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equal.cpp
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equal.cpp
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/*
https://www.interviewbit.com/problems/equal/
Equal
Given an array A of integers, find the index of values that satisfy A + B = C + D, where A,B,C & D are integers values in the array
Note:
1) Return the indices `A1 B1 C1 D1`, so that
A[A1] + A[B1] = A[C1] + A[D1]
A1 < B1, C1 < D1
A1 < C1, B1 != D1, B1 != C1
2) If there are more than one solutions,
then return the tuple of values which are lexicographical smallest.
Assume we have two solutions
S1 : A1 B1 C1 D1 ( these are values of indices int the array )
S2 : A2 B2 C2 D2
S1 is lexicographically smaller than S2 iff
A1 < A2 OR
A1 = A2 AND B1 < B2 OR
A1 = A2 AND B1 = B2 AND C1 < C2 OR
A1 = A2 AND B1 = B2 AND C1 = C2 AND D1 < D2
Example:
Input: [3, 4, 7, 1, 2, 9, 8]
Output: [0, 2, 3, 5] (O index)
If no solution is possible, return an empty list.
*/
vector<int> Solution::equal(vector<int> &A) {
map <int ,pair<int,int> > m;
map <int ,pair<int,int> > :: iterator it;
int n,i,j;
n=A.size();
//sf("%d",&n);
//int a[n];
//for(i=0;i<n;i++) sf("%d",&a[i]);
//vector <int> ans ;
vector < tuple <int,int,int,int > > ans ;
set <int> s;
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
int sum = A[i]+A[j];
if(m.find(sum)==m.end())
{
m[sum] = make_pair(i,j);
// m[sum].first = i;
// m[sum].second = j;
}
else
{
it = m.find(sum);
int a = (it->second).first;
int b = (it->second).second;
int c = i;
int d = j;
s.insert(a);
s.insert(b);
s.insert(c);
s.insert(d);
if(s.size()==4)
{ans.push_back(tuple <int,int,int,int >( (it->second).first , (it->second).second ,i,j ) );
s.clear();}
else
s.clear();
//break;
}
}
}
sort(ans.begin(),ans.end());
vector <int> pp;
pp.push_back(get<0>(ans[0]));
pp.push_back(get<1>(ans[0]));
pp.push_back(get<2>(ans[0]));
pp.push_back(get<3>(ans[0]));
return pp;
}