Counting specific digits used in squares of numbers using JavaScript

We need to write a JavaScript function that takes an integer n (n >= 0) and a digit d (0

Problem Breakdown

For example, if n = 25 and d = 1, we need to:

• Square all numbers from 0 to 25: 0², 1², 2², ..., 25²
• Count occurrences of digit "1" in all these squares
• Return the total count

Solution

const n = 25;
const d = 1;

const countDigits = (n, d) => {
    let k = 0, count = 0;
    d = d.toString();
    
    while (k <= n) {
        let a = 0;
        let s = (k * k).toString();
        
        for (let i = 0; i < s.length; i++) {
            if (s[i] == d) {
                a++;
            }
        }
        
        if (a > 0) {
            count += a;
        }
        
        k++;
    }
    
    return count;
};

console.log(countDigits(n, d));

11

How It Works

The function works by:

1. Converting the target digit to a string for comparison
2. Iterating through all numbers from 0 to n
3. Squaring each number and converting to string
4. Counting occurrences of the target digit in each square
5. Accumulating the total count

Example Walkthrough

// Let's trace through a smaller example: n = 5, d = 1
const traceExample = (n, d) => {
    console.log(`Finding digit ${d} in squares from 0² to ${n}²:`);
    
    for (let k = 0; k <= n; k++) {
        const square = k * k;
        const squareStr = square.toString();
        const occurrences = squareStr.split(d.toString()).length - 1;
        
        if (occurrences > 0) {
            console.log(`${k}² = ${square} contains ${occurrences} occurrence(s) of ${d}`);
        }
    }
};

traceExample(5, 1);

Finding digit 1 in squares from 0² to 5²:
1² = 1 contains 1 occurrence(s) of 1
4² = 16 contains 1 occurrence(s) of 1

Optimized Version

const countDigitsOptimized = (n, d) => {
    let count = 0;
    const digitStr = d.toString();
    
    for (let k = 0; k <= n; k++) {
        const square = (k * k).toString();
        
        for (let char of square) {
            if (char === digitStr) {
                count++;
            }
        }
    }
    
    return count;
};

console.log("Original result:", countDigits(25, 1));
console.log("Optimized result:", countDigitsOptimized(25, 1));

Original result: 11
Optimized result: 11

Conclusion

This function efficiently counts specific digits in squared numbers by converting each square to a string and iterating through its characters. The optimized version eliminates unnecessary variables while maintaining the same functionality.