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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,],]
Updated on: 29-Apr-2020
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