Probability of choosing a random pair with maximum sum in an array (original) (raw)

Last Updated : 25 Sep, 2022

Given an array of N integers, You have to find the probability of choosing a random pair(i, j), i < j such that A[i] + A[j] is maximum.

Examples :

Input : A[] = {3, 3, 3, 3} Output : 1.0000 Explanation : Here, maximum sum we can get by selecting any pair is 6. Total number of pairs possible = 6. Pairs with maximum sum = 6. Probability = 6/6 = 1.0000

Input : A[] = {1, 2, 2, 3} Output : 0.3333 Explanation : Here, maximum sum we can get by selecting a pair is 5. Total number of pairs possible = 6. Pairs with maximum sum = {2, 3} and {2, 3} = 2. Probability = 2/6 = 0.3333

Probability(event) = Number of favorable outcomes / Total number of outcomes

Naive approach : We can solve this problem using brute force solution overall pair (i, j), i < j to get the maximum value possible and then again do a brute force to calculate the number of times the maximum is attained.

Efficient Approach : Observe that we get maximum pair sum only when the pairs consists of first and second maximum elements of the array. So, the problem is to calculate the number of occurrences of those elements and calculate the favorable outcomes using a formula.

Favorable outcomes = f2 (frequency of second maximum element(f2), if maximum element occurs only once). or Favorable outcomes = f1 * (f1 - 1) / 2, (when frequency of maximum element(f1) is greater than 1).

Implementation:

C++ `

// CPP program of choosing a random pair // such that A[i]+A[j] is maximum. #include <bits/stdc++.h> using namespace std;

// Function to get max first and second int countMaxSumPairs(int a[], int n) { int first = INT_MIN, second = INT_MIN; for (int i = 0; i < n; i++) {

    /* If current element is smaller than
      first,  then update both first and 
      second */
    if (a[i] > first) {
        second = first;
        first = a[i];
    }

    /* If arr[i] is in between first and 
    second then update second */
    else if (a[i] > second && a[i] != first)
        second = a[i];
}

int cnt1 = 0, cnt2 = 0;
for (int i = 0; i < n; i++) {
    if (a[i] == first)
        cnt1++; // frequency of first maximum
    if (a[i] == second)
        cnt2++; // frequency of second maximum
}
if (cnt1 == 1) 
    return cnt2;

if (cnt1 > 1) 
    return cnt1 * (cnt1 - 1) / 2;

}

// Returns probability of choosing a pair with // maximum sum. float findMaxSumProbability(int a[], int n) { int total = n * (n - 1) / 2; int max_sum_pairs = countMaxSumPairs(a, n); return (float)max_sum_pairs/(float)total; }

// Driver Code int main() { int a[] = { 1, 2, 2, 3 }; int n = sizeof(a) / sizeof(a[0]); cout << findMaxSumProbability(a, n); return 0; }

Java

// Java program of choosing a random pair // such that A[i]+A[j] is maximum. import java.util.Scanner; import java.io.*;

class GFG {

// Function to get max first and second
static int countMaxSumPairs(int a[], int n)
{
    int first = Integer.MIN_VALUE, second = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++) {

        /* If current element is smaller than
        first, then update both first and 
        second */
        if (a[i] > first) 
        {
            second = first;
            first = a[i];
        }

        /* If arr[i] is in between first and 
        second then update second */
        else if (a[i] > second && a[i] != first)
            second = a[i];
    }

    int cnt1 = 0, cnt2 = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] == first)
            
            // frequency of first maximum
            cnt1++; 
        if (a[i] == second)
            
            // frequency of second maximum
            cnt2++; 
    }
    if (cnt1 == 1) 
        return cnt2;
    
    if (cnt1 > 1) 
        return cnt1 * (cnt1 - 1) / 2; 
        return 0;
}

// Returns probability of choosing a pair with
// maximum sum.
static float findMaxSumProbability(int a[], int n)
{
    int total = n * (n - 1) / 2;
    int max_sum_pairs = countMaxSumPairs(a, n);
    return (float)max_sum_pairs/(float)total;
}

// Driver Code
public static void main (String[] args) {

    
    int a[] = { 1, 2, 2, 3 };
    int n = a.length;;
    System.out.println(findMaxSumProbability(a, n));

// This code is contributed by ajit } }

Python 3

Python 3 program of choosing a random

pair such that A[i]+A[j] is maximum.

Function to get max first and second

def countMaxSumPairs(a, n):

first = 0
second = 0
for i in range(n):

    # If current element is smaller than
    # first, then update both first and 
    # second 
    if (a[i] > first) :
        second = first
        first = a[i]

    # If arr[i] is in between first and 
    # second then update second 
    elif (a[i] > second and a[i] != first):
        second = a[i]

cnt1 = 0
cnt2 = 0
for i in range(n):
    if (a[i] == first):
        cnt1 += 1 # frequency of first maximum
    if (a[i] == second):
        cnt2 += 1 # frequency of second maximum

if (cnt1 == 1) :
    return cnt2

if (cnt1 > 1) :
    return cnt1 * (cnt1 - 1) / 2

Returns probability of choosing a pair

with maximum sum.

def findMaxSumProbability(a, n):

total = n * (n - 1) / 2
max_sum_pairs = countMaxSumPairs(a, n)
return max_sum_pairs / total

Driver Code

if name == "main":

a = [ 1, 2, 2, 3 ]
n = len(a)
print(findMaxSumProbability(a, n))

This code is contributed by ita_c

C#

// C# program of choosing a random pair // such that A[i]+A[j] is maximum. using System;

public class GFG{

// Function to get max first and second
static int countMaxSumPairs(int []a, int n)
{
    int first = int.MinValue, second = int.MinValue;
    for (int i = 0; i < n; i++) {

        /* If current element is smaller than
        first, then update both first and 
        second */
        if (a[i] > first) 
        {
            second = first;
            first = a[i];
        }

        /* If arr[i] is in between first and 
        second then update second */
        else if (a[i] > second && a[i] != first)
            second = a[i];
    }

    int cnt1 = 0, cnt2 = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] == first)
            
            // frequency of first maximum
            cnt1++; 
        if (a[i] == second)
            
            // frequency of second maximum
            cnt2++; 
    }
    if (cnt1 == 1) 
        return cnt2;
    
    if (cnt1 > 1) 
        return cnt1 * (cnt1 - 1) / 2; 
        return 0;
}

// Returns probability of choosing a pair with
// maximum sum.
static float findMaxSumProbability(int []a, int n)
{
    int total = n * (n - 1) / 2;
    int max_sum_pairs = countMaxSumPairs(a, n);
    return (float)max_sum_pairs/(float)total;
}

// Driver Code
static public void Main ()
{
    int []a = { 1, 2, 2, 3 };
    int n = a.Length;;
    Console.WriteLine(findMaxSumProbability(a, n));
    
}

} // This code is contributed by vt_m.

PHP

i<i < i<n; $i++) { // If current element is // smaller than first, // then update both first // and second if ($a[$i] > $first) { second=second = second=first; first=first = first=a[$i]; } // If arr[i] is in between // first and second then // update second else if ($a[$i] > $second && a[a[a[i] != $first) second=second = second=a[$i]; } $cnt1 = 0; $cnt2 = 0; for ($i = 0; i<i < i<n; $i++) { if ($a[$i] == $first) // frequency of first maximum $cnt1++; if ($a[$i] == $second) // frequency of second maximum $cnt2++; } if ($cnt1 == 1) return $cnt2; if ($cnt1 > 1) return cnt1∗(cnt1 * (cnt1∗(cnt1 - 1) / 2; } // Returns probability of // choosing a pair with // maximum sum. function findMaxSumProbability($a, $n) { total=total = total=n * ($n - 1) / 2; maxsumpairs=countMaxSumPairs(max_sum_pairs = countMaxSumPairs(maxs​ump​airs=countMaxSumPairs(a, $n); return (float)$max_sum_pairs / (float) $total; } // Driver Code $a= array (1, 2, 2, 3 ); n=sizeof(n = sizeof(n=sizeof(a); echo findMaxSumProbability($a, $n); // This code is contributed by ajit ?>

JavaScript

`

Time complexity: O(n) where n is the size of the given array.
Auxiliary space: O(1)