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++

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// C++ program to find the count
// of numbers in Array ending
// with digits of number N
  
#include <bits/stdc++.h>
using namespace std;
  
// Array to keep the
// track of digits occurred
// Initially all are 0(false)
int digit[10] = { 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;
}

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Java

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// 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

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Python3

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# 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 

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C#

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// 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[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;
    }
    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

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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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