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 probabilitydouble 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 codeint 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 pairimport 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 codea = [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 pairusing 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
<?php// PHP program for Probability of // a random pair being the maximum // weighted pair// Function to return probabilityfunction probability($a, $b,
$size1, $size2)
{ // Count occurrences of maximum
// element in A[]
$max1 = PHP_INT_MIN; $count1 = 0;
for ($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[]
$max2 = PHP_INT_MIN; $count2 = 0;
for ($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
$a = array(1, 2, 3);
$b = array(1, 3, 3);
$size1 = sizeof($a);
$size2 = sizeof($b);
echo probability($a, $b,
$size1, $size2);
// This code is contributed by ajit?> |
Javascript
<script>// JavaScript program to find Probability // of a random pair being the maximum// weighted pair// Function to return probabilityfunction probability(a, b, size1, size2)
{ // Count occurrences of maximum
// element in A[]
let max1 = Number.MIN_VALUE, count1 = 0;
for(let 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[]
let max2 = Number.MIN_VALUE, count2 = 0;
for(let i = 0; i < size2; i++)
{
if (b[i] > max2)
{
max2 = b[i];
count2 = 1;
}
else if (b[i] == max2)
{
count2++;
}
}
// Returning probability
return (count1 * count2) /
(size1 * size2);
}// Driver Codelet a = [ 1, 2, 3 ];let b = [ 1, 3, 3 ];let size1 = a.length;let size2 = b.length;document.write(probability(a, b, size1, size2));
// This code is contributed by code_hunt</script> |
Output
0.222222
Time Complexity: O(n).
Auxiliary Space: O(1).