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Count frequency of digit K in given Array

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  • Last Updated : 07 Dec, 2021

Given an array arr[] of integers of size N and a single digit integer K. The task is to find the total count of occurrences of the digit K in the array

Examples:

Input: arr[] = {15, 66, 26, 91}, K = 6
Output: 3
Explanation: Occurrences of 6 in each array elements are: 0, 2, 1, 0 respectively.
Therefore, total occurrences = 0 + 2 + 1 + 0 = 3

Input: arr[] = {20, 21, 0}, K = 0
Output: 2

Approach: The idea is to traverse the array and for every individual array element, count the occurrences of the digit K in it and update the total count. Follow the steps below to solve the problem:

  1. Initialize the total occurrences of digit K, say count, as 0.
  2. Traverse the given array from the start element till the end.
  3. For every traversed element, find the frequency of digit K in that element.
  4. Add the count to the total sum.
  5. Return the total sum as answer.

Below is the implementation of the above approach:

C++




// C++ code to count frequency
// of digit K in given Array
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count occurrences
// in an element
int countOccurrences(int num, int K)
{
    // If num is 0
    if (K == 0 && num == 0)
        return 1;
 
    // Initialize count
    int count = 0;
 
    // Count for occurrences of digit K
    while (num > 0) {
        if (num % 10 == K)
            count++;
        num /= 10;
    }
    return count;
}
 
// Function to return sum of
// total occurrences of digit K
int totalOccurrences(int arr[], int N, int K)
{
 
    // Initialize sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        sum += countOccurrences(arr[i], K);
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 15, 66, 26, 91 };
    int K = 6;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << totalOccurrences(arr, N, K);
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
class GFG {
 
  // Function to count occurrences
  // in an element
  static int countOccurrences(int num, int K)
  {
 
    // If num is 0
    if (K == 0 && num == 0)
      return 1;
 
    // Initialize count
    int count = 0;
 
    // Count for occurrences of digit K
    while (num > 0) {
      if (num % 10 == K)
        count++;
      num /= 10;
    }
    return count;
  }
 
  // Function to return sum of
  // total occurrences of digit K
  static int totalOccurrences(int arr[], int N, int K)
  {
 
    // Initialize sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
      sum += countOccurrences(arr[i], K);
    }
 
    return sum;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
     
    int arr[] = { 15, 66, 26, 91 };
    int K = 6;
    int N = arr.length;
    System.out.println( totalOccurrences(arr, N, K));
  }
}
 
// This code is contributed by Potta Lokesh


Python3




# Python3 code to count frequency
# of digit K in given array
 
# Function to count occurrences
# in an element
def countOccurrences(num, K):
 
    # If num is 0
    if (K == 0 and num == 0):
        return 1
 
    # Initialize count
    count = 0
 
    # Count for occurrences of digit K
    while (num > 0):
        if (num % 10 == K):
            count += 1
             
        num = (num // 10)
         
    return count
 
# Function to return sum of
# total occurrences of digit K
def totalOccurrences(arr, N, K):
 
    # Initialize sum
    sum = 0
 
    # Traverse the array
    for i in range(N):
        sum += countOccurrences(arr[i], K)
 
    return sum
 
# Driver Code
arr = [ 15, 66, 26, 91 ]
K = 6
 
N = len(arr)
 
print(totalOccurrences(arr, N, K))
 
# This code is contributed by saurabh_jaiswal.


C#




// C# program for the above approach
using System;
class GFG {
 
  // Function to count occurrences
  // in an element
  static int countOccurrences(int num, int K)
  {
 
    // If num is 0
    if (K == 0 && num == 0)
      return 1;
 
    // Initialize count
    int count = 0;
 
    // Count for occurrences of digit K
    while (num > 0) {
      if (num % 10 == K)
        count++;
      num /= 10;
    }
    return count;
  }
 
  // Function to return sum of
  // total occurrences of digit K
  static int totalOccurrences(int []arr, int N, int K)
  {
 
    // Initialize sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
      sum += countOccurrences(arr[i], K);
    }
 
    return sum;
  }
 
  // Driver Code
  public static void Main ()
  {
     
    int []arr = { 15, 66, 26, 91 };
    int K = 6;
    int N = arr.Length;
    Console.Write( totalOccurrences(arr, N, K));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
 
// Javascript code to count frequency
// of digit K in given Array
 
// Function to count occurrences
// in an element
function countOccurrences(num, K)
{
 
    // If num is 0
    if (K == 0 && num == 0)
        return 1;
 
    // Initialize count
    let count = 0;
 
    // Count for occurrences of digit K
    while (num > 0) {
        if (num % 10 == K)
            count++;
        num = Math.floor(num / 10);
    }
    return count;
}
 
// Function to return sum of
// total occurrences of digit K
function totalOccurrences(arr, N, K) {
 
    // Initialize sum
    let sum = 0;
 
    // Traverse the array
    for (let i = 0; i < N; i++) {
        sum += countOccurrences(arr[i], K);
    }
 
    return sum;
}
 
// Driver Code
let arr = [15, 66, 26, 91];
let K = 6;
 
let N = arr.length
document.write(totalOccurrences(arr, N, K));
 
// This code is contributed by saurabh_jaiswal.
</script>


Output

3

Time Complexity: O(N*D) where D is the maximum number of digit in an array element
Auxiliary Space: O(1)


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