Thousand Separator - Practice Coding Problems

Thousand Separator - Problem

String Easy

You're given an integer n and need to format it as a human-readable number by adding dots (".") as thousand separators. This is similar to how large numbers are displayed in many European countries or financial systems.

For example, the number 1234567 should become "1.234.567" - much easier to read at a glance!

Goal: Transform any integer into a properly formatted string with dots separating every three digits from right to left.

Examples:
• n = 987 → "987" (no separator needed)
• n = 1234 → "1.234"
• n = 123456789 → "123.456.789"

Input & Output

example_1.py — Python

$ Input: n = 987

› Output: "987"

💡 Note: The number is less than 1000, so no thousand separator is needed. We simply return the number as a string.

example_2.py — Python

$ Input: n = 1234

› Output: "1.234"

💡 Note: We need to separate the thousands: 1 (thousand) and 234 (remainder), resulting in '1.234'.

example_3.py — Python

$ Input: n = 123456789

› Output: "123.456.789"

💡 Note: This large number needs multiple separators: 123 (millions), 456 (thousands), and 789 (remainder).

Constraints

0 ≤ n ≤ 2³¹ - 1
n is a non-negative integer
The result should use dot (.) as separator

Visualization

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Understanding the Visualization

Start with Number

Take the input number 1234567

Convert to String

Transform to string '1234567' for easy manipulation

Group from Right

Count positions from right: 567 (pos 0-2), 234 (pos 3-5), 1 (pos 6)

Insert Separators

Add dots between groups: 1.234.567

Key Takeaway

🎯 Key Insight: Process digits from right to left and insert separators every 3 positions to create readable number formatting

Asked in

G Google 12 a Amazon 8 ⊞ Microsoft 6 Apple 4

The optimal approach is to convert the number to a string and process it from right to left, inserting dots every three digits. This requires O(n) time complexity where n is the number of digits, making it both simple and efficient for the given constraints.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Character by Character)	O(n)	O(n)	Convert to string and manually insert separators by counting positions

Brute Force (Character by Character) — Algorithm Steps

Convert the integer to a string
Iterate through the string from right to left
For every group of 3 digits, insert a dot separator
Handle the leftmost group which might have fewer than 3 digits
Return the formatted string

Visualization

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Step-by-Step Walkthrough

Convert to String

Transform number 1234567 to string '1234567'

Process Right to Left

Start from rightmost digit, count positions

Insert Separators

Add dots at every 3rd position: 1.234.567

Code -

solution.c — C

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

char* thousandSeparator(int n) {
    char s[20];
    sprintf(s, "%d", n);
    int len = strlen(s);
    
    char* result = (char*)malloc(30 * sizeof(char));
    int idx = 0;
    
    for (int i = 0; i < len; i++) {
        int pos = len - 1 - i;
        if (i > 0 && i % 3 == 0) {
            memmove(result + 1, result, idx);
            result[0] = '.';
            idx++;
        }
        memmove(result + 1, result, idx);
        result[0] = s[pos];
        idx++;
    }
    
    result[idx] = '\0';
    return result;
}

int main() {
    int n;
    scanf("%d", &n);
    printf("%s\n", thousandSeparator(n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Where n is the number of digits - we process each digit once

✓ Linear Growth

Space Complexity

O(n)

We create new strings to store the result

⚡ Linearithmic Space

28.5K Views

Medium Frequency

~8 min Avg. Time

842 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Character by Character) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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