Count Pairs With Given Sum | Array Interview

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Time Lines
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0:05 Introduction
2:19 Brute Force Solution
3:21 Best Solution
Count pairs with given sum is one of the array interview question, I have explained both the brute force and the best way to solve it.
First approach can be using two different loops and just have the comparison of all the pairs possible.Second approach or the best approach is to use an unordered_map and keep the record of occurrences.
#array #dsa #datastructure #Algorithms

[@MuthuKumar-vl8of]

I think, in unordered_map you wrongly added -1 instead of 7

[@FANSasFRIENDS]

Good speed and voice, keep it up!

[@CppNuts]

Thanks man

[@ergingokkaya8323]

Excellent

[@nquanta1548]

👍nice

[@its_dsa]

I think -1 entry will not be there . implemeted this
int main(void)
{
int sum = 6,count = 0;
vector v = {1,5,7,1,1};
unordered_map m;
for ( auto i = 0 ; i < v.size(); i++)
{
if ( m.find(sum-v[i]) == m.end() )
{
m.insert(make_pair(v[i],1));
}
else
{
count = count + m[sum-v[i]];
m.insert(make_pair(v[i],1));
}
}
for ( auto i : m)
cout

[@CppNuts]

Hi guys sorry for the confusion, you guys are correct, i don't know why i said that that doesn't make sense to put (sum-Arr[i]) in umap, i wanted to tell that 7 will be entered in the umap.

[@its_dsa]

@Suraj Kumar modified code for better modularity :
#include
#include
#include
using namespace std;
vector v = {1,1,1,1};
int Findpairsum(int sum)
{
int count = 0;
unordered_map m;
for ( auto i = 0 ; i < v.size(); i++)
{
if ( m.find(sum-v[i]) == m.end() )
{
m.insert(make_pair(v[i],1));
}
else
{
count = count + m[sum-v[i]];
m.insert(make_pair(v[i],1));
}
}
for ( auto i : m)
cout

[@CppNuts]

Output should be 6 not 3.
As i explained in video..

[@unoshihaam]

@@CppNutsi have tried to implement algo so that the output for arr = {1,1,1,1} is 6 but all the time i am getting three . sir can u please give the logic to implement the code

[@AmitKumar-tp4fv]

#include
// //[1,7,3,5,4]
// using namespace std;
// void countsum(int arr[],int n,int target)
// {
// int cnt=0;
// setmp;
// for(int i=0;iarr[i];
// }
// int target;
// cin>>target;
// countsum(arr,n,target);
// }