密钥K出现在数组中的概率

给定一个数组A[],数组的大小为N,另一个密钥为K。任务是找到密钥K出现在数组中的概率。 例如:

null
Input : N = 6        A[] = { 4, 7, 2, 0, 8, 7, 5 }        K = 3Output :0Since 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 = 5Output :0.2

可使用以下公式计算出发生故障的概率:

Probability = total number of K present /                          size of array.

首先,计算K的数量,然后概率将是K的数量除以N,即计数/N。 以下是上述方法的实施情况:

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 probability
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 Code
int 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 probability
def kPresentProbability(a, n, k) :
count = a.count(k)
# find probability upto
# 2 decimal places
return round (count / n , 2 )
# Driver Code
if __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 array
class GFG
{
// Function to find the probability
static 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 Code
public 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 array
using System;
class GFG
{
// Function to find the probability
static 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 Code
public 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 probability
function 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 probability
function 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>


输出

0

时间复杂度:O(N)

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