What is the time complexity of Delete Columns to Make Sorted?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Delete Columns to Make Sorted?

This problem uses Array, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Delete Columns to Make Sorted?

Insight 1: We need to iterate through each column of the grid independently to check if it's sorted. Insight 2: To check if a column is sorted, we compare adjacent characters within the column. If we find any adjacent characters where the lower one is lexicographically smaller than the upper one, the column is not sorted.

Which companies ask LeetCode 944. Delete Columns to Make Sorted in interviews?

This problem is frequently asked at Garmin.

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Delete Columns to Make Sorted - LeetCode 944 Solution

Delete Columns to Make Sorted - Complete Solution Guide

Delete Columns to Make Sorted is LeetCode problem 944, a Easy 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 an array of n strings strs , all of the same length. The strings can be arranged such that there is one on each line, making a grid. For example, strs = ["abc", "bce", "cae"] can be arranged as follows: abc bce cae You want to delete the columns that are not sorted lexicographically . In the above example ( 0-indexed ), columns 0 ( 'a' , 'b' , 'c' ) and 2 ( 'c' , 'e' , 'e' ) are sorted, while column 1 ( 'b' , 'c' , 'a' ) is not, so you would delete column 1. Return the number of co

Detailed Explanation

The problem asks us to analyze a grid of characters formed by an array of strings, all of the same length. We need to determine the number of columns that are *not* sorted lexicographically (i.e., in ascending order). A column is considered sorted if, reading from top to bottom, the characters in that column are in non-decreasing order. The goal is to count how many columns need to be deleted to make the remaining columns sorted.

Solution Approach

The solution iterates through each column of the input array of strings. For each column, it iterates through the rows, comparing adjacent characters to check if they are in non-decreasing order. If it finds a pair of characters that are not in order, it increments a counter (representing the number of columns to delete) and breaks out of the inner loop (because it's confirmed that the column is not sorted). Finally, it returns the total count of columns to delete.

Step-by-Step Algorithm

Step 1: Initialize a counter `ans` (or `count`) to 0. This will store the number of columns to delete.
Step 2: Iterate through the columns of the input array `strs` using a loop (from `j = 0` to `length of strs[0] - 1`).
Step 3: For each column `j`, iterate through the rows of `strs` using another loop (from `i = 1` to `length of strs - 1`).
Step 4: Inside the inner loop, compare `strs[i][j]` with `strs[i - 1][j]`. If `strs[i][j] < strs[i - 1][j]`, it means the column is not sorted.
Step 5: If the column is not sorted, increment the `ans` (or `count`) and break out of the inner loop (because we've already determined that the column needs to be deleted).
Step 6: After iterating through all the columns, return the final value of `ans` (or `count`).

Key Insights

Insight 1: We need to iterate through each column of the grid independently to check if it's sorted.
Insight 2: To check if a column is sorted, we compare adjacent characters within the column. If we find any adjacent characters where the lower one is lexicographically smaller than the upper one, the column is not sorted.
Insight 3: The problem requires a simple counting process, incrementing the count only if a column is found to be unsorted.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: Array, String.

Companies

Asked at: Garmin.

Delete Columns to Make Sorted | LeetCode 944

Easy

Array String

Time: O(n*m)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minDeletionSize(String[] strs) {
        int n = strs.length;
        int len = strs[0].length();
        int count = 0;
        for (int j = 0; j < len; j++) {
            for (int i = 1; i < n; i++) {
                if (strs[i].charAt(j) < strs[i - 1].charAt(j)) {
                    count++;
                    break;
                }
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 944: Delete Columns to Make Sorted - Detailed explanation and solution approaches.

Topics

Array String

Companies

Garmin

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:944

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

6 steps

1
Step 1: Initialize a counter `ans` (or `count`) to 0. This will store the number of columns to delete.
2
Step 2: Iterate through the columns of the input array `strs` using a loop (from `j = 0` to `length of strs[0] - 1`).
3
Step 3: For each column `j`, iterate through the rows of `strs` using another loop (from `i = 1` to `length of strs - 1`).
4
Step 4: Inside the inner loop, compare `strs[i][j]` with `strs[i - 1][j]`. If `strs[i][j] < strs[i - 1][j]`, it means the column is not sorted.
5
Step 5: If the column is not sorted, increment the `ans` (or `count`) and break out of the inner loop (because we've already determined that the column needs to be deleted).
6
Step 6: After iterating through all the columns, return the final value of `ans` (or `count`).

💡

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(n*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the nested loops that iterate through the grid of characters. The outer loop iterates `m` times, and the inner loop iterates `n-1` times. Therefore, the time complexity is proportional to `n*m`. The space complexity is constant because we are only using a fixed number of variables to store the count and loop indices.

📖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

Early Exit Optimization

A slight optimization is to keep track of whether all characters are equal in a column so far. If they are, we don't need to check the remaining characters. However, this only provides a very slight performance improvement in some cases.

Time: O(n*m) in the worst case, but potentially faster if there are many columns with early exits.Space: O(1)

Trade-offs: This approach might provide a slight performance improvement in some test cases, but the overall time complexity remains the same. It may not be worth the extra complexity.

✨

Key Insights

1Insight 1: We need to iterate through each column of the grid independently to check if it's sorted.
2Insight 2: To check if a column is sorted, we compare adjacent characters within the column. If we find any adjacent characters where the lower one is lexicographically smaller than the upper one, the column is not sorted.
3Insight 3: The problem requires a simple counting process, incrementing the count only if a column is found to be unsorted.

📚

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!

Edge Cases

Edge case 1: Input array `strs` is empty. The code handles this implicitly because the outer loops won't execute. It will return 0 since the count is initialized to 0.
Edge case 2: All strings in `strs` have length 1. The code handles this correctly because the inner loop will run only once and simply compare adjacent characters in each column.
Edge case 3: The strings are already sorted. The algorithm correctly returns 0 in this case, as the count won't be incremented.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that if *any* character in a column is out of order, the entire column is sorted. It's crucial to check for *every* adjacent pair of characters.

✗Common mistake 2: Failing to break out of the inner loop once a column is determined to be unsorted. Continuing to iterate through the inner loop after finding an unsorted pair is unnecessary and inefficient.

✅

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: Data validation. Imagine a system that stores data in columnar format. This algorithm could be used to check if certain columns are sorted according to a defined order. If a column is not sorted, it could indicate an issue with the data or the sorting process.

🏦Real-world scenario 2: Lexicographical sorting of tabular data. In database management systems, ensuring data integrity and correct sorting based on multiple columns is crucial. This problem represents a simplified version of validating column-wise lexicographical order.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: July 15, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Delete Columns to Make Sorted"?

For this easy-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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