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++ 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 than 3
// then second largest product pair
// does 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;
} |
/*package whatever //do not write package name here */ import java.util.*;
public class GFG
{ static class Tuple<K,V>{
K key;
V val;
Tuple(K k, V v){
key = k;
val = v;
}
}
// Function to find second largest
// product of pairs
static void secondLargerstPair( int [] arr, int N)
{
// If size of array is less than 3
// then second largest product pair
// does not exits.
if (N < 3 )
return ;
// Declaring set of pairs which
// contains possible pairs of array
// and their products
TreeSet<Tuple<Integer, Tuple<Integer, Integer>>> s = new TreeSet<>((t1,t2)->{
int a1 = t1.key, a2 = t1.val.key, a3 = t1.val.val;
int b1 = t2.key, b2 = t2.val.key, b3 = t2.val.val;
if (a1 < b1)
return - 1 ;
if (a1 > b1)
return 1 ;
if (a2 < b2)
return - 1 ;
if (a2 > b2)
return 1 ;
if (a3 < b3)
return - 1 ;
if (a3 > b3)
return 1 ;
return 0 ;
});
// Declaring vector of pairs
List<Tuple<Integer, Integer> > v = new ArrayList<>();
for ( int i = 0 ; i < N; ++i) {
for ( int j = i + 1 ; j < N; ++j) {
// Inserting a set
s.add( new Tuple(arr[i] * arr[j], new Tuple(arr[i],arr[j])));
}
}
// Traverse set of pairs
for (var i : s){
// Inserting values in vector
// of pairs
v.add( new Tuple((i.val).key,(i.val).val));
}
Collections.sort(v,(a,b)->a.key - b.key);
int vsize = v.size();
// Printing the result
System.out.println(v.get(vsize - 2 ).key + " " + v.get(vsize - 2 ).val);
}
// Driver Code
public static void main(String[] args)
{
// Given Array
int [] arr = { 5 , 2 , 67 , 45 , 160 , 78 };
// Size of Array
int N = arr.length;
// Function Call
secondLargerstPair(arr, N);
}
} // This code is contributed by aadityaburujwale. |
# Python3 program for the above approach # Function to find second largest # product of pairs def secondLargerstPair(arr, N):
# If size of array is less than 3
# then second largest product pair
# does not exits.
if (N < 3 ):
return ;
# Declaring set of pairs which
# contains possible pairs of array
# and their products
s = set ()
# Declaring vector of pairs
v = [];
for i in range (N):
for j in range (i + 1 , N):
# Inserting a set
s.add((arr[i] * arr[j], (arr[i], arr[j])))
# Traverse set of pairs
for i in sorted (s):
# Inserting values in vector
# of pairs
v.append((i[ 1 ][ 0 ], i[ 1 ][ 1 ]));
vsize = len (v)
# Printing the result
print (v[vsize - 2 ][ 0 ], v[vsize - 2 ][ 1 ]);
# Driver Code # Given Array arr = [ 5 , 2 , 67 , 45 , 160 , 78 ];
# Size of Array N = len (arr)
# Function Call secondLargerstPair(arr, N); # This code is contributed by phasing17 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to find second largest
// product of pairs
static void secondLargerstPair( int [] arr, int N)
{
// If size of array is less than 3
// then second largest product pair
// does not exits.
if (N < 3)
return ;
// Declaring set of pairs which
// contains possible pairs of array
// and their products
HashSet<Tuple< int , Tuple< int , int > > > s = new HashSet<Tuple< int , Tuple< int , int > > >();
// Declaring vector of pairs
List<Tuple< int , int > > v = new List<Tuple< int , int > >();
for ( int i = 0; i < N; ++i) {
for ( int j = i + 1; j < N; ++j) {
// Inserting a set
s.Add(Tuple.Create(arr[i] * arr[j], Tuple.Create(arr[i],
arr[j])));
}
}
// Traverse set of pairs
foreach ( var i in s) {
// Inserting values in vector
// of pairs
v.Add(
Tuple.Create((i.Item2).Item1,
(i.Item2).Item2));
}
v.Sort();
int vsize = v.Count;
// Printing the result
Console.WriteLine(v[vsize - 2].Item1 + " "
+ v[vsize - 2].Item2);
}
// Driver Code
public static void Main( string [] args)
{
// Given Array
int [] arr = { 5, 2, 67, 45, 160, 78 };
// Size of Array
int N = arr.Length;
// Function Call
secondLargerstPair(arr, N);
}
} // This code is contributed by phasing17 |
// JS program for the above approach // Function to find second largest // product of pairs function secondLargerstPair(arr, N)
{ // If size of array is less than 3
// then second largest product pair
// does not exits.
if (N < 3)
return ;
// Declaring set of pairs which
// contains possible pairs of array
// and their products
let s = new Set()
// Declaring vector of pairs
let v = [];
for ( var i = 0; i < N; i++)
for ( var j = i + 1; j < N; j++)
// Inserting a set
s.add([arr[i] * arr[j], arr[i], arr[j]].join( '#' ))
// Traverse set of pairs
s = Array.from(s)
s.sort( function (a, b)
{
a = a.split( '#' )
b = b.split( '#' )
let a1 = parseInt(a[0]), a2 = parseInt(a[1]), a3 = parseInt(a[2])
let b1 = parseInt(b[0]), b2 = parseInt(b[1]), b3 = parseInt(b[2])
if (a1 < b1)
return -1
if (a1 > b1)
return 1
if (a2 < b2)
return -1
if (a2 > b2)
return 1
if (a3 < b3)
return -1
if (a3 > b3)
return 1
return 0
})
for (let i of s)
{
i = i.split( '#' )
// Inserting values in vector
// of pairs
v.push([i[1], i[2]])
}
let vsize = v.length
// Printing the result
console.log(v[vsize - 2].join( ' ' ))
} // Driver Code // Given Array let arr = [5, 2, 67, 45, 160, 78]; // Size of Array let N = arr.length // Function Call secondLargerstPair(arr, N); // This code is contributed by phasing17 |
67 160
Time Complexity: O(N2)
Auxiliary Space: O(N), since n extra space has been taken.
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++ 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 does not largest but
// larger than 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;
} |
// 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 does not largest but
// larger than 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 |
# 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 does not largest but
# larger than 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 |
// 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 does not largest but
// larger than 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 |
<script> // JavaScript program for the above approach // Function to find second largest // product pair in arr[0..n-1] function maxProduct(arr, N)
{ // No pair exits
if (N < 3) {
return ;
}
// Initialize max product pair
let a = arr[0], b = arr[1];
let c = 0, d = 0;
// Traverse through every possible pair
// and keep track of largest product
for (let i = 0; i < N; i++)
for (let 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 does not largest but
// larger than second largest
if (arr[i] * arr[j] < a * b
&& arr[i] * arr[j] > c * d)
c = arr[i], d = arr[j];
}
// Print the pairs
document.write(c + " " + d);
} // Driver Code // Given array
let arr = [ 5, 2, 67, 45, 160, 78 ];
let N = arr.length;
// Function Call
maxProduct(arr, N);
// This code is contributed by Surbhi Tyagi. </script> |
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++ 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;
} |
// 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 |
# 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 ];
# Print second 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 |
// 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 |
<script> // JavaScript program for the above approach // Function to find second largest // product pair in arr[0..n-1] function maxProduct(arr, N)
{ // No pair exits
if (N < 3)
{
return ;
}
// Sort the array
arr.sort((a, b) => a - b)
// Initialize smallest element
// of the array
let smallest1 = arr[0];
let smallest3 = arr[2];
// Initialize largest element
// of the array
let largest1 = arr[N - 1];
let largest3 = arr[N - 3];
// Print second largest product pair
if (smallest1 * smallest3 >=
largest1 * largest3)
{
document.write(smallest1 + " " +
smallest3);
}
else
{
document.write(largest1 + " " +
largest3);
}
} // Driver Code // Given array
let arr = [ 5, 2, 67, 45, 160, 78 ];
let N = arr.length;
// Function Call
maxProduct(arr, N);
</script> |
160 67
Time Complexity: O(N*log N)
Auxiliary Space: O(1), since no extra space has been taken.