JavaScript: Count total number of 1s from 1 to N

JavaScript Math: Exercise-102 with Solution

Count Digit 1 in Range

Write a JavaScript program to count the number of times the digit 1 appears in 1 to a given number.

Example : 11, 1 appears 4 time between 1 to 11 [1->1, 10->1 ,11-.2]

Test Data:
(11) -> 4
(305) -> 161
(0) -> false

Sample Solution:

Solution-1

JavaScript Code:

/**
 * Function to count the total number of occurrences of digit 1 from 1 to n.
 * @param {number} num - The upper limit.
 * @returns {number} - The total number of occurrences of digit 1.
 */
function test(num) {
    if (num < 1) // If the input is less than 1, return false
        return false;
    var ctr = 0; // Initialize the counter for occurrences of digit 1
    for (var i = 1; i <= num; i *= 10) { // Loop through each digit place
        var d = i * 10; // Calculate the next digit place
        // Count the occurrences of digit 1 in the current digit place
        ctr += parseInt(num / d) * i + Math.min(Math.max(num % d - i + 1, 0), i);
    }
    return ctr; // Return the total count
}
n = 11;
console.log("n = " + n);
console.log("Total number of 1s from 1 to n: " + test(n));

n = 305;
console.log("n = " + n);
console.log("Total number of 1s from 1 to n: " + test(n));

n = 0;
console.log("n = " + n);
console.log("Total number of 1s from 1 to n: " + test(n));

Output:

n = 11
Total number of 1s from 1 to n: 4
n = 305
Total number of 1s from 1 to n: 161
n = 0
Total number of 1s from 1 to n: false

Flowchart:

Live Demo:

See the Pen javascript-math-exercise-102 by w3resource (@w3resource) on CodePen.

Solution-2

JavaScript Code:

// Define a function named test that counts the occurrences of the digit 1 from 1 to n.
function test(n) {
    // If n is less than 1, return false.
    if (n < 1)
        return false;
    // Initialize a counter variable.
    var ctr = 0;
    // Iterate through powers of 10 until i exceeds n.
    for (var i = 1; i <= n; i *= 10) {
        // Calculate the range of digits for the current power of 10.
        var d = i * 10;
        // Update the counter with the number of occurrences of digit 1 in the current range.
        ctr += parseInt(n / d) * i +
            Math.min(Math.max(n % d - i + 1, 0), i);
    }
    // Return the total count of occurrences of digit 1.
    return ctr;
}
// Test the function with different values of n.
n = 11;
console.log("n = " + n);
console.log("Total number of 1s from 1 to n: " + test(n));
n = 305;
console.log("n = " + n);
console.log("Total number of 1s from 1 to n: " + test(n));
n = 0;
console.log("n = " + n);
console.log("Total number of 1s from 1 to n: " + test(n));

Output:

n = 11
Total number of 1s from 1 to n: 4
n = 305
Total number of 1s from 1 to n: 161
n = 0
Total number of 1s from 1 to n: false

Flowchart:

Live Demo:

See the Pen javascript-math-exercise-102-1 by w3resource (@w3resource) on CodePen.

For more Practice: Solve these Related Problems:

• Write a JavaScript function that counts the number of times the digit '1' appears in all numbers from 1 to a given number using string conversion.
• Write a JavaScript function that iterates over each number in the range and counts the occurrences of '1' in its decimal representation.
• Write a JavaScript function that optimizes the count by analyzing each digit position across the range.
• Write a JavaScript function that validates the input and returns an error for non-positive numbers before counting.

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