数组中存在密钥K的概率

给定一个数组A []，数组的大小为N，另一个为键K。任务是找到键K在数组中出现的概率。
例子：

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

可以使用以下公式找出的概率：

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count /N。
下面是上述方法的实现：

C++

// C++ code to find the probability of
// search key K present in array
#include 
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

Javascript

// 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
 

输出

0

时间复杂度：O(N)