Given an array arr[] of N integers, the task is to find the probability of getting the maximum sum pair (arr[i], arr[j]) from the array when a random pair is chosen.
Examples:
Input: arr[] = {3, 3, 3, 3}
Output: 1
All the pairs will give the maximum sum i.e. 6.
Input: arr[] = {1, 1, 1, 2, 2, 2}
Output: 0.2
Only the pairs (2, 2), (2, 2) and (2, 2) will give
the maximum sum out of 15 pairs.
3 / 15 = 0.2
Approach: Run two nested loops to get the sum for every single pair, keep the maximum sum for any pair and its count (i.e. the number of pairs that give this sum). Now, the probability of getting this sum will be (count / totalPairs) where totalPairs = (n * (n – 1)) / 2.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the probability // of getting the maximum pair sum // when a random pair is chosen // from the given array float findProb( int arr[], int n)
{ // Initialize the maximum sum, its count
// and the count of total pairs
long maxSum = INT_MIN, maxCount = 0, totalPairs = 0;
// For every single pair
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
// Get the sum of the current pair
int sum = arr[i] + arr[j];
// If the sum is equal to the current
// maximum sum so far
if (sum == maxSum) {
// Increment its count
maxCount++;
}
// If the sum is greater than
// the current maximum
else if (sum > maxSum) {
// Update the current maximum and
// re-initialize the count to 1
maxSum = sum;
maxCount = 1;
}
totalPairs++;
}
}
// Find the required probability
float prob = ( float )maxCount / ( float )totalPairs;
return prob;
} // Driver code int main()
{ int arr[] = { 1, 1, 1, 2, 2, 2 };
int n = sizeof (arr) / sizeof ( int );
cout << findProb(arr, n);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to return the probability // of getting the maximum pair sum // when a random pair is chosen // from the given array static float findProb( int arr[], int n)
{ // Initialize the maximum sum, its count
// and the count of total pairs
long maxSum = Integer.MIN_VALUE,
maxCount = 0 , totalPairs = 0 ;
// For every single pair
for ( int i = 0 ; i < n - 1 ; i++)
{
for ( int j = i + 1 ; j < n; j++)
{
// Get the sum of the current pair
int sum = arr[i] + arr[j];
// If the sum is equal to the current
// maximum sum so far
if (sum == maxSum)
{
// Increment its count
maxCount++;
}
// If the sum is greater than
// the current maximum
else if (sum > maxSum)
{
// Update the current maximum and
// re-initialize the count to 1
maxSum = sum;
maxCount = 1 ;
}
totalPairs++;
}
}
// Find the required probability
float prob = ( float )maxCount /
( float )totalPairs;
return prob;
} // Driver code public static void main(String args[])
{ int arr[] = { 1 , 1 , 1 , 2 , 2 , 2 };
int n = arr.length;
System.out.println(findProb(arr, n));
} } // This code is contributed by Rajput-Ji |
# Python3 implementation of the approach import sys
# Function to return the probability # of getting the maximum pair sum # when a random pair is chosen # from the given array def findProb(arr, n) :
# Initialize the maximum sum, its count
# and the count of total pairs
maxSum = - (sys.maxsize - 1 );
maxCount = 0 ;
totalPairs = 0 ;
# For every single pair
for i in range (n - 1 ) :
for j in range (i + 1 , n) :
# Get the sum of the current pair
sum = arr[i] + arr[j];
# If the sum is equal to the current
# maximum sum so far
if ( sum = = maxSum) :
# Increment its count
maxCount + = 1 ;
# If the sum is greater than
# the current maximum
elif ( sum > maxSum) :
# Update the current maximum and
# re-initialize the count to 1
maxSum = sum ;
maxCount = 1 ;
totalPairs + = 1 ;
# Find the required probability
prob = maxCount / totalPairs;
return prob;
# Driver code if __name__ = = "__main__" :
arr = [ 1 , 1 , 1 , 2 , 2 , 2 ];
n = len (arr);
print (findProb(arr, n));
# This code is contributed by AnkitRai01 |
// C# implementation of above approach using System;
class GFG
{ // Function to return the probability // of getting the maximum pair sum // when a random pair is chosen // from the given array static float findProb( int []arr, int n)
{ // Initialize the maximum sum, its count
// and the count of total pairs
long maxSum = int .MinValue,
maxCount = 0, totalPairs = 0;
// For every single pair
for ( int i = 0; i < n - 1; i++)
{
for ( int j = i + 1; j < n; j++)
{
// Get the sum of the current pair
int sum = arr[i] + arr[j];
// If the sum is equal to the current
// maximum sum so far
if (sum == maxSum)
{
// Increment its count
maxCount++;
}
// If the sum is greater than
// the current maximum
else if (sum > maxSum)
{
// Update the current maximum and
// re-initialize the count to 1
maxSum = sum;
maxCount = 1;
}
totalPairs++;
}
}
// Find the required probability
float prob = ( float )maxCount /
( float )totalPairs;
return prob;
} // Driver code public static void Main(String []args)
{ int []arr = { 1, 1, 1, 2, 2, 2 };
int n = arr.Length;
Console.WriteLine(findProb(arr, n));
} } // This code is contributed by 29AjayKumar |
<script> // Javascript implementation of the approach // Function to return the probability // of getting the maximum pair sum // when a random pair is chosen // from the given array function findProb(arr, n)
{ // Initialize the maximum sum, its count
// and the count of total pairs
var maxSum = -100000000, maxCount = 0, totalPairs = 0;
// For every single pair
for ( var i = 0; i < n - 1; i++) {
for ( var j = i + 1; j < n; j++) {
// Get the sum of the current pair
var sum = arr[i] + arr[j];
// If the sum is equal to the current
// maximum sum so far
if (sum == maxSum) {
// Increment its count
maxCount++;
}
// If the sum is greater than
// the current maximum
else if (sum > maxSum) {
// Update the current maximum and
// re-initialize the count to 1
maxSum = sum;
maxCount = 1;
}
totalPairs++;
}
}
// Find the required probability
var prob = maxCount / totalPairs;
return prob;
} // Driver code var arr = [ 1, 1, 1, 2, 2, 2 ]
var n = arr.length;
document.write(findProb(arr, n)); // This code is contributed by rutvik_56. </script> |
0.2
Time Complexity: O(N2)
Auxiliary Space: O(1)