1556. Thousand Separator

EasyString
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Problem Description

The given problem requires us to take an integer n and format it as a string that represents the number with a dot . inserted every three digits from the right. This is commonly known as adding a "thousand separator." For example, if n is 123456789, the output should be "123.456.789". If n is less than 1000, it should be returned without change, since there are no thousands to separate.

Intuition

To solve this problem, one intuitive approach is to process the integer digit by digit from right to left (least significant digit to most significant digit). We can achieve that by continuously dividing the number by 10 and getting the remainder, which represents the current rightmost digit.

Each time we extract a digit, we append it to a result list. We also need to keep a count of how many digits we've added so that we know when to insert a dot. For this problem, we insert a dot every three digits. When there are no more digits left (i.e., the remaining number is 0), we stop the process. Finally, we reverse the result list since we built the number from right to left, join the elements to form a string, and return the formatted number.

Here are the specific steps of the process:

  1. Initialize a count cnt to 0. This will keep track of the number of digits processed.
  2. Initialize an empty list ans to build the answer from individual digits and dots.
  3. Enter a loop that will run until there are no more digits to process in the number n. Within the loop:
    • Divide n by 10, separate the quotient and the remainder. The remainder is the current digit to add to the answer, while the quotient is the reduced number for the next iteration.
    • Convert the remainder to a string and append it to ans.
    • Increment the count cnt.
    • If cnt is equal to 3 and there are still digits in n (i.e., n is not 0), append a . to ans and reset cnt to 0.
  4. Break the loop when n is reduced to 0.
  5. Reverse the list ans since we built it backwards.
  6. Join the elements in ans to form the final string.
  7. Return the resulting string.

This method ensures that we are adding a dot every three digits while taking the basic rules of string formatting into account.

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Solution Approach

The solution to this problem involves a straightforward implementation of the intuition described. The solution uses simple arithmetic operations and a list as the primary data structure to build the answer. No particular algorithmic pattern is necessary other than following this numeric processing method:

  1. The solution class Solution contains the method thousandSeparator, which takes an integer n as an argument and returns a string with the formatted number.

  2. A counter cnt is initialized to 0. This counter is used to track the number of digits added to the answer list ans since the last dot was added (or since the beginning if no dot has been added yet).

  3. An empty list ans is created to accumulate the digits and dots in reverse order, as we will be processing the digits from least significant to most significant.

  4. The while loop while 1: ensures that the loop continues until explicitly broken.

  5. Inside the loop, n, v = divmod(n, 10) divides n by 10, storing the quotient back in n for the next iteration and the remainder in v. The remainder represents the current least significant digit to be added to the ans list.

  6. ans.append(str(v)) adds the current digit to the answer list as a string, since we want the final output to be a string.

  7. The counter cnt is incremented by 1 each time a digit is added to ans.

  8. A conditional if n == 0: checks whether the number n has been fully processed. If n is 0, the loop is terminated by executing a break.

  9. Another conditional if cnt == 3: checks if three digits have been added to ans since the last dot or since the start. If true, a dot '.' is appended to ans, and cnt is reset to 0 to count the next three digits.

  10. Once the loop is broken, return ''.join(ans[::-1]) is executed. This joins the elements of ans together into a string, and ans[::-1] reverses the list since we built the number from the least significant digit to the most significant.

By maintaining a counter and using the divmod function to split the integer into digits, the approach avoids string-to-integer conversions except when appending digits to the answer list. The use of a list to construct the answer in reverse order helps efficiently build the result since lists in Python have a time complexity of O(1) for typical append operations. Once the number is fully converted into a list of characters, the reversal and join operations form the final string to be returned.

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Example Walkthrough

Let's walk through the solution with a small example. Supposing the input integer n is 12345. We need to format it as a string with a dot inserted every three digits from the right. So, our expected output is "12.345".

Here are the steps demonstrating how the algorithm would process this input:

  1. Initialize the counter cnt to 0 and the list ans to an empty list.

  2. Enter the loop since n is non-zero.

  3. In the first iteration, divmod(n, 10) gives us n = 1234 and v = 5. We append '5' to ans and increment cnt by 1.

  4. In the second iteration, now n = 123 and v = 4. We append '4' to ans and increment cnt by 1.

  5. In the third iteration, n = 12 and v = 3. We append '3' to ans. cnt is now 3, so we append a '.' to ans and reset cnt to 0.

  6. In the fourth iteration, n = 1 and v = 2. We append '2' to ans and increment cnt by 1.

  7. In the fifth iteration, n = 0 and v = 1. We append '1' to ans. Since n is now 0, we break out of the loop.

  8. Now ans is ['5', '4', '3', '.', '2', '1']. We need to reverse the list to get the digits in the correct order.

  9. After reversing, ans is ['1', '2', '.', '3', '4', '5'].

  10. Joining the elements in ans with '', we get the final result "12.345".

  11. We return the resulting string "12.345".

Note: If n had been smaller than 1000, say n = 12, the process would be the same without the insertion of a dot, resulting in "12".

Solution Implementation

1class Solution:
2    def thousandSeparator(self, value: int) -> str:
3        # Initialize a counter to track the number of digits processed
4        digit_counter = 0
5      
6        # Initialize a list to build the answer incrementally
7        result_parts = []
8      
9        # Loop until the entire number has been processed
10        while True:
11            # Divide the number by 10 to get the next digit and the remainder
12            value, remainder = divmod(value, 10)
13          
14            # Convert the remainder (a digit) to a string and append to the list
15            result_parts.append(str(remainder))
16          
17            # Increment the digit counter
18            digit_counter += 1
19          
20            # Check if the number has been completely divided
21            if value == 0:
22                break
23          
24            # If three digits have been processed, insert a period and reset counter
25            if digit_counter == 3:
26                result_parts.append('.')
27                digit_counter = 0
28      
29        # Since digits are processed in reverse order, reverse the list and join the parts into a string
30        formatted_number = ''.join(result_parts[::-1])
31      
32        return formatted_number
33
1class Solution {
2    // This method converts an integer to a string with dot separators for every three digits
3    public String thousandSeparator(int number) {
4        int count = 0; // Initialize a counter to keep track of every three digits
5        StringBuilder formattedNumber = new StringBuilder(); // Use StringBuilder to efficiently manipulate strings
6      
7        // Loop until the entire number has been processed
8        while (true) {
9            int digit = number % 10; // Extract the last digit of the number
10            number /= 10; // Remove the last digit from the number
11            formattedNumber.append(digit); // Append the digit to the StringBuilder
12            count++; // Increment the counter
13
14            // If the number is reduced to zero, break out of the loop
15            if (number == 0) {
16                break;
17            }
18          
19            // After every third digit, append a dot to the StringBuilder
20            if (count == 3) {
21                formattedNumber.append('.');
22                count = 0; // Reset the counter after appending the dot
23            }
24        }
25      
26        // Reverse the StringBuilder content to maintain the correct order
27        // and convert it to a String before returning
28        return formattedNumber.reverse().toString();
29    }
30}
31
1class Solution {
2public:
3    // Function to add a thousand separator in the given integer.
4    // It inserts a period '.' every three digits from right to left
5    string thousandSeparator(int n) {
6        int count = 0;             // Initialize a counter to keep track of the number of digits
7        string formattedNumber;    // Initialize an empty string to build the result
8
9        // Proceed to iterate until the entire number has been processed
10        do {
11            int digit = n % 10;    // Extract the rightmost digit
12            n /= 10;               // Remove the rightmost digit from the number
13            formattedNumber += to_string(digit); // Append the digit to the result string
14            count++;               // Increment the digit counter
15
16            // Insert a period after every third digit from the right, but only if more digits are left to process
17            if (count == 3 && n != 0) { 
18                formattedNumber += '.';
19                count = 0;         // Reset the digit counter after inserting a period
20            }
21        } while (n != 0);          // Continue as long as there are digits left
22
23        // Since the digits were added in reverse order, reverse the string to get the correct format
24        reverse(formattedNumber.begin(), formattedNumber.end());
25      
26        return formattedNumber;   // Return the properly formatted number
27    }
28};
29
1/**
2 * Function to add a thousand separator in the given integer.
3 * It inserts a period '.' every three digits from right to left.
4 * @param {number} n - The number in which the thousand separator must be added.
5 * @returns {string} - The number as a string with thousand separators added.
6 */
7function thousandSeparator(n: number): string {
8    let count = 0;               // Initialize a counter to keep track of the number of digits.
9    let formattedNumber = '';    // Initialize an empty string to build the result.
10
11    // Proceed to iterate until the entire number has been processed.
12    do {
13        const digit = n % 10;    // Extract the rightmost digit.
14        n = Math.floor(n / 10);  // Remove the rightmost digit from the number.
15        formattedNumber += digit.toString(); // Append the digit to the result string.
16        count++;                 // Increment the digit counter.
17
18        // Insert a period after every third digit from the right, but only if more digits are left to process.
19        if (count === 3 && n !== 0) {
20            formattedNumber += '.';
21            count = 0;           // Reset the digit counter after inserting a period.
22        }
23    } while (n !== 0);            // Continue as long as there are digits left.
24
25    // Since the digits were added in reverse order, reverse the string to get the correct format.
26    formattedNumber = reverseString(formattedNumber);
27
28    return formattedNumber; // Return the properly formatted number.
29}
30
31/**
32 * Helper function to reverse a string.
33 * @param {string} str - The string to be reversed.
34 * @returns {string} - The reversed string.
35 */
36function reverseString(str: string): string {
37    return str.split('').reverse().join('');
38}
39
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Time and Space Complexity

The time complexity of the provided code is O(d), where d is the number of digits in the integer n. This is because the while loop runs once for each digit of n until n becomes 0, performing a constant amount of work inside the loop for each digit (division, modulo, and counter operations).

The space complexity is also O(d), as the main additional space used is the list ans which stores each digit as a character. In the worst case, for every three digits, there is an additional period character, resulting in d/3 period characters. Thus, the total space used is proportional to the number of digits d.

Learn more about how to find time and space complexity quickly using problem constraints.

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