Thousand Separator

Thousand Separator - Problem

String Easy

Given an integer n, add a dot (".") as the thousands separator and return it in string format.

The thousands separator should be placed every three digits from the right.

Examples:

1234 becomes "1.234"
987654321 becomes "987.654.321"
123 becomes "123" (no separator needed)

Input & Output

Example 1 — Basic Case

$ Input: n = 987

› Output: "987"

💡 Note: Number has 3 digits, so no separator needed

Example 2 — Four Digits

$ Input: n = 1234

› Output: "1.234"

💡 Note: Insert dot after first digit: 1 and 234

Example 3 — Large Number

$ Input: n = 123456789

› Output: "123.456.789"

💡 Note: Two dots needed: 123, 456, and 789 are each 3-digit groups

Constraints

0 ≤ n ≤ 2³¹ - 1

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 10

The key insight is to work from left to right, inserting dots when the remaining digits form groups of 3. The optimal approach uses direct construction with position calculation. Time: O(n), Space: O(n)

Common Approaches

✓ Direct Construction

⏱️ Time: O(n) Space: O(n)

Calculate exactly where dots should be placed based on the length of the number, then build the result string in a single pass without needing to reverse.

String Manipulation

⏱️ Time: O(n) Space: O(n)

Convert the number to a string, then iterate from right to left, inserting a dot every three digits. This approach directly manipulates the string representation.

Direct Construction — Algorithm Steps

Calculate total result length including dots
Iterate through original digits
Insert dots at calculated positions
Build result efficiently

Visualization

Tap to expand

Step-by-Step Walkthrough

Calculate Positions

Determine where dots should go

Single Pass

Build result left to right

Final Result

Complete formatted string

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(int n) {
    char temp[20];
    sprintf(temp, "%d", n);
    int length = strlen(temp);
    
    if (length <= 3) {
        char* result = (char*)malloc((length + 1) * sizeof(char));
        strcpy(result, temp);
        return result;
    }
    
    // Calculate result length
    int dots = (length - 1) / 3;
    char* result = (char*)malloc((length + dots + 1) * sizeof(char));
    
    int pos = 0;
    for (int i = 0; i < length; i++) {
        // Add dot if not at start and remaining digits divisible by 3
        int remaining = length - i;
        if (i > 0 && remaining % 3 == 0) {
            result[pos++] = '.';
        }
        result[pos++] = temp[i];
    }
    result[pos] = '\0';
    
    return result;
}

int main() {
    int n;
    scanf("%d", &n);
    
    char* result = solution(n);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the digits with constant time operations

✓ Linear Growth

Space Complexity

O(n)

Space for the result string only

⚡ Linearithmic Space

28.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Construction — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler