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4_Sum_Problem.cpp
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4_Sum_Problem.cpp
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/*
Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]]
such that:
0 <= a, b, c, d < n
a, b, c, and d are distinct.
nums[a] + nums[b] + nums[c] + nums[d] == target
Example 1:
Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Example 2:
Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]
Constraints:
1 <= nums.length <= 200
-109 <= nums[i] <= 109
-109 <= target <= 109
*/
#include<iostream>
#include<vector>
#include<set>
#include<algorithm>
using namespace std;
class Four_sum
{
private:
vector<int> nums;
int target;
public:
Four_sum(vector<int>& vec,int t)
{
nums=vec; //the nums vector containing the elements of the given array
target=t; //the target is the sum which we want to achieve by 4sum
}
vector<vector<int>> four_sum_solution()
{
if(nums.size()<4) //handling the corner case when the size of the given array is less than 4
{
return {};
}
sort(nums.begin(),nums.end()); //first the elements are sorted, so we can apply an approach of two pointer
set<vector<int>> res; //first storing the solution vectors in the set to avoid duplicates
for(int i=0;i<nums.size()-3;i++) //the 1st element of the quadriple, starting from first element to 4th last element
{
for(int j=i+1;j<nums.size()-2;j++) //the 2nd element, starting from i+1
{
int left=j+1; //3rd element, starting from right of j, which is j+1
int right=nums.size()-1; //4th element, starting from right most element of the given array
while(left<right) //applying the two pointer approach with these two variables
{
long long sum=(long long)nums[i]+nums[j]+nums[left]+nums[right];
if(sum==target) //when we get the sum value as equal to the target value,...
{
vector<int> vec; //crete a temporary vector is created and the elements are inserted
vec.push_back(nums[i]);
vec.push_back(nums[j]);
vec.push_back(nums[left]);
vec.push_back(nums[right]);
res.insert(vec);
}
if(sum<target) //if the sum is less than the target, then we should move the left pointer to one position in right
{
left++; //as we have already sorted the array, so it is gurranteed that the we would get larger value when we go left++
}
else
{
right--; //if the sum is greater than the target, then we should move the right pointer to one position in left
}
}
}
}
vector<vector<int>> res_vec;
for(auto i:res)
{
res_vec.push_back(i); //from the set of the vectors, the resulting vectors are inserted into the vector<vector<int>>
}
return res_vec;
}
};
int main()
{
int n;
cin>>n;
vector<int> vec;
int x;
for(int i=0;i<n;i++)
{
cin>>x;
vec.push_back(x);
}
int target;
cin>>target;
Four_sum obj(vec,target);
vector<vector<int>> res=obj.four_sum_solution();
if(res.size()!=0)
{
for(auto i:res)
{
for(auto j:i)
{
cout<<j<<" ";
}
cout<<endl;
}
}
else
{
cout<<"The given array does not contain 4 elements"<<endl;
}
return 0;
}