Counting Integers Less Than n Containing the Digit 4: Techniques and Optimization

In computer science and mathematics, there are various interesting problems that deal with specific properties of numbers. One such intriguing problem is counting the number of integers less than a given number n that contain the digit 4. This topic is not only fascinating but also provides a good platform to understand and implement different algorithmic techniques, including brute force and more optimized solutions like dynamic programming.

Steps to Count Integers Containing the Digit 4

Here’s a systematic approach to solve the problem:

Steps to Count Integers Containing the Digit 4

1. Identify the Range: The goal is to count all integers from 0 to n-1.

2. Iterate Through the Range: Loop through each integer in this range and check if it contains the digit 4.

3. Check for the Digit: Convert each integer to a string and check if 4 is in that string.

4. Count Matches: Maintain a count of how many integers meet the criteria.

Example Code in Python

Here’s a simple Python function to accomplish this:

def count_integers_with_digit_4(n):
    count  0
    for i in range(n):
        if '4' in str(i):
            count   1
    return count

Example Usage

For example, if n 50:

n  50
result  count_integers_with_digit_4(n)
print(result)

Explanation of the Code:

Loop from 0 to n-1: The for loop iterates through all integers less than n.

Convert to String: Each integer i is converted to a string to check for the presence of the digit 4.

Count Matching Integers: If 4 is found, increment the count.

Performance Consideration

This approach works well for smaller values of n. However, for very large values of n, this brute-force method may become inefficient. In such cases, you might consider more advanced techniques such as combinatorial counting or digit dynamic programming to count numbers more efficiently without iterating through every integer.

Solution: Brute Force and Dynamic Programming Solutions

Let's explore an optimized dynamic programming approach to solve this problem.

Dynamic Programming Solution

The dynamic programming solution uses memoization to store previously computed results, which significantly improves the performance by avoiding redundant calculations. Here's a C implementation:

#include iostream
using namespace std;
int table[1000]  {-1};
int fourcountDP(int i) {
    if (i  4) return 1;
    if (table[i] ! -1) return table[i];
    if (i % 10  4) table[i]  1   fourcountDP(i / 10);
    else table[i]  fourcountDP(i / 10);
    return table[i];
}
int fourcount(int i) {
    int count  0;
    if (i  4) return 1;
    while(i  0) {
        if (i % 10  4) count  ;
        i  i / 10;
    }
    return count;
}
int main() {
    for (int i  0; i  100; i  ) {
        if (fourcount(i)  0)
            cout  i  endl;
    }
    return 0;
}

In the above code:

fourcountDP function: This function is a recursive function with memoization to store and reuse previously computed results.

fourcount function: This is a simple iterative function to count the number of times 4 appears in the integer.

This dynamic programming approach is significantly more efficient for large values of n, offering a substantial improvement in runtime over the brute-force method.

By understanding and implementing both brute force and dynamic programming solutions, you can tackle various similar problems in computer science with efficient and optimized algorithms.