Thousand Separator - Practice Coding Problems

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

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Understanding the Visualization

Input

Integer number like 1234567

Process

Group digits by 3 from right, insert dots

Output

Formatted string "1.234.567"

Key Takeaway

🎯 Key Insight: Group digits from right to left in sets of 3, inserting separators between groups

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

Approach	Time	Space	Notes
✓ String Manipulation	O(n)	O(n)	Convert to string and insert dots by iterating from right to left
Direct Construction	O(n)	O(n)	Build result string directly by calculating dot positions mathematically

String Manipulation — Algorithm Steps

Convert integer to string
Iterate from right to left
Insert dot every 3 digits
Build result string

Visualization

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Step-by-Step Walkthrough

Convert to String

Transform number into string format

Process Right to Left

Add dots every 3 digits from the right

Build Result

Construct final 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 len = strlen(temp);
    
    // Calculate result length
    int dots = (len - 1) / 3;
    char* result = (char*)malloc((len + dots + 1) * sizeof(char));
    
    int pos = 0;
    for (int i = len - 1, count = 0; i >= 0; i--, count++) {
        if (count > 0 && count % 3 == 0) {
            result[pos++] = '.';
        }
        result[pos++] = temp[i];
    }
    result[pos] = '\0';
    
    // Reverse the result
    int resultLen = pos;
    for (int i = 0; i < resultLen / 2; i++) {
        char t = result[i];
        result[i] = result[resultLen - 1 - i];
        result[resultLen - 1 - i] = t;
    }
    
    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 of the number

✓ Linear Growth

Space Complexity

O(n)

Store the result string and temporary variables

⚡ Linearithmic Space

28.0K Views

Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

String Manipulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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