# Probability of choosing a random pair with maximum sum in an array

• Difficulty Level : Medium
• Last Updated : 22 Apr, 2021

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).

## C++

 // CPP program of choosing a random pair// such that A[i]+A[j] is maximum.#include using namespace std; // Function to get max first and secondint 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 Codeint 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 seconddef 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 Codeif __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

 \$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];    }     \$cnt1 = 0;    \$cnt2 = 0;    for (\$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;} // Returns probability of// choosing a pair with// maximum sum.function findMaxSumProbability(\$a, \$n){    \$total = \$n * (\$n - 1) / 2;    \$max_sum_pairs = countMaxSumPairs(\$a, \$n);    return (float)\$max_sum_pairs / (float) \$total;}     // Driver Code    \$a= array (1, 2, 2, 3 );    \$n = sizeof(\$a);    echo findMaxSumProbability(\$a, \$n); // This code is contributed by ajit?>

## Javascript



Output :

0.333333

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