<<"]"; } int main() { vector<int> nums1{1,7,11}; vector<int> nums2{2,4,6}; int k = 3; Solution ob1; vector<vector<int>> numsRet; numsRet = ob1.kSmallestPairs(nums1, nums2, k); cout<<"["; for (vector<int> x : numsRet) { print(x); cout<<","; } cout<<"]"<<endl; return 0; }
Input
[1,7,11] [2,4,6] 3 vector<int> nums1{1,7,11}; vector<int> nums2{2,4,6}; int k = 3; Solution ob1; vector<vector<int>> numsRet; numsRet = ob1.kSmallestPairs(nums1, nums2, k);
Output
[[1,2,],[1,4,],[1,6,],]
Advertisements