Convert Object to JSON String

Convert Object to JSON String - Problem

JavaScript Medium

Given a value, return a valid JSON string representation of that value. The value can be a string, number, array, object, boolean, or null.

The returned string should not include extra spaces and the order of keys should be the same as the order returned by Object.keys().

Important: You cannot use the built-in JSON.stringify method.

Input & Output

Example 1 — Basic String

$ Input: value = "hello"

› Output: "\"hello\""

💡 Note: String values need to be wrapped in quotes and escaped properly

Example 2 — Simple Array

$ Input: value = [1, 2, 3]

› Output: [1,2,3]

💡 Note: Array elements are converted recursively and joined with commas, no spaces

Example 3 — Object with Mixed Types

$ Input: value = {"name": "John", "age": 30, "active": true}

› Output: {"name":"John","age":30,"active":true}

💡 Note: Object keys and values are converted recursively, maintaining key order. The result is a JSON object format without outer quotes.

Constraints

The value can be any valid JavaScript data type
Objects will have at most 1000 keys
Arrays will have at most 1000 elements
Maximum nesting depth is 100 levels
String values will not exceed 10000 characters

Visualization

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Asked in

G Google 25 f Meta 20 a Amazon 18 M Microsoft 15

The key insight is to use recursion to naturally handle nested structures. The recursive approach elegantly processes each data type and calls itself for nested objects and arrays. Time: O(n), Space: O(d) where d is nesting depth.

Common Approaches

✓ Brute Force Type Checking

⏱️ Time: O(n) Space: O(1)

Check the type of value using typeof and instanceof, then manually format each type with string concatenation. Handle objects and arrays with separate loops.

Recursive Approach

⏱️ Time: O(n) Space: O(d)

Create a recursive function that calls itself for nested objects and arrays. This naturally handles any depth of nesting and reduces code duplication.

Brute Force Type Checking — Algorithm Steps

Step 1: Check if value is null, boolean, number, or string
Step 2: Handle arrays by iterating and converting each element
Step 3: Handle objects by iterating keys and converting each value

Visualization

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Step-by-Step Walkthrough

Type Check

Determine if input is null, boolean, number, string, array, or object

Format

Apply specific formatting rules for each type

Result

Return properly formatted JSON string

Code -

solution.c — C

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

bool areAlmostEqual(int num1, int num2) {
    char s1[20], s2[20];
    sprintf(s1, "%d", num1);
    sprintf(s2, "%d", num2);
    
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    
    // Check if s1 can become s2 with one swap
    for (int i = 0; i < len1; i++) {
        for (int j = i + 1; j < len1; j++) {
            char temp = s1[i];
            s1[i] = s1[j];
            s1[j] = temp;
            if (strcmp(s1, s2) == 0) return true;
            // Swap back
            s1[j] = s1[i];
            s1[i] = temp;
        }
    }
    
    // Check if s2 can become s1 with one swap
    for (int i = 0; i < len2; i++) {
        for (int j = i + 1; j < len2; j++) {
            char temp = s2[i];
            s2[i] = s2[j];
            s2[j] = temp;
            if (strcmp(s2, s1) == 0) return true;
            // Swap back
            s2[j] = s2[i];
            s2[i] = temp;
        }
    }
    
    // Check if already equal
    return strcmp(s1, s2) == 0;
}

int solution(int* nums, int numsSize) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (areAlmostEqual(nums[i], nums[j])) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int size = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Visit each element/property once where n is total number of values

✓ Linear Growth

Space Complexity

O(1)

Only uses variables for string building, no extra data structures

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Type Checking — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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