LeetCode 1556: Thousand Separator Solution

Master LeetCode problem 1556 (Thousand Separator), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1556. Thousand Separator

Easy

String

Problem Explanation

Explanation:

To solve this problem, we can convert the integer to a string and then insert a dot character every 3 digits from the right. We need to handle the edge case where the number of digits is not divisible by 3.

Convert the given integer n to a string.
Initialize an empty string result.
Iterate over the characters in the string representation of n from right to left.
- Prepend the current character to result.
- If the current index is not the last character and the distance from the current index to the end of the string is divisible by 3, insert a dot character at the beginning of result.
Return the final result string.

Time Complexity:

The time complexity of this approach is O(log(n)), where n is the input integer.

Space Complexity:

The space complexity is O(log(n)) to store the string representation of the integer.

Solution Code

class Solution {
    public String thousandSeparator(int n) {
        String str = String.valueOf(n);
        StringBuilder result = new StringBuilder();
        
        for (int i = str.length() - 1, count = 0; i >= 0; i--) {
            result.insert(0, str.charAt(i));
            count++;
            if (i != 0 && count % 3 == 0) {
                result.insert(0, '.');
            }
        }
        
        return result.toString();
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 1556 (Thousand Separator)?

The time complexity for LeetCode 1556 (Thousand Separator) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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