Method Chaining - Practice Coding Problems

Method Chaining - Problem

Databse Easy

Given a DataFrame animals with columns name, species, age, and weight, write a solution to list the names of animals that weigh strictly more than 100 kilograms.

Return the animals sorted by weight in descending order.

Input & Output

Example 1 — Basic Filtering

$ Input: animals = [{"name":"Lion","species":"Panthera leo","age":5,"weight":80},{"name":"Elephant","species":"Loxodonta africana","age":10,"weight":150}]

› Output: ["Elephant"]

💡 Note: Only Elephant weighs more than 100kg (150 > 100), so return ["Elephant"]

Example 2 — Multiple Heavy Animals

$ Input: animals = [{"name":"Rhino","species":"Rhinoceros","age":8,"weight":120},{"name":"Hippo","species":"Hippopotamus","age":6,"weight":180},{"name":"Elephant","species":"Elephantidae","age":12,"weight":150}]

› Output: ["Hippo","Elephant","Rhino"]

💡 Note: All three animals weigh > 100kg. Sorted by weight descending: Hippo(180), Elephant(150), Rhino(120)

Example 3 — No Heavy Animals

$ Input: animals = [{"name":"Tiger","species":"Panthera tigris","age":4,"weight":90},{"name":"Lion","species":"Panthera leo","age":3,"weight":85}]

› Output: []

💡 Note: No animals weigh more than 100kg (Tiger: 90kg, Lion: 85kg), so return empty array

Constraints

1 ≤ animals.length ≤ 10⁴
1 ≤ weight ≤ 1000
name contains only letters and spaces

Visualization

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Understanding the Visualization

Input

DataFrame with animal records

Filter

Select animals with weight > 100kg

Sort & Extract

Sort by weight descending, return names

Key Takeaway

🎯 Key Insight: Use DataFrame method chaining to filter, sort, and select in one expression

Asked in

M Meta 25 G Google 20 a Amazon 15

The key insight is to use DataFrame operations for efficient filtering and sorting. Filter animals with weight > 100, sort by weight descending, then select names. Best approach uses method chaining for readable code. Time: O(n log n), Space: O(k)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Iteration	O(n log n)	O(k)	Manually iterate through each row, collect heavy animals, then sort
Method Chaining	O(n log n)	O(k)	Use DataFrame operations to filter and sort in a single chained expression

Brute Force Iteration — Algorithm Steps

Step 1: Create empty result list
Step 2: Loop through all animals
Step 3: Check weight > 100 condition
Step 4: Sort by weight descending

Visualization

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Step-by-Step Walkthrough

Filter

Check each animal's weight > 100

Collect

Gather heavy animals

Sort

Sort by weight descending

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <cjson/cJSON.h>

int compare_weights(const void *a, const void *b) {
    cJSON *animalA = *(cJSON**)a;
    cJSON *animalB = *(cJSON**)b;
    int weightA = cJSON_GetObjectItem(animalA, "weight")->valueint;
    int weightB = cJSON_GetObjectItem(animalB, "weight")->valueint;
    return weightB - weightA; // Descending order
}

char* solution(char* animals_json) {
    cJSON *json = cJSON_Parse(animals_json);
    int size = cJSON_GetArraySize(json);
    
    cJSON **heavy_animals = malloc(size * sizeof(cJSON*));
    int count = 0;
    
    for (int i = 0; i < size; i++) {
        cJSON *animal = cJSON_GetArrayItem(json, i);
        int weight = cJSON_GetObjectItem(animal, "weight")->valueint;
        if (weight > 100) {
            heavy_animals[count++] = animal;
        }
    }
    
    // Sort by weight descending
    qsort(heavy_animals, count, sizeof(cJSON*), compare_weights);
    
    // Create result array
    cJSON *result = cJSON_CreateArray();
    for (int i = 0; i < count; i++) {
        char *name = cJSON_GetObjectItem(heavy_animals[i], "name")->valuestring;
        cJSON_AddItemToArray(result, cJSON_CreateString(name));
    }
    
    char *result_str = cJSON_Print(result);
    
    free(heavy_animals);
    cJSON_Delete(json);
    cJSON_Delete(result);
    
    return result_str;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    char* result = solution(line);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

One pass to filter (O(n)) plus sorting filtered results (O(k log k) where k ≤ n)

⚡ Linearithmic

Space Complexity

O(k)

Space for filtered results list containing k animals where k ≤ n

✓ Linear Space

31.5K Views

Medium Frequency

~8 min Avg. Time

850 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Iteration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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