How do you solve Flip Columns For Maximum Number of Equal Rows on LeetCode?

Flip Columns For Maximum Number of Equal Rows (LeetCode #1072) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m) space complexity. Key insight: Flipping columns can be seen as normalizing the rows based on the first column's value.

What is the brute force solution for Flip Columns For Maximum Number of Equal Rows?

The brute force approach for Flip Columns For Maximum Number of Equal Rows is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible combination of column flips and count how many rows become equal. This method is straightforward but inefficient because it explores all combinations. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Flip Columns For Maximum Number of Equal Rows?

Flip Columns For Maximum Number of Equal Rows uses the following concepts: Array, Hash Table, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Flip Columns For Maximum Number of Equal Rows?

Always think about how to reduce the problem size — here, we reduced it by normalizing rows. Practice identifying patterns in problems, such as using HashMaps for counting. Explain your thought process clearly; it helps the interviewer understand your reasoning.

What are common mistakes when solving Flip Columns For Maximum Number of Equal Rows?

Not normalizing rows correctly before counting. Overlooking the need to consider the first element for normalization.

#1072

Flip Columns For Maximum Number of Equal Rows

Medium

ArrayHash TableMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

Instead of checking all combinations, we can use a HashMap to count the frequency of each unique row pattern after normalizing them based on the first column. This way, we can directly count how many rows can be made equal with the same flips.

⚙️

Algorithm

4 steps

1Step 1: Create a HashMap to store the frequency of each unique row pattern.
2Step 2: For each row, create a normalized version by flipping it based on the first element.
3Step 3: Count the occurrences of each normalized row in the HashMap.
4Step 4: The maximum value in the HashMap is the answer.

solution.py9 lines

1from collections import defaultdict
2
3def maxEqualRowsAfterFlips(matrix):
4    count = defaultdict(int)
5    for row in matrix:
6        # Normalize row based on the first element
7        normalized = tuple((cell ^ row[0]) for cell in row)
8        count[normalized] += 1
9    return max(count.values())

ℹ

Complexity note: The time complexity is O(m * n) because we process each row and each element in the row once. The space complexity is O(m) due to the storage of the HashMap for unique row patterns.

1Flipping columns can be seen as normalizing the rows based on the first column's value.
2Using a HashMap allows us to efficiently count and compare row patterns.

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