DataFrame animals
+-------------+--------+
| Column Name | Type |
+-------------+--------+
| name | object |
| species | object |
| age | int |
| weight | int |
+-------------+--------+
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.
The result format is in the following example.
Example 1:
Input: DataFrame animals: +----------+---------+-----+--------+ | name | species | age | weight | +----------+---------+-----+--------+ | Tatiana | Snake | 98 | 464 | | Khaled | Giraffe | 50 | 41 | | Alex | Leopard | 6 | 328 | | Jonathan | Monkey | 45 | 463 | | Stefan | Bear | 100 | 50 | | Tommy | Panda | 26 | 349 | +----------+---------+-----+--------+ Output: +----------+ | name | +----------+ | Tatiana | | Jonathan | | Tommy | | Alex | +----------+ Explanation: All animals weighing more than 100 should be included in the results table. Tatiana's weight is 464, Jonathan's weight is 463, Tommy's weight is 349, and Alex's weight is 328. The results should be sorted in descending order of weight.
In Pandas, method chaining enables us to perform operations on a DataFrame without breaking up each operation into a separate line or creating multiple temporary variables.
Can you complete this task in just one line of code using method chaining?
The key idea behind Method Chaining is to design methods so that multiple operations can be executed in a single statement. This is typically achieved by returning the current object instance (this) from each method. When a method returns the same object, the next method can be immediately invoked on that returned object, forming a chain of calls.
To solve this problem, structure your class so that every operation updates the internal state and then returns the object itself. This enables expressions like obj.method1().method2().method3(). The main challenge is ensuring each method performs the correct update before returning the instance.
Since each method performs a constant amount of work and simply returns the instance, the operations run in O(1) time. The solution also uses O(1) space because no additional data structures are required beyond the object's internal state.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Return Object Instance for Chaining | O(1) per method call | O(1) |
NeetCode
This approach leverages a hash map (or dictionary) to solve the problem efficiently. By taking advantage of the average O(1) time complexity for insert and lookup operations in hash maps, we can create a mapping between elements and their indices (or frequencies, depending on the problem requirements). This method not only offers efficient retrieval but also makes it easier to track elements as we iterate through the data structure.
Time Complexity: O(n), where n is the number of elements in the input list.
Space Complexity: O(n), due to the storage of numbers in the hash map.
1import java.util.HashMap;
2import java.util.Map;
3
4public int[] twoSum(int[] nums, intThis Java solution uses a HashMap to store elements and their indices. During each iteration, the complement is calculated, and the map is checked for its presence. If the complement is found, the indices are returned.
This approach utilizes a two-pointer technique which is particularly effective when the input is sorted (or can be sorted) without significantly impacting performance. By using two pointers to traverse the array from both ends, we can efficiently find the pair of elements that sum to the target. Note that this approach is based on the assumption that sorting the input is feasible and will not exceed time limits.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) because we store tuples of indices and numbers.
1using System;
2using System.Linq;
public class Solution {
public int[] TwoSum(int[] nums, int target) {
var numsWithIndex = nums.Select((num, index) => new { num, index }).ToArray();
Array.Sort(numsWithIndex, (a, b) => a.num.CompareTo(b.num));
int left = 0, right = numsWithIndex.Length - 1;
while (left < right) {
int sum = numsWithIndex[left].num + numsWithIndex[right].num;
if (sum == target) {
return new int[] { numsWithIndex[left].index, numsWithIndex[right].index };
} else if (sum < target) {
left++;
} else {
right--;
}
}
return new int[] {};
}
}Watch expert explanations and walkthroughs
Jot down your thoughts, approach, and key learnings
Yes, method chaining concepts can appear in coding interviews, especially in object-oriented design questions. Interviewers may test whether candidates understand returning object instances and designing fluent interfaces. It is also common in real-world libraries and APIs.
Method chaining usually does not require a special data structure. Instead, it relies on object-oriented design where the class maintains internal state and each method returns the object instance. The internal structure depends on the operations the class performs.
Returning 'this' allows the current object instance to be reused for the next method call in the chain. Without returning the instance, the chain would break after the first method call. This design pattern improves readability and fluent API design.
The optimal approach is to design each method so it performs its operation and then returns the same object instance, typically using 'this'. This allows multiple methods to be called sequentially in a single statement. It keeps the code concise and readable.
In this C# solution, the numbers and their indices are paired using LINQ and then sorted. Two pointers are then managed to detect the target pair.