Find a pair in Array with second largest product
Given an array arr[] of N integers, where N > 2, the task is to find the second largest product pair from the given array.
Examples:
Input: arr[] = {10, 20, 12, 40, 50}
Output: 20 50
Explanation:
A pair of array elements = [(10, 20), (10, 12), (10, 40), (10, 50), (20, 12), (20, 40), (20, 50), (12, 40), (12, 50), (40, 50)]
If do product of each pair will get the largest pair as (40, 50) and second largest pair (20, 50)Input: arr[] = {5, 2, 67, 45, 160, 78}
Output: 67 160
Naive Approach: The naive approach is to generate all possible pairs from the given array and insert the product with the pair into the set of pairs. After inserting all the pair products in the set print the second last product of the set. Below are the steps:
- Make a set of pairs and their products by the given array.
- Insert all the pairs in vector of pairs.
- If vector size is 1 then print this pair otherwise print the pair at (total vector size – 2)th position of vector.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find second largest // product of pairs void secondLargerstPair(int arr[], int N) { // If size of array is less then 3 // then second largest product pair // dose not exits. if (N < 3) return; // Declaring set of pairs which // contains possible pairs of array // and their products set<pair<int, pair<int, int> > > s; // Declaring vector of pairs vector<pair<int, int> > v; for (int i = 0; i < N; ++i) { for (int j = i + 1; j < N; ++j) { // Inserting a set s.insert(make_pair(arr[i] * arr[j], make_pair(arr[i], arr[j]))); } } // Traverse set of pairs for (auto i : s) { // Inserting values in vector // of pairs v.push_back( make_pair((i.second).first, (i.second).second)); } int vsize = v.size(); // Printing the result cout << v[vsize - 2].first << " " << v[vsize - 2].second << endl; } // Driver Code int main() { // Given Array int arr[] = { 5, 2, 67, 45, 160, 78 }; // Size of Array int N = sizeof(arr) / sizeof(arr[0]); // Function Call secondLargerstPair(arr, N); return 0; } |
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Output:
67 160
Time Complexity: O(N2)
Auxiliary Space: O(N)
Better Solution: A better solution is to traverse all the pairs of the array and while traversing store the largest and second-largest product pairs. After traversal print the pairs with second-largest pairs stored.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find second largest // product pair in arr[0..n-1] void maxProduct(int arr[], int N) { // No pair exits if (N < 3) { return; } // Initialize max product pair int a = arr[0], b = arr[1]; int c = 0, d = 0; // Traverse through every possible pair // and keep track of largest product for (int i = 0; i < N; i++) for (int j = i + 1; j < N; j++) { // If pair is largest if (arr[i] * arr[j] > a * b) { // Second largest c = a, d = b; a = arr[i], b = arr[j]; } // If pair dose not largest but // larger then second largest if (arr[i] * arr[j] < a * b && arr[i] * arr[j] > c * d) c = arr[i], d = arr[j]; } // Print the pairs cout << c << " " << d; } // Driver Code int main() { // Given array int arr[] = { 5, 2, 67, 45, 160, 78 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call maxProduct(arr, N); return 0; } |
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Java
// Java program for the above approach class GFG{ // Function to find second largest // product pair in arr[0..n-1] static void maxProduct(int arr[], int N) { // No pair exits if (N < 3) { return; } // Initialize max product pair int a = arr[0], b = arr[1]; int c = 0, d = 0; // Traverse through every possible pair // and keep track of largest product for (int i = 0; i < N; i++) for (int j = i + 1; j < N-1; j++) { // If pair is largest if (arr[i] * arr[j] > a * b) { // Second largest c = a; d = b; a = arr[i]; b = arr[j]; } // If pair dose not largest but // larger then second largest if (arr[i] * arr[j] < a * b && arr[i] * arr[j] > c * d) c = arr[i]; d = arr[j]; } // Print the pairs System.out.println(c + " " + d); } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 5, 2, 67, 45, 160, 78 }; int N = arr.length; // Function Call maxProduct(arr, N); } } // This code is contributed by Ritik Bansal |
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Python3
# Python3 program for the above approach # Function to find second largest # product pair in arr[0..n-1] def maxProduct(arr, N): # No pair exits if (N < 3): return; # Initialize max product pair a = arr[0]; b = arr[1]; c = 0; d = 0; # Traverse through every possible pair # and keep track of largest product for i in range(0, N, 1): for j in range(i + 1, N - 1, 1): # If pair is largest if (arr[i] * arr[j] > a * b): # Second largest c = a; d = b; a = arr[i]; b = arr[j]; # If pair dose not largest but # larger then second largest if (arr[i] * arr[j] < a * b and arr[i] * arr[j] > c * d): c = arr[i]; d = arr[j]; # Print the pairs print(c, " ", d); # Driver Code if __name__ == '__main__': # Given array arr = [ 5, 2, 67, 45, 160, 78]; N = len(arr); # Function call maxProduct(arr, N); # This code is contributed by Amit Katiyar |
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C#
// C# program for the above approach using System; class GFG{ // Function to find second largest // product pair in arr[0..n-1] static void maxProduct(int []arr, int N) { // No pair exits if (N < 3) { return; } // Initialize max product pair int a = arr[0], b = arr[1]; int c = 0, d = 0; // Traverse through every possible pair // and keep track of largest product for(int i = 0; i < N; i++) for(int j = i + 1; j < N - 1; j++) { // If pair is largest if (arr[i] * arr[j] > a * b) { // Second largest c = a; d = b; a = arr[i]; b = arr[j]; } // If pair dose not largest but // larger then second largest if (arr[i] * arr[j] < a * b && arr[i] * arr[j] > c * d) c = arr[i]; d = arr[j]; } // Print the pairs Console.WriteLine(c + " " + d); } // Driver Code public static void Main(String[] args) { // Given array int []arr = { 5, 2, 67, 45, 160, 78 }; int N = arr.Length; // Function call maxProduct(arr, N); } } // This code is contributed by 29AjayKumar |
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Output:
67 160
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach:
- Sort the array.
- Find first and third smallest elements for handling negative numbers.
- Find the first and third largest elements for handling positive numbers.
- Compare the product of the smallest pair and largest pair.
- Return the largest one of them.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find second largest // product pair in arr[0..n-1] void maxProduct(int arr[], int N) { // No pair exits if (N < 3) { return; } // Sort the array sort(arr, arr + N); // Initialize smallest element // of the array int smallest1 = arr[0]; int smallest3 = arr[2]; // Initialize largest element // of the array int largest1 = arr[N - 1]; int largest3 = arr[N - 3]; // Print second largest product pair if (smallest1 * smallest3 >= largest1 * largest3) { cout << smallest1 << " " << smallest3; } else { cout << largest1 << " " << largest3; } } // Driver Code int main() { // Given array int arr[] = { 5, 2, 67, 45, 160, 78 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call maxProduct(arr, N); return 0; } |
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Java
// Java program for the above approach import java.util.*; class GFG{ // Function to find second largest // product pair in arr[0..n-1] static void maxProduct(int arr[], int N) { // No pair exits if (N < 3) { return; } // Sort the array Arrays.sort(arr); // Initialize smallest element // of the array int smallest1 = arr[0]; int smallest3 = arr[2]; // Initialize largest element // of the array int largest1 = arr[N - 1]; int largest3 = arr[N - 3]; // Print second largest product pair if (smallest1 * smallest3 >= largest1 * largest3) { System.out.print(smallest1 + " " + smallest3); } else { System.out.print(largest1 + " " + largest3); } } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 5, 2, 67, 45, 160, 78 }; int N = arr.length; // Function Call maxProduct(arr, N); } } // This code is contributed by gauravrajput1 |
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Python3
# Python3 program for the above approach # Function to find second largest # product pair in arr[0..n-1] def maxProduct(arr, N): # No pair exits if (N < 3): return; # Sort the array arr.sort(); # Initialize smallest element # of the array smallest1 = arr[0]; smallest3 = arr[2]; # Initialize largest element # of the array largest1 = arr[N - 1]; largest3 = arr[N - 3]; # Prsecond largest product pair if (smallest1 * smallest3 >= largest1 * largest3): print(smallest1 , " " , smallest3); else: print(largest1 , " " , largest3); # Driver Code if __name__ == '__main__': # Given array arr = [5, 2, 67, 45, 160, 78]; N = len(arr); # Function Call maxProduct(arr, N); # This code is contributed by sapnasingh4991 |
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C#
// C# program for the above approach using System; class GFG{ // Function to find second largest // product pair in arr[0..n-1] static void maxProduct(int []arr, int N) { // No pair exits if (N < 3) { return; } // Sort the array Array.Sort(arr); // Initialize smallest element // of the array int smallest1 = arr[0]; int smallest3 = arr[2]; // Initialize largest element // of the array int largest1 = arr[N - 1]; int largest3 = arr[N - 3]; // Print second largest product pair if (smallest1 * smallest3 >= largest1 * largest3) { Console.Write(smallest1 + " " + smallest3); } else { Console.Write(largest1 + " " + largest3); } } // Driver Code public static void Main(String[] args) { // Given array int []arr = { 5, 2, 67, 45, 160, 78 }; int N = arr.Length; // Function Call maxProduct(arr, N); } } // This code is contributed by Rohit_ranjan |
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Output:
160 67
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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