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

Delete Columns to Make Sorted II (LeetCode #955) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(m) space complexity. Key insight: Understanding lexicographic order is crucial.

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

The brute force approach for Delete Columns to Make Sorted II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of deletion indices and checking if the remaining strings are in lexicographic order. This is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n * m).

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

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

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

Explain your thought process clearly. Consider edge cases, like strings of the same characters. Practice identifying patterns in string manipulation problems.

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

Not considering all columns for deletion. Overlooking the need to check adjacent strings.

#955

Delete Columns to Make Sorted II

Medium

ArrayStringGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal approach uses a greedy strategy to determine which columns to delete by comparing adjacent strings. If a column causes disorder, we mark it for deletion. This avoids the need for generating combinations and checking all possibilities.

⚙️

Algorithm

4 steps

1Step 1: Initialize a set to keep track of columns to delete.
2Step 2: Iterate through the strings, comparing each string with the next one.
3Step 3: For each pair of strings, check each column. If the current column causes disorder, mark it for deletion.
4Step 4: Return the size of the deletion set as the result.

solution.py12 lines

1# Full working Python code
2
3def minDeletionSize(strs):
4    to_delete = set()
5    n = len(strs)
6    m = len(strs[0])
7    for j in range(m):
8        for i in range(n - 1):
9            if strs[i][j] > strs[i + 1][j]:
10                to_delete.add(j)
11                break
12    return len(to_delete)

ℹ

Complexity note: The time complexity is O(n * m) because we iterate through each character in the strings, comparing adjacent strings, leading to a linear scan of the characters.

1Understanding lexicographic order is crucial.
2Greedy approaches can often simplify complex problems.

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