What is the time complexity of Query Kth Smallest Trimmed Number?

The optimal solution has a time complexity of O(L*N*D + Q).

What is the space complexity of Query Kth Smallest Trimmed Number?

The space complexity is O(N*L).

What data structures are used to solve Query Kth Smallest Trimmed Number?

This problem uses Array, String, Divide and Conquer, Sorting, Heap (Priority Queue), Radix Sort, Quickselect. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Query Kth Smallest Trimmed Number?

Insight 1: Radix sort can be efficiently used to sort the strings based on their trimmed values because the strings consist of digits only. Using memoization/caching of the sorted indices for each trim length avoids redundant computation. Insight 2: The sorting needs to be stable (i.e., preserve the original order for equal elements) to correctly handle the tie-breaking condition based on indices.

Which companies ask LeetCode 2343. Query Kth Smallest Trimmed Number in interviews?

This problem is frequently asked at DE Shaw, Goldman Sachs.

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Query Kth Smallest Trimmed Number - LeetCode 2343 Solution

Query Kth Smallest Trimmed Number - Complete Solution Guide

Query Kth Smallest Trimmed Number is LeetCode problem 2343, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed array of strings nums , where each string is of equal length and consists of only digits. You are also given a 0-indexed 2D integer array queries where queries[i] = [k i , trim i ] . For each queries[i] , you need to: Trim each number in nums to its rightmost trim i digits. Determine the index of the k i th smallest trimmed number in nums . If two trimmed numbers are equal, the number with the lower index is considered to be smaller. Reset each number in nums to its ori

Detailed Explanation

The problem requires you to process a series of queries on an array of strings (`nums`). Each query asks for the index of the k-th smallest number in `nums` after trimming each string to a certain number of rightmost digits. The trimming involves removing digits from the left until only the specified number of rightmost digits remain. The comparison of trimmed numbers considers leading zeros (e.g., '02' is considered as 2), and in case of ties, the smaller index is considered smaller. The problem needs to return the index of the k-th smallest trimmed number for each query.

Solution Approach

The provided solutions use radix sort to determine the order of indices of the strings after trimming to various lengths. A memoization technique is employed to store the sorted indices for different trim lengths. This memoized information is then used to quickly answer each query without recomputing the sorted order. The algorithm iterates through all possible trim lengths, and within each iteration, performs radix sort (implemented using bucket sort) on the strings trimmed to that length. The resulting sorted indices are stored in a dictionary/map (`memo`) keyed by the trim length. Finally, the solution iterates through each query, retrieves the corresponding sorted indices from `memo`, and returns the k-th smallest index.

Step-by-Step Algorithm

Step 1: Initialize a list of indices from 0 to n-1, representing the initial order of the strings in `nums`.
Step 2: Iterate through all possible trim lengths from 1 to the length of the strings.
Step 3: For each trim length, create buckets (lists) for each digit (0-9).
Step 4: Iterate through the current list of indices. For each index, extract the digit at the corresponding position in the trimmed string and place the index into the corresponding bucket.
Step 5: Concatenate the buckets to obtain a new sorted list of indices. This step performs one pass of radix sort.
Step 6: Update the list of indices with the new sorted list.
Step 7: Store the sorted list of indices in the `memo` dictionary/map, keyed by the trim length.
Step 8: Iterate through the queries. For each query, retrieve the sorted indices corresponding to the trim length from `memo` and return the k-th smallest index from the list (k-1 index).
Step 9: Return the list of answers for all queries.

Key Insights

Insight 1: Radix sort can be efficiently used to sort the strings based on their trimmed values because the strings consist of digits only. Using memoization/caching of the sorted indices for each trim length avoids redundant computation.
Insight 2: The sorting needs to be stable (i.e., preserve the original order for equal elements) to correctly handle the tie-breaking condition based on indices.
Insight 3: Since the problem constraints mention the possibility of leading zeros, you don't explicitly need to handle them, as string comparison already considers them correctly. Radix sort ensures stability required by the problem.

Complexity Analysis

Time Complexity: O(L*N*D + Q)

Space Complexity: O(N*L)

Topics

This problem involves: Array, String, Divide and Conquer, Sorting, Heap (Priority Queue), Radix Sort, Quickselect.

Companies

Asked at: DE Shaw, Goldman Sachs.

Query Kth Smallest Trimmed Number | LeetCode 2343

Medium

Array String Divide and Conquer Sorting Heap (Priority Queue)Radix Sort Quickselect

Time: O(L*N*D + Q)Space: O(N*L)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] smallestTrimmedNumbers(String[] nums, int[][] queries) {
        int n_nums = nums.length;
        int num_len = nums[0].length();
        
        java.util.Map<Integer, List<Integer>> memo = new java.util.HashMap<>();
        
        List<Integer> indices = new ArrayList<>();
        for (int i = 0; i < n_nums; i++) {
            indices.add(i);
        }

        for (int trim_len = 1; trim_len <= num_len; trim_len++) {
            List<List<Integer>> buckets = new ArrayList<>();
            for (int i = 0; i < 10; i++) {
                buckets.add(new ArrayList<>());
            }
            
            int digit_pos = num_len - trim_len;
            for (int original_idx : indices) {
                int digit = nums[original_idx].charAt(digit_pos) - '0';
                buckets.get(digit).add(original_idx);
            }
            
            List<Integer> sorted_indices = new ArrayList<>();
            for (List<Integer> bucket : buckets) {
                sorted_indices.addAll(bucket);
            }
            
            indices = sorted_indices;
            memo.put(trim_len, new ArrayList<>(indices));
            
        }
        
        int[] answer = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int k = queries[i][0];
            int trim = queries[i][1];
            answer[i] = memo.get(trim).get(k - 1);
        }
        
        return answer;
    }
}

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Problem Info

Description

Problem 2343: Query Kth Smallest Trimmed Number - Detailed explanation and solution approaches.

Topics

Array String Divide and Conquer Sorting Heap (Priority Queue)Radix Sort Quickselect

Companies

DE ShawGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2343

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize a list of indices from 0 to n-1, representing the initial order of the strings in `nums`.
2
Step 2: Iterate through all possible trim lengths from 1 to the length of the strings.
3
Step 3: For each trim length, create buckets (lists) for each digit (0-9).
4
Step 4: Iterate through the current list of indices. For each index, extract the digit at the corresponding position in the trimmed string and place the index into the corresponding bucket.
5
Step 5: Concatenate the buckets to obtain a new sorted list of indices. This step performs one pass of radix sort.
6
Step 6: Update the list of indices with the new sorted list.
7
Step 7: Store the sorted list of indices in the `memo` dictionary/map, keyed by the trim length.
8
Step 8: Iterate through the queries. For each query, retrieve the sorted indices corresponding to the trim length from `memo` and return the k-th smallest index from the list (k-1 index).
9
Step 9: Return the list of answers for all queries.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(L*N*D + Q)

💾

Space Complexity

O(N*L)

📝

Why?

The time complexity is derived from the nested loops in the radix sort implementation and the query processing. The outer loop iterates L times. The inner loop iterates through N strings. Retrieving element from memo takes O(1). Space complexity comes from storing sorted indices for all possible trim values of all numbers.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting with Custom Comparator

Instead of using radix sort, one can sort a list of (trimmed_string, index) tuples for each query using a custom comparator that takes into account string comparison and index-based tie-breaking. This involves creating trimmed strings for each number in `nums` for each query.

Time: O(Q * N * log N * L), where Q is the number of queries, N is the number of strings, and L is the maximum length of strings. Sorting each set of trimmed strings takes O(N log N) time, and creating the trimmed strings takes O(N*L) time per query.Space: O(N * L), where N is the number of strings and L is the maximum length of the strings, for storing trimmed strings.

Trade-offs: This approach is simpler to implement than radix sort but has a higher time complexity, especially when the number of queries is large. Use this when radix sort implementation is undesired, and query count is low.

✨

Key Insights

1Insight 1: Radix sort can be efficiently used to sort the strings based on their trimmed values because the strings consist of digits only. Using memoization/caching of the sorted indices for each trim length avoids redundant computation.
2Insight 2: The sorting needs to be stable (i.e., preserve the original order for equal elements) to correctly handle the tie-breaking condition based on indices.
3Insight 3: Since the problem constraints mention the possibility of leading zeros, you don't explicitly need to handle them, as string comparison already considers them correctly. Radix sort ensures stability required by the problem.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty nums or queries array: The given constraints ensure that nums and queries will not be empty. The code gracefully handles the empty cases by initializing the required data structures.
Edge case 2: k > nums.length: The constraints state that 1 <= k <= nums.length, so k will always be a valid index within the sorted indices list.
Edge case 3: trim == 0: The constraint 1 <= trim <= nums[i].length prevents the possibility of trim equals zero. If such a trim value were possible, we should handle it by considering the value zero as a special case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the radix sort, leading to instability. The stability of the sort is crucial to handling ties based on the original index.

✗Common mistake 2: Not memoizing the sorted indices for different trim lengths, leading to redundant computation and a higher time complexity of O(Q * N * L * log N), as sorting is repeatedly done for each query.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Sorting log files based on date/time stamps after extracting relevant parts from each log entry. Here, trimming could be used to extract specific parts of the date/time stamp.

🏦Real-world scenario 2: Ranking search results based on the relevance score. Trimming could be used to focus on the most significant digits/characters of the relevance score for sorting and ranking.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Query Kth Smallest Trimmed Number"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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