# Count of numbers in Array ending with digits of number N

Given a number N and an array arr[] consisting of K numbers, the task is to find the count of numbers in the array which ends with any of the digit present in the number N.
Examples:

Input: N = 1731 arr[] = {57, 6786}
Output:
Explanation:
For 57, the last digit is 7 and since 7 is present is N, so the count is 1.
For 6786, the last digit is 6 and since 6 is not present in N, the count remains 1.
Input: N = 1324, arr[] = {23, 25, 12, 121}
Output:

Naive Approach: The naive approach for this problem is that for every number in the array arr[], check if its last digit is equal to any of the digits in N. Increment the count for each of such number and print it at the end.
Time Complexity: O(N * K), where N is the number and K is the number of elements in the array arr[].
Efficient Approach: The efficient approach for this problem is to perform a preprocessing.

• Initially, create an array A[] of size 10.
• This array acts as a hash which stores all the digits occurred in the number N.
• After this, for every number in the array arr[], extract the last digit and check if this last digit has occurred or not in the array.
• Increment the count for each of such number and print it at the end.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the count ` `// of numbers in Array ending ` `// with digits of number N ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Array to keep the ` `// track of digits occurred ` `// Initially all are 0(false) ` `int` `digit = { 0 }; ` ` `  `// Function to initialize true ` `// if the digit is present ` `void` `digitsPresent(``int` `n) ` `{ ` `    ``// Variable to store the last digit ` `    ``int` `lastDigit; ` ` `  `    ``// Loop to iterate through every ` `    ``// digit of the number N ` `    ``while` `(n != 0) { ` `        ``lastDigit = n % 10; ` ` `  `        ``// Updating the array according ` `        ``// to the presence of the ` `        ``// digit in n at the array index ` `        ``digit[lastDigit] = ``true``; ` `        ``n /= 10; ` `    ``} ` `} ` ` `  `// Function to check if the ` `// numbers in the array ` `// end with the digits of ` `// the number N ` `int` `checkLastDigit(``int` `num) ` `{ ` ` `  `    ``// Variable to store the count ` `    ``int` `count = 0; ` ` `  `    ``// Variable to store the last digit ` `    ``int` `lastDigit; ` `    ``lastDigit = num % 10; ` ` `  `    ``// Checking the presence of ` `    ``// the last digit in N ` `    ``if` `(digit[lastDigit] == ``true``) ` `        ``count++; ` ` `  `    ``return` `count; ` `} ` ` `  `// Function to find ` `// the required count ` `void` `findCount(``int` `N, ``int` `K, ``int` `arr[]) ` `{ ` ` `  `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i < K; i++) { ` ` `  `        ``count = checkLastDigit(arr[i]) == 1 ` `                    ``? count + 1 ` `                    ``: count; ` `    ``} ` `    ``cout << count << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 1731; ` ` `  `    ``// Preprocessing ` `    ``digitsPresent(N); ` ` `  `    ``int` `K = 5; ` `    ``int` `arr[] = { 57, 6786, ` `                  ``1111, 3, 9812 }; ` ` `  `    ``findCount(N, K, arr); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the count ` `// of numbers in Array ending ` `// with digits of number N ` `class` `GFG{ ` ` `  `// Array to keep the ` `// track of digits occurred  ` `// Initially all are 0(false) ` `public` `static` `int``[] digit = ``new` `int``[``10``]; ` ` `  `// Function to initialize 1(true) ` `// if the digit is present ` `public` `static` `void` `digitsPresent(``int` `n) ` `{ ` `     `  `    ``// Variable to store the last digit ` `    ``int` `lastDigit; ` ` `  `    ``// Loop to iterate through every ` `    ``// digit of the number N ` `    ``while` `(n != ``0``) ` `    ``{ ` `        ``lastDigit = n % ``10``; ` ` `  `        ``// Updating the array according ` `        ``// to the presence of the ` `        ``// digit in n at the array index ` `        ``digit[lastDigit] = ``1``; ` `        ``n /= ``10``; ` `    ``} ` `} ` ` `  `// Function to check if the ` `// numbers in the array ` `// end with the digits of ` `// the number N ` `public` `static` `int` `checkLastDigit(``int` `num) ` `{ ` ` `  `    ``// Variable to store the count ` `    ``int` `count = ``0``; ` ` `  `    ``// Variable to store the last digit ` `    ``int` `lastDigit; ` `    ``lastDigit = num % ``10``; ` ` `  `    ``// Checking the presence of ` `    ``// the last digit in N ` `    ``if` `(digit[lastDigit] == ``1``) ` `        ``count++; ` ` `  `    ``return` `count; ` `} ` ` `  `// Function to find ` `// the required count ` `public` `static` `void` `findCount(``int` `N, ``int` `K, ` `                             ``int` `arr[]) ` `{ ` `    ``int` `count = ``0``; ` ` `  `    ``for``(``int` `i = ``0``; i < K; i++)  ` `    ``{ ` `       ``count = checkLastDigit(arr[i]) == ``1` `?  ` `               ``count + ``1` `: count; ` `    ``} ` `    ``System.out.println(count); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``1731``; ` ` `  `    ``// Preprocessing ` `    ``digitsPresent(N); ` ` `  `    ``int` `K = ``5``; ` `    ``int` `arr[] = { ``57``, ``6786``, ``1111``, ``3``, ``9812` `}; ` ` `  `    ``findCount(N, K, arr); ` `} ` `} ` ` `  `// This code is contributed by Sayantan Pal `

## Python3

 `# Python3 program to find the count  ` `# of numbers in Array ending  ` `# with digits of number N  ` ` `  `# Array to keep the  ` `# track of digits occurred  ` `# Initially all are 0(false)  ` `digit ``=` `[``0``] ``*` `10` ` `  `# Function to initialize true  ` `# if the digit is present ` `def` `digitsPresent(n): ` `     `  `    ``# Variable to store the last digit ` `    ``lastDigit ``=` `0``; ` ` `  `    ``# Loop to iterate through every  ` `    ``# digit of the number N ` `    ``while` `(n !``=` `0``): ` `        ``lastDigit ``=` `n ``%` `10``; ` `         `  `        ``# Updating the array according  ` `        ``# to the presence of the  ` `        ``# digit in n at the array index  ` `        ``digit[``int``(lastDigit)] ``=` `1``; ` `        ``n ``/``=` `10``; ` ` `  `# Function to check if the numbers  ` `# in the array end with the digits  ` `# of the number N ` `def` `checkLastDigit(num): ` ` `  `    ``# Variable to store the count ` `    ``count ``=` `0``; ` ` `  `    ``# Variable to store the last digit ` `    ``lastDigit ``=` `0``; ` `    ``lastDigit ``=` `num ``%` `10``; ` ` `  `    ``# Checking the presence of  ` `    ``# the last digit in N ` `    ``if` `(digit[``int``(lastDigit)] ``=``=` `1``): ` `        ``count ``+``=` `1` ` `  `    ``return` `count; ` ` `  `# Function to find the required count ` `def` `findCount(N, K, arr): ` ` `  `    ``count ``=` `0``; ` `    ``for` `i ``in` `range``(K): ` `        ``if` `checkLastDigit(arr[i]) ``=``=` `1``: ` `            ``count ``+``=` `1` `        ``else``: ` `            ``count ` `             `  `    ``print``(count) ` ` `  `# Driver code ` `N ``=` `1731``; ` ` `  `# Preprocessing ` `digitsPresent(N); ` ` `  `K ``=` `5``; ` `arr ``=` `[ ``57``, ``6786``, ``1111``, ``3``, ``9812` `]; ` ` `  `findCount(N, K, arr); ` ` `  `# This code is contributed by grand_master  `

## C#

 `// C# program to find the count ` `// of numbers in Array ending ` `// with digits of number N ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Array to keep the track of digits occurred  ` `// Initially all are 0(false) ` `public` `static` `int` `[]digit = ``new` `int``; ` ` `  `// Function to initialize 1(true) ` `// if the digit is present ` `public` `static` `void` `digitsPresent(``int` `n) ` `{ ` ` `  `    ``// Variable to store the last digit ` `    ``int` `lastDigit; ` ` `  `    ``// Loop to iterate through every ` `    ``// digit of the number N ` `    ``while` `(n != 0) ` `    ``{ ` `        ``lastDigit = n % 10; ` ` `  `        ``// Updating the array according to the  ` `        ``// presence of the digit in n at the ` `        ``// array index ` `        ``digit[lastDigit] = 1; ` `        ``n /= 10; ` `    ``} ` `} ` ` `  `// Function to check if the numbers in the ` `// array end with the digits of the number N ` `public` `static` `int` `checkLastDigit(``int` `num) ` `{ ` ` `  `    ``// Variable to store the count ` `    ``int` `count = 0; ` ` `  `    ``// Variable to store the last digit ` `    ``int` `lastDigit; ` `    ``lastDigit = num % 10; ` ` `  `    ``// Checking the presence of ` `    ``// the last digit in N ` `    ``if` `(digit[lastDigit] == 1) ` `        ``count++; ` ` `  `    ``return` `count; ` `} ` ` `  `// Function to find the required count ` `public` `static` `void` `findCount(``int` `N, ``int` `K, ` `                             ``int` `[]arr) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``for``(``int` `i = 0; i < K; i++)  ` `    ``{ ` `        ``count = checkLastDigit(arr[i]) == 1 ? ` `                ``count + 1 : count; ` `    ``} ` `    ``Console.WriteLine(count); ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main() ` `{ ` `    ``int` `N = 1731; ` ` `  `    ``// Preprocessing ` `    ``digitsPresent(N); ` ` `  `    ``int` `K = 5; ` `    ``int` `[]arr = { 57, 6786, 1111, 3, 9812 }; ` ` `  `    ``findCount(N, K, arr); ` `} ` `} ` ` `  `// This code is contributed by piyush3010 `

Output:

```3
```

Time Complexity:

• O(N), where N is the given number for preprocessing.
• O(K), where K is the number of queries to find answers for the queries.

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