Lucky Numbers in a Matrix

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Steps

setup

Initialize matrix dimensions

Determine the number of rows (m) and columns (n) in the matrix to prepare for processing.

💡 Knowing matrix size is essential to iterate correctly over rows and columns.

Line:

m, n = len(matrix), len(matrix[0])

💡 Establishing matrix dimensions sets the bounds for all subsequent loops.

fill_row

Compute minimum of row 0

Find the minimum value in the first row [3,7,8], which is 3.

💡 Row minimums help identify candidates for lucky numbers in each row.

Line:

row_mins = [min(row) for row in matrix]

💡 The minimum of row 0 is 3, recorded as the first element of row_mins.

fill_row

Compute minimum of row 1

Find the minimum value in the second row [9,11,13], which is 9.

💡 Continuing to find minimums for each row builds the row_mins array.

Line:

row_mins = [min(row) for row in matrix]

💡 The minimum of row 1 is 9, added as the second element of row_mins.

fill_row

Compute minimum of row 2

Find the minimum value in the third row [15,16,17], which is 15.

💡 Completing row minimums prepares for the next step of column maximums.

Line:

row_mins = [min(row) for row in matrix]

💡 The minimum of row 2 is 15, completing the row_mins array as [3,9,15].

fill_row

Compute maximum of column 0

Find the maximum value in the first column [3,9,15], which is 15.

💡 Column maximums identify candidates for lucky numbers from the column perspective.

Line:

col_maxs = [max(matrix[i][j] for i in range(m)) for j in range(n)]

💡 The maximum of column 0 is 15, recorded as the first element of col_maxs.

fill_row

Compute maximum of column 1

Find the maximum value in the second column [7,11,16], which is 16.

💡 Continuing to find maximums for each column builds the col_maxs array.

Line:

col_maxs = [max(matrix[i][j] for i in range(m)) for j in range(n)]

💡 The maximum of column 1 is 16, added as the second element of col_maxs.

fill_row

Compute maximum of column 2

Find the maximum value in the third column [8,13,17], which is 17.

💡 Completing column maximums prepares for the final lucky number search.

Line:

col_maxs = [max(matrix[i][j] for i in range(m)) for j in range(n)]

💡 The maximum of column 2 is 17, completing col_maxs as [15,16,17].

compare

Check element at (0,0): 3

Check if matrix[0][0] = 3 equals row_mins[0] = 3 and col_maxs[0] = 15.

💡 This step tests if the element is a lucky number by matching both conditions.

Line:

if matrix[i][j] == row_mins[i] == col_maxs[j]:

💡 An element must be both row minimum and column maximum to be lucky.

compare

Check element at (2,0): 15

Check if matrix[2][0] = 15 equals row_mins[2] = 15 and col_maxs[0] = 15.

💡 This element matches both conditions, so it is a lucky number.

Line:

if matrix[i][j] == row_mins[i] == col_maxs[j]:

💡 This confirms the lucky number definition and how it is found.

compare

Check remaining elements for lucky numbers

Check all other elements; none match both row minimum and column maximum conditions.

💡 Verifying all elements ensures no lucky numbers are missed.

Line:

if matrix[i][j] == row_mins[i] == col_maxs[j]:

💡 Only one lucky number exists in this matrix.

reconstruct

Return lucky numbers list

Return the list of lucky numbers found: [15].

💡 The final step outputs the result of the algorithm.

Line:

return [matrix[i][j] for i in range(m) for j in range(n) if matrix[i][j] == row_mins[i] == col_maxs[j]]

💡 The algorithm successfully identifies and returns all lucky numbers.

from typing import List

def luckyNumbers(matrix: List[List[int]]) -> List[int]:
    m, n = len(matrix), len(matrix[0])  # STEP 1: Initialize dimensions
    row_mins = [min(row) for row in matrix]  # STEP 2-4: Compute row minimums
    col_maxs = [max(matrix[i][j] for i in range(m)) for j in range(n)]  # STEP 5-7: Compute column maximums
    # STEP 8-10: Find lucky numbers by checking each element
    return [matrix[i][j] for i in range(m) for j in range(n) if matrix[i][j] == row_mins[i] == col_maxs[j]]  # STEP 11: Return result

if __name__ == '__main__':
    mat = [[3,7,8],[9,11,13],[15,16,17]]
    print(luckyNumbers(mat))  # Output: [15]

📊

Lucky Numbers in a Matrix - Watch the Algorithm Execute, Step by Step

Watching each step helps you understand how precomputing row minimums and column maximums simplifies the search for lucky numbers, avoiding repeated scanning.

setup

0

1

2

Result: []

compare

i

3

0

7

1

8

2

Result: [3]

compare

9

0

i

11

1

13

2

Result: [3,9]

compare

15

0

16

1

i

17

2

Result: [3,9,15]

compare

i

0

1

2

Result: [15]

compare

i

0

j

1

2

Result: [15,16]

compare

i

0

1

j

2

Result: [15,16,17]

compare

j

0

1

2

Result: []

record

j

0

1

i

2

Result: [15]

compare

0

1

j

2

Result: [15]

record

0

1

2

Result: [15]

Key Takeaways

✓ Precomputing row minimums and column maximums allows a single pass to identify lucky numbers efficiently.

This insight is hard to see from code alone because it requires understanding the benefit of separating concerns before the final check.

✓ The algorithm only needs to check each element once against precomputed arrays, avoiding repeated scanning.

Visualizing pointer movements clarifies how the algorithm avoids nested repeated searches.

✓ A lucky number must simultaneously be the smallest in its row and largest in its column, a rare condition clearly highlighted by the trace.

Seeing the exact comparisons and decisions helps concretize this definition beyond abstract code.

Practice

1. You are given an n x n matrix representing an image. The task is to rotate the image by 90 degrees clockwise in-place, without using extra space. Which of the following approaches guarantees an optimal in-place solution with O(1) extra space and O(n²) time complexity?

easy

A. Perform a layer-by-layer rotation by swapping four elements at a time in-place for each layer of the matrix.

B. Apply a dynamic programming approach to store intermediate rotations and build the final rotated matrix.

C. Use a brute force approach by creating a new matrix and copying elements to their rotated positions.

D. Transpose the matrix and then reverse each column to achieve the rotation.

Solution

1. Step 1: Understand the problem constraints

   The problem requires rotating the matrix in-place with O(1) extra space and O(n²) time.

2. Step 2: Evaluate approaches

   Brute force uses extra space, DP is unrelated, and transposing then reversing columns rotates counterclockwise, not clockwise. Layer-by-layer four-way swaps correctly rotate clockwise in-place.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Layer-by-layer swaps rotate in-place with correct direction and complexity [OK]

Hint: Layer-by-layer four-way swaps rotate in-place clockwise [OK]

Common Mistakes:

• Confusing transpose+reverse columns with clockwise rotation
• Using extra space when in-place is required

2. You are given a square matrix and asked to transform it so that rows become columns and columns become rows without using extra space. Which approach guarantees an optimal in-place solution?

easy

A. Use a brute force approach by creating a new matrix and copying elements accordingly.

B. Swap elements symmetrically across the diagonal only once, iterating over the upper triangle of the matrix.

C. Apply a greedy approach by swapping elements row-wise without considering the diagonal symmetry.

D. Use dynamic programming to store intermediate transposed states and build the final matrix.

Solution

1. Step 1: Understand the problem constraints

   The problem requires an in-place transpose of a square matrix, meaning no extra matrix can be used.

2. Step 2: Identify the correct approach

   Swapping elements symmetrically across the diagonal once (upper triangle only) ensures each pair is swapped exactly once, achieving in-place transpose efficiently.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Swapping only upper triangle elements avoids double swaps and extra space [OK]

Hint: Swap only upper triangle elements once [OK]

Common Mistakes:

• Using extra space unnecessarily
• Swapping elements twice by iterating entire matrix
• Ignoring diagonal symmetry leading to incorrect results

3. What is the time complexity of the optimal in-place boundary shrinking spiral traversal algorithm for an m x n matrix?

medium

A. O(m * n * min(m, n)) due to nested loops over shrinking boundaries

B. O(m + n) because each boundary shrinks linearly

C. O(m * n) because every element is visited exactly once

D. O(m * n + m + n) combining traversal and boundary updates

Solution

1. Step 1: Identify the number of elements visited

   The algorithm visits each element exactly once, so total operations scale with m*n.

2. Step 2: Analyze loops and boundary updates

   Though boundaries shrink, the total number of elements processed remains m*n, no nested multiplicative overhead beyond that.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Each element processed once -> O(m*n) [OK]

Hint: Count total elements visited, not loop nesting [OK]

Common Mistakes:

• Confusing boundary shrinking loops as nested multiplicative loops
• Ignoring that each element is visited once
• Adding linear terms incorrectly

4. Suppose the Game of Life rules are modified so that cells can be in one of three states: dead, alive, or dormant, and the next state depends on neighbors' states with more complex transitions. Which modification to the in-place encoding approach is necessary to correctly update the board?

hard

A. Use a larger integer encoding scheme with more states to represent current and next states simultaneously.

B. Switch to copying the entire board each iteration to avoid encoding complexity.

C. Update cells immediately without encoding since more states simplify neighbor counting.

D. Ignore dormant state neighbors during counting to reduce complexity.

Solution

1. Step 1: Understand the need for more states

   With three possible states and complex transitions, encoding must represent both current and next states simultaneously.

2. Step 2: Choose encoding approach

   Using a larger integer encoding scheme (e.g., 6 or more states) allows in-place updates without losing original state info.

3. Step 3: Reject naive approaches

   Copying the board increases space; immediate updates corrupt neighbor info; ignoring dormant neighbors breaks rules.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   More states require richer encoding to preserve info [OK]

Hint: More states require more encoding bits, not naive copying [OK]

Common Mistakes:

• Switching to copying board wastes space
• Immediate updates corrupt neighbor counts
• Ignoring certain neighbors breaks rules

5. Suppose the problem is modified so that the matrix is rectangular (not necessarily square). Which of the following statements about adapting the in-place rotation algorithm is true?

hard

A. The layer-by-layer four-way swap algorithm works unchanged for rectangular matrices.

B. In-place rotation is impossible for non-square matrices; extra space or a different approach is required.

C. Transposing and reversing rows can rotate rectangular matrices in-place without issues.

D. Rotating rectangular matrices by 90 degrees clockwise can be done by swapping elements only in the first row.

Solution

1. Step 1: Understand matrix shape constraints

   In-place rotation by 90 degrees requires the matrix to be square because dimensions swap (rows become columns).

2. Step 2: Evaluate options

   Layer-by-layer swaps rely on equal row and column counts; rectangular matrices require extra space or new matrix to hold rotated result.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Non-square matrices cannot be rotated in-place without dimension mismatch [OK]

Hint: In-place 90° rotation requires square matrix due to dimension swap [OK]

Common Mistakes:

• Assuming in-place rotation works for rectangular matrices
• Confusing transpose with rotation for non-square matrices