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

• Difficulty Level : Medium
• 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 ` `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

 ` ``\$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`

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

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