Given an array arr of integers of size N and a target number, the task is to find all unique quadruplets in it, whose sum is equal to the target number.
Examples:
Input: arr[] = {4, 1, 2, -1, 1, -3], target = 1
Output: [[-3, -1, 1, 4], [-3, 1, 1, 2]]
Explanation:
Both the quadruplets sum equals to target.Input: arr[] = {2, 0, -1, 1, -2, 2}, target = 2
Output: [[-2, 0, 2, 2], [-1, 0, 1, 2]]
Two-Pointers approach:
This problem follows the Two Pointers pattern and shares similarities with Triplet Sum to Zero.
We can follow a similar approach to iterate through the array, taking one number at a time. At every step during the iteration, we will search for the quadruplets similar to Triplet Sum to Zero whose sum is equal to the given target.
C++
// C++ program to find four // elements that sum to a given value #include <bits/stdc++.h> using namespace std; class QuadrupleSumToTarget { public: // Function to find quadruplets static vector<vector<int> > searchQuadruplets( vector<int>& arr, int target) { sort(arr.begin(), arr.end()); vector<vector<int> > quadruplets; for (int i = 0; i < arr.size() - 3; i++) { if (i > 0 && arr[i] == arr[i - 1]) { // Skip same element // to avoid duplicate continue; } for (int j = i + 1; j < arr.size() - 2; j++) { if (j > i + 1 && arr[j] == arr[j - 1]) { // Skip same element // to avoid duplicate quad continue; } searchPairs( arr, target, i, j, quadruplets); } } return quadruplets; } private: // Function to search Quadruplets static void searchPairs( const vector<int>& arr, int targetSum, int first, int second, vector<vector<int> >& quadruplets) { int left = second + 1; int right = arr.size() - 1; while (left < right) { int sum = arr[first] + arr[second] + arr[left] + arr[right]; if (sum == targetSum) { // Found the quadruplet quadruplets .push_back( { arr[first], arr[second], arr[left], arr[right] }); left++; right--; // Skip same element to avoid // duplicate quadruplets while (left < right && arr[left] == arr[left - 1]) { left++; } // Skip same element to avoid // duplicate quadruplets while (left < right && arr[right] == arr[right + 1]) { right--; } } // We need a pair // with a bigger sum else if (sum < targetSum) { left++; } // We need a pair // with a smaller sum else { right--; } } } }; void printQuad( vector<int>& vec, int target) { // Function call auto result = QuadrupleSumToTarget:: searchQuadruplets( vec, target); // Print Quadruples for (int j = 0; j < result.size(); j++) { vector<int> vec = result[j]; if (j == 0) cout << "["; for (int i = 0; i < vec.size(); i++) { if (i == 0) cout << "["; cout << vec[i]; if (i != vec.size() - 1) cout << ", "; else cout << "]"; } if (j != result.size() - 1) cout << ", "; else cout << "]"; } } // Driver code int main(int argc, char* argv[]) { vector<int> vec = { 4, 1, 2, -1, 1, -3 }; int target = 1; printQuad(vec, target); return 0; } |
[[-3, -1, 1, 4], [-3, 1, 1, 2]]
Time complexity: Sorting the array will take O(N*logN). Overall searchQuadruplets() will take O(N * logN + N^3), which is asymptotically equivalent to O(N^3).
Auxiliary Space complexity: The auxiliary space complexity of the above algorithm will be O(N) which is required for sorting.
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- Find four elements that sum to a given value | Set 3 (Hashmap)
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