How do you solve Delete Columns to Make Sorted on LeetCode?

Delete Columns to Make Sorted (LeetCode #944) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Columns must be checked independently for sorting.

What is the brute force solution for Delete Columns to Make Sorted?

The brute force approach for Delete Columns to Make Sorted is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each column one by one to see if it is sorted. If any character in a column is greater than the character below it, that column is considered unsorted and marked for deletion. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Delete Columns to Make Sorted?

Delete Columns to Make Sorted uses the following concepts: Array, String. The recommended patterns to study are: Array, String.

What are the interview tips for Delete Columns to Make Sorted?

Explain your thought process clearly while coding. Consider edge cases, such as single-column inputs. Practice similar problems to build confidence.

What are common mistakes when solving Delete Columns to Make Sorted?

Not breaking out of the loop after finding an unsorted column. Confusing row and column indices.

#944

Delete Columns to Make Sorted

Easy

ArrayStringArrayString

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses the same logic as the brute force but focuses on reducing unnecessary checks by breaking early when an unsorted column is found. This approach is efficient and avoids redundant comparisons.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter for unsorted columns.
2Step 2: Loop through each column index from 0 to the length of the first string.
3Step 3: For each column, loop through each row index from 0 to n-2.
4Step 4: If the current character is greater than the next character, increment the counter and break out of the inner loop.
5Step 5: Return the counter after checking all columns.

solution.py10 lines

1# Full working Python code
2
3def minDeletionSize(strs):
4    unsorted_count = 0
5    for col in range(len(strs[0])):
6        for row in range(len(strs) - 1):
7            if strs[row][col] > strs[row + 1][col]:
8                unsorted_count += 1
9                break
10    return unsorted_count

ℹ

Complexity note: The time complexity is O(n) because we are only iterating through each column once and breaking early when we find an unsorted column. The space complexity remains O(1) as we are using a constant amount of extra space.

1Columns must be checked independently for sorting.
2Breaking early can save unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.