Probability of a key K present in array
Given an array A[] and size of array is N and one another key K. The task is to find the probability that the Key K present in array.
Examples:
Input : N = 6
A[] = { 4, 7, 2, 0, 8, 7, 5 }
K = 3
Output :0
Since value of k = 3 is not present in array,
hence the probability of 0.
Input :N = 10
A[] = { 2, 3, 5, 1, 9, 8, 0, 7, 6, 5 }
K = 5
Output :0.2
The probability of can be found out using the below formula:
Probability = total number of K present /
size of array.First, count the number of K’s and then the probability will be the number of K’s divided by N i.e. count / N.
Below is the implementation of the above approach:
C++
// C++ code to find the probability of// search key K present in array#include <bits/stdc++.h>using namespace std;// Function to find the probabilityfloat kPresentProbability(int a[], int n, int k){ float count = 0; for (int i = 0; i < n; i++) if (a[i] == k) count++; // find probability return count / n;}// Driver Codeint main(){ int A[] = { 4, 7, 2, 0, 8, 7, 5 }; int K = 3; int N = sizeof(A) / sizeof(A[0]); cout << kPresentProbability(A, N, K); return 0;} |
Python3
# Python3 code to find the# probability of search key# K present in 1D-array (list).# Function to find the probabilitydef kPresentProbability(a, n, k) : count = a.count(k) # find probability upto # 2 decimal places return round(count / n , 2)# Driver Codeif __name__ == "__main__" : A = [ 4, 7, 2, 0, 8, 7, 5 ] K = 2 N = len(A) print(kPresentProbability( A, N, K))# This code is contributed# by AnkitRai1 |
Java
// Java code to find the probability// of search key K present in arrayclass GFG{// Function to find the probabilitystatic float kPresentProbability(int a[], int n, int k){ float count = 0; for (int i = 0; i < n; i++) if (a[i] == k) count++; // find probability return count/ n;}// Driver Codepublic static void main(String[] args){ int A[] = { 4, 7, 2, 0, 8, 7, 5 }; int K = 2; int N = A.length; double n = kPresentProbability(A, N, K); double p = (double)Math.round(n * 100) / 100; System.out.println(p);}}// This code is contributed// by ChitraNayal |
C#
// C# code to find the probability// of search key K present in arrayusing System;class GFG{// Function to find the probabilitystatic float kPresentProbability(int[] a, int n, int k){ float count = 0; for (int i = 0; i < n; i++) if (a[i] == k) count++; // find probability return count/ n;}// Driver Codepublic static void Main(){ int[] A = { 4, 7, 2, 0, 8, 7, 5 }; int K = 2; int N = A.Length; double n = kPresentProbability(A, N, K); double p = (double)Math.Round(n * 100) / 100; Console.Write(p);}}// This code is contributed// by ChitraNayal |
PHP
<?php// PHP code to find the probability// of search key K present in array// Function to find the probabilityfunction kPresentProbability(&$a, $n, $k){ $count = 0; for ($i = 0; $i < $n; $i++) if ($a[$i] == $k) $count++; // find probability return $count / $n;}// Driver Code$A = array( 4, 7, 2, 0, 8, 7, 5 );$K = 2;$N = sizeof($A);echo round(kPresentProbability($A, $N, $K), 2);// This code is contributed// by ChitraNayal?> |
Javascript
<script>// JavaScript code to find the probability of// search key K present in array// Function to find the probabilityfunction kPresentProbability(a,n,k){ let count = 0; for (let i = 0; i < n; i++) if (a[i] == k) count+=1; // find probability return count / n;}// Driver Code let A = [ 4, 7, 2, 0, 8, 7, 5 ]; let K = 3; let N = A.length; document.write(kPresentProbability(A, N,K)); // This code contributed by Rajput-Ji</script> |
Output
0
Time Complexity: O(N)
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