SPOJ SUMFOUR.....TLE on test case 9 - c++

Question

I am trying to solve SPOJ Problem SUMFOUR....I am geting TLE on test case 9 http://www.spoj.com/problems/SUMFOUR/
So,Which part of my code has to be edited and how?Here N<=4000
#include <iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<map>
#include<vector>
using namespace std;
int main()
{
int a[4005][5],n;
cin>>n;
for(int i=1;i<=n;i++)
for(int j=1;j<=4;j++)
scanf("%d",&a[i][j]);
int k=0;
for(int i=1;i<=n;i++)
{ int p=a[i][1];
for(int j=1;j<=n;j++)
{ b.push_back(p+a[j][2]);
k++;
}
}
k=0;
for(int i=1;i<=n;i++)
{ int p=a[i][3];
for(int j=1;j<=n;j++)
{ c.push_back(p+a[j][4]);
k++;
}
}
sort(b.begin(),b.end());
int cnt=0;
for(int j=0;j<k;j++)
if(find(b.begin(),b.end(),-c[j])!=b.end() )
cnt=cnt+count(b.begin(),b.end(),-c[j]) ;
printf("%d\n",cnt);
return 0;
}

Answer 1

The problem is here:
for(int j=0;j<k;j++)
if(find(b.begin(),b.end(),-c[j])!=b.end() )
cnt=cnt+count(b.begin(),b.end(),-c[j]) ;
for n = 4000, so there are 4000^2 elements in b and c. So, the time complexity for this loop is 4000^4, as find and count time complexity is O(n), which of course will cause you time limit exceed.
So, how you can reduce the time? You can use binary search to faster the count process, which reduce the time complexity of the above loop to O(n^2 log n), as I notice you already sort b.
Or , you can use map to count and store the frequency of each element in b and c.
map<long long, int> b;
map<long long, int> c;
for(int i=1;i<=n;i++)
{ long long p=a[i][1];
for(int j=1;j<=n;j++)
{
long long tmp =p + a[j][2];
b[tmp] = b[tmp] + 1;
}
}
// Similar for c
map <long long, int>::iterator it;
long long result;
for (it = c.begin(); it != c.end(); ++it)
result += c[it->first]*b[-(it->first)];
For your new update, please change this:
for(int j=1;j<=n;j++)
{ if( b.count(a[i][1]+a[j][2]) )
{ b[a[i][1]+a[j][2]]+=1;
c[a[i][3]+a[j][4]]+=1;
}
else
{ b[a[i][1]+a[j][2]]=1;
c[a[i][3]+a[j][4]]=1;
}
}
into this:
for(int j=1;j<=n;j++)
{
b[a[i][1]+a[j][2]]+=1;
c[a[i][3]+a[j][4]]+=1;
}
The condition check if( b.count(a[i][1]+a[j][2]) ) is for b only, and you use it for c, which make c incorrect.
Update: After trying to get accepted in SPOJ, it turns out that map is not fast enough, so I make a change into binary search, and got accepted.
My accepted code

Answer 2

Please Don't Use Map as its worst Case Complexity Can be O(log(n)) .
SO instead You can just Use two sorted arrays and for every element as in the
first array , Binary Search for its -ve agent in the Second Cumulative array .
Just Change the find method in Last lines to Binary search(c.begin(),c.end(),key) and find the repititons till the end with that index as it gives the lower_bound index .
That Total Sum gives the answer and its expected Complexity is
O(n^2log(n)).