What is the brute force solution for Minimum Number of Operations to Satisfy Conditions?

The brute force approach for Minimum Number of Operations to Satisfy Conditions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each cell in the grid and making adjustments to satisfy the conditions. This method is straightforward but inefficient, as it may require multiple passes over the grid. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Satisfy Conditions?

Minimum Number of Operations to Satisfy Conditions uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: .

What are the interview tips for Minimum Number of Operations to Satisfy Conditions?

Think about how to minimize changes by leveraging existing values. Practice identifying patterns in matrix problems. Be prepared to explain your thought process clearly.

What are common mistakes when solving Minimum Number of Operations to Satisfy Conditions?

Overlooking the need to adjust for adjacent cells. Not considering the most frequent values in columns.

#3122

Minimum Number of Operations to Satisfy Conditions

Medium

ArrayDynamic ProgrammingMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the idea of grouping values in columns and ensuring that adjacent cells in rows are different. This reduces unnecessary operations by focusing on the relationships between cells rather than individual adjustments.

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Algorithm

3 steps

1Step 1: Create a frequency map for each column to count occurrences of each number.
2Step 2: For each column, determine the most frequent number and calculate the number of operations needed to make all cells in that column equal to this number.
3Step 3: For each row, ensure that adjacent cells are different by adjusting values based on the previous column's most frequent number.

solution.py12 lines

1# Full working Python code
2
3def min_operations(grid):
4    from collections import Counter
5    m, n = len(grid), len(grid[0])
6    operations = 0
7    for j in range(n):
8        count = Counter(grid[i][j] for i in range(m))
9        most_common = count.most_common(1)[0][1]
10        operations += m - most_common
11    return operations
12

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Complexity note: The time complexity is O(n) because we only need to iterate through each column once to count occurrences. The space complexity is O(n) due to the frequency map used to store counts of each number in the columns.

1Focus on column relationships to minimize operations.
2Utilize frequency counts to determine optimal adjustments.

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