Search in 2D Matrix in DSA Python - Build from Scratch

Search in 2D Matrix

📖 Scenario: You are working with a grid of numbers, like a map of heights or temperatures arranged in rows and columns. You want to find out if a certain number exists in this grid.

🎯 Goal: Build a program that searches for a specific number in a 2D matrix and tells if it is found or not.

📋 What You'll Learn

Create a 2D list called matrix with exact values

Create a variable called target with the exact number to search

Use nested for loops with variables row and col to search the matrix

Print Found if the target number is in the matrix, otherwise print Not Found

💡 Why This Matters

🌍 Real World

Searching in 2D grids is common in games, maps, image processing, and spreadsheets.

💼 Career

Understanding how to search in 2D data structures is important for software developers working with matrices, tables, or grid-based data.

1

Create the 2D matrix

Create a 2D list called matrix with these exact rows and columns:
[[1, 3, 5], [7, 9, 11], [13, 15, 17]]

DSA Python

# Create the 2D list called matrix with the given values
# Your code here

Hint

Use square brackets to create a list of lists. Each inner list is a row.

2

Set the target number to search

Create a variable called target and set it to the number 9.

DSA Python

matrix = [[1, 3, 5], [7, 9, 11], [13, 15, 17]]
# Create the variable target and set it to 9
# Your code here

Hint

Just assign the number 9 to the variable named target.

3

Search the matrix for the target

Use nested for loops with variables row and col to go through each element in matrix. Create a variable called found and set it to False before the loops. If you find target, set found to True and break out of the loops.

DSA Python

matrix = [[1, 3, 5], [7, 9, 11], [13, 15, 17]]
target = 9
# Search for target in matrix using nested for loops
# Your code here

Hint

Use two loops: one for each row, and one for each element in that row. Use break to stop searching when found.

4

Print the search result

Print Found if found is True, otherwise print Not Found.

DSA Python

matrix = [[1, 3, 5], [7, 9, 11], [13, 15, 17]]
target = 9
found = False
for row in matrix:
    for col in row:
        if col == target:
            found = True
            break
    if found:
        break
# Print 'Found' if found is True else 'Not Found'
# Your code here

Hint

Use an if statement to check found and print the correct message.

Practice

1. What is the main idea behind searching an element in a 2D matrix where each row and column is sorted in ascending order?

easy

A. Start from top-right corner and move left or down based on comparison

B. Start from bottom-left corner and move left or down based on comparison

C. Search each element one by one from left to right, top to bottom

D. Use binary search on each row independently

Solution

1. Step 1: Understand matrix properties and use top-right corner

   Each row and column is sorted ascending, so top-right element is largest in its row and smallest in its column. If target is less than current element, move left; if greater, move down.

2. Final Answer:

   Start from top-right corner and move left or down based on comparison -> Option A

3. Quick Check:

   Top-right start = efficient search [OK]

Hint: Start at top-right to decide direction quickly [OK]

Common Mistakes:

• Starting from bottom-left without adjusting logic
• Searching all elements linearly
• Using binary search on each row without considering columns

2. Which of the following Python code snippets correctly initializes a 2D matrix named matrix with 3 rows and 4 columns filled with zeros?

easy

A. matrix = [[0]*4]*3

B. matrix = [[0]*3 for _ in range(4)]

C. matrix = [0]*3*[0]*4

D. matrix = [[0 for _ in range(4)] for _ in range(3)]

Solution

1. Step 1: Understand list comprehension for 2D matrix and check options

   Use nested list comprehension [[0 for _ in range(4)] for _ in range(3)] to create independent 3 rows with 4 zeros each. A shares references, B wrong dimensions, C syntax invalid.

2. Final Answer:

   matrix = [[0 for _ in range(4)] for _ in range(3)] -> Option D

3. Quick Check:

   Nested list comprehension = correct matrix init [OK]

Hint: Use nested list comprehension for independent rows [OK]

Common Mistakes:

• Using [[0]*4]*3 causes shared references
• Multiplying lists incorrectly
• Swapping rows and columns in comprehension

3. What will be the output of the following code?

matrix = [
  [1, 3, 5, 7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3

row, col = 0, len(matrix[0]) - 1
found = False
while row < len(matrix) and col >= 0:
    if matrix[row][col] == target:
        found = True
        break
    elif matrix[row][col] > target:
        col -= 1
    else:
        row += 1
print(found)

medium

A. IndexError

B. False

C. True

D. SyntaxError

Solution

1. Step 1: Trace starting position and comparisons until target found

   Start at top-right: matrix[0][3] = 7 > 3, move left to col=2; matrix[0][2] = 5 > 3, move left col=1; matrix[0][1] = 3 == target, found = True.

2. Final Answer:

   True -> Option C

3. Quick Check:

   Found target 3 = True [OK]

Hint: Start top-right, move left/down until found or out of bounds [OK]

Common Mistakes:

• Confusing row and column indices
• Not updating row or col correctly
• Assuming linear search needed

4. The following code is intended to search for a target in a sorted 2D matrix but does not work correctly. What is the error?

def search_matrix(matrix, target):
    row = 0
    col = len(matrix[0])
    while row < len(matrix) and col >= 0:
        if matrix[row][col] == target:
            return True
        elif matrix[row][col] > target:
            col -= 1
        else:
            row += 1
    return False

medium

A. col should start at len(matrix[0]) - 1, not len(matrix[0])

B. row should start at len(matrix) - 1

C. The while loop condition should use 'or' instead of 'and'

D. The function should return null instead of false

Solution

1. Step 1: Check initial column index and correct start

   col is set to len(matrix[0]) which is out of range (indexing starts at 0). It should be len(matrix[0]) - 1 to start at last valid column index.

2. Final Answer:

   col should start at len(matrix[0]) - 1, not len(matrix[0]) -> Option A

3. Quick Check:

   Index out of range fix = col start - 1 [OK]

Hint: Start col at last index: length minus one [OK]

Common Mistakes:

• Starting col at length causes IndexError
• Changing row start incorrectly
• Using wrong loop condition

5. Given a 2D matrix where each row and column is sorted ascending, write a function to count how many times a target number appears. Which approach is most efficient?

hard

A. Traverse entire matrix and count matches (O(m*n))

B. Use binary search on each row and sum counts (O(m*log n))

C. Start from top-right and move left/down counting matches (O(m+n))

D. Flatten matrix to 1D list and use built-in count method

Solution

1. Step 1: Analyze time complexities and correct approach

   Full traversal O(m*n), top-right single pass misses duplicates off the path, flatten O(m*n); binary search on each row finds leftmost/rightmost target positions to compute count per row, total O(m*log n) efficient and correct.

2. Final Answer:

   Use binary search on each row and sum counts (O(m*log n)) -> Option B

3. Quick Check:

   Binary per row = efficient count [OK]

Hint: Binary search range in each row [OK]

Common Mistakes:

• Using full traversal wasting time
• Flattening matrix loses sorted structure
• Top-right movement misses off-path duplicates