# Count numbers with unit digit k in given range

Here given a range from low to high and given a number k.You have to find out the number of count which a number has same digit as k
Examples:

Input: low = 2, high = 35, k = 2
Output: 4
Numbers are 2, 12, 22, 32

Input: low = 3, high = 30, k = 3
Output: 3
Numbers are  3, 13, 23

A naive approach is to traverse through all numbers in given range and check last digit of every number and increment result if last digit is equal to k.

## C++

 // Simple CPP program to count numbers with // last digit as k in given range. #include using namespace std;   // Returns count of numbers with k as last // digit. int counLastDigitK(int low, int high, int k) {     int count = 0;     for (int i = low; i <= high; i++)         if (i % 10 == k)             count++;       return count; }   // Driver Program int main() {     int low = 3, high = 35, k = 3;     cout << counLastDigitK(low, high, k);     return 0; }

## Java

 // Simple Java program to count numbers with // last digit as k in given range. import java.util.*; import java.lang.*;   public class GfG {     // Returns count of numbers with     // k as last digit.     public static int counLastDigitK(int low,                                 int high, int k)     {         int count = 0;         for (int i = low; i <= high; i++)             if (i % 10 == k)                 count++;         return count;     }           // driver function     public static void main(String args[])     {         int low = 3, high = 35, k = 3;         System.out.println(counLastDigitK(low, high, k));     } }   // This code is contributed by Sagar Shukla

## Python3

 # Simple python program to count numbers with # last digit as k in given range.   # Returns count of numbers with k as last # digit. def counLastDigitK(low, high, k):     count = 0     for i in range(low, high+1):         if (i % 10 == k):             count+=1     return count     # Driver Program low = 3 high = 35 k = 3 print(counLastDigitK(low, high, k))   # This code is contributed by # Smitha Dinesh Semwal

## C#

 // Simple C# program to count numbers with // last digit as k in given range. using System;   public class GfG {     // Returns count of numbers with     // k as last digit.     public static int counLastDigitK(int low,                                 int high, int k)     {         int count = 0;         for (int i = low; i <= high; i++)             if (i % 10 == k)                 count++;         return count;     }           // Driver function     public static void Main()     {         int low = 3, high = 35, k = 3;         Console.WriteLine(counLastDigitK(low, high, k));     } }   // This code is contributed by vt_m

## PHP



Output:

4

Time Complexity: O(high – low)

An efficient solution is based on the fact that every digit appears once as the last digit in every 10 consecutive numbers.

## C++

 // Efficient CPP program to count numbers  // with last digit as k in given range. #include using namespace std;   // Returns count of numbers with k as last // digit. int countLastDigitK(long long low,         long long high, long long K) {     long long mlow = 10 * ceil(low/10.0);   long long mhigh = 10 * floor(high/10.0);     int count = (mhigh - mlow)/10;   if (high % 10 >= K)     count++;   if (low % 10 <=K && (low%10))     count++;     return count; }   // Driver Code int main() {     int low = 3, high = 35, k = 3;     cout << countLastDigitK(low, high, k);     return 0; }

## Java

 // Efficient Java program to count numbers // with last digit as k in given range. import java.util.*; import java.lang.*;   public class GfG {     // Returns count of numbers with     // k as last digit.     public static int counLastDigitK(int low,                              int high, int k)     {         int mlow = 10 * (int)                       Math.ceil(low/10.0);         int mhigh = 10 * (int)                       Math.floor(high/10.0);         int count = (mhigh - mlow)/10;         if (high % 10 >= k)             count++;         if (low % 10 <= k && (low%10) > 0)             count++;         return count;     }           // driver function     public static void main(String argc[])     {         int low = 3, high = 35, k = 3;         System.out.println(counLastDigitK(low, high, k));     } }   // This code is contributed by Sagar Shukla

## Python3

 import math # Efficient python program to count numbers # with last digit as k in given range.   # Returns count of numbers with k as last # digit. def counLastDigitK(low, high, k):     mlow = 10 * math.ceil(low/10.0)     mhigh = 10 * int(high/10.0)           count = (mhigh - mlow)/10     if (high % 10 >= k):         count += 1     if (low % 10 <= k and \         (low%10) > 0):         count += 1     return int(count)     # Driver Code low = 3 high = 35 k = 3 print(counLastDigitK(low, high, k))   # This code is contributed by # Smitha Dinesh Semwal

## C#

 // Efficient Java program to count numbers // with last digit as k in given range. using System;   public class GfG {     // Returns count of numbers with     // k as last digit.     public static int counLastDigitK(int low,                                 int high, int k)     {         int mlow = 10 * Convert.ToInt32(                   Math.Ceiling(low/10.0));         int mhigh = 10 * Convert.ToInt32(                   Math.Floor(high/10.0));         int count = (mhigh - mlow) / 10;         if (high % 10 >= k)             count++;         if (low % 10 <= k && (low%10) > 0)             count++;         return count;     }           // Driver function     public static void Main()     {         int low = 3, high = 35, k = 3;         Console.WriteLine(           counLastDigitK(low, high, k));     } }   // This code is contributed by vt_m

Output

4

Time Complexity : O(1)

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