Probability of a random pair being the maximum weighted pair (original) (raw)

Last Updated : 27 Jul, 2022

Given two arrays A and B, a random pair is picked having an element from array A and another from array B. Output the probability of the pair being maximum weighted.

Examples:

Input : A[] = 1 2 3 B[] = 1 3 3 Output : 0.222 Explanation : Possible pairs are : {1, 1}, {1, 3}, {1, 3}, {2, 1}, {2, 3}, {2, 3}, {3, 1}, {3, 3}, {3, 3} i.e. 9. The pair with maximum weight is {3, 3} with frequency 2. So, the probability of random pair being maximum is 2/9 = 0.2222.

• Brute Force Method : Generate all possible pairs in N^2 time complexity and count
  maximum weighted pairs.
• Better Method : Sort both the arrays and count the last (max) elements from A and B. No. of maximum weighted pairs will be product of both counts. The probability will be
  (product of counts) / sizeof(A) * sizeof(B)
• Best Method Best approach will be to traverse both the arrays and count the maximum element. No. of maximum weighted pairs will be product of both counts. The probability will be (product of counts) / sizeof(A) * sizeof(B)

Below is the implementation:

C++ `

#include <bits/stdc++.h> using namespace std;

// Function to return probability double probability(int a[], int b[], int size1, int size2) { // Count occurrences of maximum element // in A[] int max1 = INT_MIN, count1 = 0; for (int i = 0; i < size1; i++) { if (a[i] > max1) { max1 = a[i]; count1 = 1; } else if (a[i] == max1) { count1++; } }

// Count occurrences of maximum element 
// in B[]
int max2 = INT_MIN, count2 = 0;
for (int i = 0; i < size2; i++) {
    if (b[i] > max2) {
        max2 = b[i];
        count2 = 1;
    }
    else if (b[i] == max2) {
        count2++;
    }
}

// Returning probability
return (double)(count1 * count2) / 
              (size1 * size2);

}

// Driver code int main() { int a[] = { 1, 2, 3 }; int b[] = { 1, 3, 3 };

int size1 = sizeof(a) / sizeof(a[0]);
int size2 = sizeof(b) / sizeof(b[0]);

cout << probability(a, b, size1, size2);
return 0;

}

Java

// Java program to find Probability // of a random pair being the maximum // weighted pair import java.io.*;

class GFG {

// Function to return probability
static double probability(int a[], int b[], 
                        int size1,int size2)
{
    // Count occurrences of maximum 
    // element in A[]
    int max1 = Integer.MIN_VALUE,  count1 = 0;
    for (int i = 0; i < size1; i++) {
        if (a[i] > max1) {
            max1 = a[i];
            count1 = 1;
        }
        else if (a[i] == max1) {
            count1++;
        }
    }
 
    // Count occurrences of maximum 
    // element in B[]
    int max2 = Integer.MIN_VALUE, count2 = 0;
    for (int i = 0; i < size2; i++) {
        if (b[i] > max2) {
            max2 = b[i];
            count2 = 1;
        }
        else if (b[i] == max2) {
            count2++;
        }
    }
 
    // Returning probability
    return (double)(count1 * count2) / (size1 * size2);
}
 
// Driver code
public static void main(String args[])
{
    int a[] = { 1, 2, 3 };
    int b[] = { 1, 3, 3 };
 
    int size1 = a.length;
    int size2 = b.length;
 
    System.out.println(probability(a, b, 
                        size1, size2));
}

}

/This code is contributed by Nikita Tiwari./

Python3

import sys

Function to return probability

def probability(a, b, size1, size2):

# Count occurrences of maximum
# element in A[] 
max1 = -(sys.maxsize - 1)
count1 = 0
for i in range(size1):
    if a[i] > max1:
        count1 = 1
    elif a[i] == max1:
        count1 += 1

# Count occurrences of maximum 
# element in B[] 
max2 = -(sys.maxsize - 1)
count2 = 0
for i in range(size2):
    if b[i] > max2:
        max2 = b[i]
        count2 = 1
    elif b[i] == max2:
        count2 += 1

# Returning probability 
return round((count1 * count2) / 
             (size1 * size2), 6)

Driver code

a = [1, 2, 3] b = [1, 3, 3] size1 = len(a) size2 = len(b) print(probability(a, b, size1, size2))

This code is contributed

by Shrikant13

C#

// C# program to find Probability of a random // pair being the maximum weighted pair using System;

class GFG {

// Function to return probability
static float probability(int []a, int []b, 
                      int size1,int size2)
{
    
    // Count occurrences of maximum 
    // element in A[]
    int max1 = int.MinValue, count1 = 0;
    
    for (int i = 0; i < size1; i++) {
        if (a[i] > max1) {
            max1 = a[i];
            count1 = 1;
        }
        else if (a[i] == max1) {
            count1++;
        }
    }

    // Count occurrences of maximum 
    // element in B[]
    int max2 = int.MinValue, count2 = 0;
    
    for (int i = 0; i < size2; i++) {
        if (b[i] > max2) {
            max2 = b[i];
            count2 = 1;
        }
        else if (b[i] == max2) {
            count2++;
        }
    }

    // Returning probability
    return (float)(count1 * count2) / 
                        (size1 * size2);
}

// Driver code
public static void Main()
{
    int []a = { 1, 2, 3 };
    int []b = { 1, 3, 3 };

    int size1 = a.Length;
    int size2 = b.Length;

    Console.WriteLine(probability(a, b, 
                        size1, size2));
}

}

/* This code is contributed by vt_m.*/

PHP

JavaScript

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Time Complexity: O(n).
Auxiliary Space: O(1).