# TLE in SUM OF THREE VALUES

Can anyone please pinpoint the reason of TLE in this code.
https://cses.fi/paste/764a138ddfa322b81b912c/
Problem Name - CSES SUM OF THREE VALUES.

The code is O(N^{3}) . Even though n<=5000 , I don’t think O(N^{3}) will work. An O(N^{2}) solution might be accepted.

3rd for loop will only work for once so in my knowledge i think it will work for O(n^2);

`unordered_map` access is O(n) in the worst case, not O(1).

After using this resource I decided to change my hashmap function but there is still TLE.
So after that I just switch to sliding window technique to solve that question in O(n^2).But thanks for the resource.
Also congratulation on solving 6 question in march challenge.

Alternatively, you can fix the middle element and then use two pointers to solve this.

Code
``````#include <iostream>
#include <algorithm>
#include <vector>

/*#pragma GCC target ("avx2")
#pragma GCC optimization ("Ofast")
#pragma GCC optimization ("unroll-loops")
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")*/

using namespace std;

void usaco()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
}

#define int long long

int32_t main()
{
usaco();
int n, x;
cin >> n >> x;
vector <pair <int, int>> a(n);
for (int i = 0; i < n; ++i)
{
cin >> a[i].first;
a[i].second = i+1;
}
if (n < 3)
{
cout << "IMPOSSIBLE\n";
return 0;
}
sort(begin(a), end(a));
for (int k = 0; k < n; ++k)
{
int temp = x-a[k].first;
int i = 0, j = n-1;
while (i < j)
{
if (i == k) ++i;
if (j == k) --j;
if (i >= j) break;
if (a[i].first + a[j].first == temp)
{
cout << a[i].second << " " << a[j].second << " " << a[k].second;
return 0;
}
if (a[i].first + a[j].first > temp) --j;
else ++i;
}
}
cout << "IMPOSSIBLE\n";
}
``````

But two problems are from div 3 