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Asked: 2024-09-24T21:51:02+05:30 In:

You are provided with an integer, n, and a digit, k. Your task is to determine how many integers from 1 to n (inclusive) end with the digit k. Write a function that efficiently counts these occurrences and returns the total number for a given n and k. For example, if n is 100 and k is 3, your function should return the count of numbers from 1 to 100 that have 3 as their last digit.

I’ve got a fun little problem that I’m sure will tickle your brain cells! Imagine you have a number, let’s say n, and you also have a single digit k. Your mission, should you choose to accept it, is to figure out how many integers between 1 and n (inclusive) have the digit k as their last digit.

For example, let’s say n is 100, and k is 3. You’re trying to find out how many numbers from 1 to 100 end with the digit 3. So, if you think about it, the numbers that match are 3, 13, 23, 33, 43, 53, 63, 73, 83, and 93. Can you guess how many there are? If you said 10, you’d be spot on!

Now, here’s the twist for you: I want you to write a function that can compute this in a super efficient way. We’re not just counting the numbers; we need something that can handle larger values of n quickly because, let’s face it, nobody wants to sit around counting each one individually!

While you’re at it, let’s throw in some variations. What if n is 1000 and k is 7? How about 5000 with k as 1? I’m curious how you would adjust your function to handle those larger scales while still keeping it efficient!

The logic here can be simple: you could iterate through all numbers from 1 to n, check each one for whether it ends with k, and count them up. But I’m wondering if you can think of a way to do this without looping through every single number, maybe using some clever math or patterns that come into play with how digits work.

So, what would your function look like? And more importantly, can you explain your thought process as you solve this? I’m really excited to see how you’d tackle this problem!

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    2 Answers



    1. Count Last Digit Function

      Finding Numbers Ending with a Specific Digit

      Okay, so let’s break this problem down! We need to find out how many numbers from 1 to n end with a specific digit, k. Instead of checking each number one by one, there’s a simpler way to think about it. Let’s dive into the logic!

      Understanding the Problem

      When you look at the pattern of numbers ending with a certain digit, you can see that:

      • Every 10 numbers, you’ll see each digit appear once as the last digit.
      • So, if k = 3, the sequence will look like: 3, 13, 23, …, and so on.

      Quick Calculation

      If n is larger than 10, you can easily derive how many numbers there are:

      1. First, find how many complete groups of ten are in n. You can do this with Math.floor(n / 10).
      2. Then, add one more if the last digit of n is greater than or equal to k.

      Proposed Function

          
    function countNumbersEndingWith(n, k) {
        if (k < 0 || k > 9) return 0; // In case k is not a single digit
        
        let count = Math.floor(n / 10); // Count of complete groups of ten
        
        // Check if we need to add an extra number
        if (n % 10 >= k) {
            count++; // This means we have one more number that ends with k
        }
        
        // Check if k is 0 but n < 10
        if (k === 0 && n < 10) {
            return count > 0 ? count - 1 : 0; // Adjust if k is 0 and single digit count
        }

        return count;
    }

    // Example calls
    console.log(countNumbersEndingWith(100, 3)); // Output: 10
    console.log(countNumbersEndingWith(1000, 7)); // Output: 100
    console.log(countNumbersEndingWith(5000, 1)); // Output: 500

      Testing the Function

      Just by calling the function with different values like above, you can quickly see how many numbers fit your criteria without counting each one. This makes it super efficient!

      Conclusion

      This method saves a lot of time, especially with larger numbers! It’s kind of a neat little math trick that lets us avoid unnecessary loops. Hope this helps!


    2. The task at hand is to find out how many integers from 1 to n have a specific digit k as their last digit. Instead of iterating through every number, we can utilize a more mathematical approach. The patterns in number formation show us that the last digits of numbers from 1 to n repeat every 10 numbers. Thus, for any integer n, you can count how many complete sets of 10 fit into n, which is given by the integer division of n by 10 (i.e., n // 10). Each of these complete sets contributes one occurrence of k as the last digit, except for the case when k is less than or equal to the remainder of n when divided by 10 (i.e., n % 10). If k is less than or equal to that, we add one more to our count since it would appear among the smaller range of numbers. This efficient counting lets us handle larger values of n seamlessly without looping through each number.

      Here’s a Python function that implements this logic efficiently:

      
def count_numbers_with_last_digit(n, k):
    # k should be a single digit between 0 and 9
    if k < 0 or k > 9:
        return 0
    
    complete_sets = n // 10
    result = complete_sets
    
    # Check if k is less than or equal to the remainder
    if k <= n % 10:
        result += 1
        
    return result

      Using this function, for n = 1000 and k = 7, the function returns 100, since there are 100 numbers between 1 and 1000 that end with a 7. Similarly, for n = 5000 and k = 1, the result would be 500. This method is not only efficient but also scalable, allowing you to compute the result in constant time complexity, O(1), regardless of how large n gets.

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