Search in 2D Matrix in DSA C++ - Time & Space Complexity

Time Complexity: Search in 2D Matrix

O(m x n)

Understanding Time Complexity

We want to understand how the time needed to find a number in a 2D matrix changes as the matrix size grows.

Specifically, how does searching through rows and columns affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

bool searchMatrix(std::vector<std::vector<int>>& matrix, int target) {
    int rows = matrix.size();
    int cols = matrix[0].size();
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            if (matrix[i][j] == target) {
                return true;
            }
        }
    }
    return false;
}
    

This code checks every element in the matrix one by one to find the target number.

Identify Repeating Operations

• Primary operation: Checking each element in the matrix.
• How many times: The outer loop runs once per row, and the inner loop runs once per column, so total checks = rows x columns.

How Execution Grows With Input

As the matrix gets bigger, the number of checks grows by multiplying rows and columns.

Input Size (rows x cols)	Approx. Operations
10 x 10	100 checks
100 x 100	10,000 checks
1000 x 1000	1,000,000 checks

Pattern observation: Doubling rows and columns causes the total work to grow by the square of the size increase.

Final Time Complexity

Time Complexity: O(m x n)

This means the time to search grows proportionally with the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Since the matrix is sorted, searching is always fast like binary search."

[OK] Correct: This code checks every element without using the sorted order, so it still takes time proportional to all elements.

Interview Connect

Understanding how nested loops affect time helps you explain and improve search methods in matrices during interviews.

Self-Check

"What if we use a search method that starts from the top-right corner and moves left or down based on comparisons? How would the time complexity change?"

Practice

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

easy

A. Use binary search on each row independently

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

C. Search each element row by row until the target is found

D. Start from the bottom-left corner and move right or up based on comparison

Solution

1. Step 1: Understand matrix properties

   Each row and column is sorted ascending, so top-right element is a good start to decide direction.

2. Step 2: Use top-right corner to guide search

   If target is less than current element, move left; if greater, move down.

3. Final Answer:

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

4. Quick Check:

   Top-right corner search = Start from the top-right corner and move left or down based on comparison [OK]

Hint: Start at top-right, move left/down based on target comparison [OK]

Common Mistakes:

• Starting from bottom-left without adjusting moves
• Searching all elements without using sorting
• Using binary search on unsorted columns

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

easy

A. std::vector matrix(4, std::vector<int>(3, 0));

B. std::vector<int> matrix(3, std::vector<int>(4, 0));

C. std::vector matrix(3, std::vector<int>(4, 0));

D. std::vector matrix(3, 4, 0);

Solution

1. Step 1: Understand vector initialization syntax

   To create a 2D vector with 3 rows and 4 columns, use vector of vectors with inner vector size 4.

2. Step 2: Check each option

   std::vector matrix(3, std::vector<int>(4, 0)); correctly uses vector with 3 rows and inner vector of size 4 filled with 0.

3. Final Answer:

   std::vector matrix(3, std::vector<int>(4, 0)); -> Option C

4. Quick Check:

   Correct 2D vector init = std::vector matrix(3, std::vector<int>(4, 0)); [OK]

Hint: Use vector(rows, vector<int>(cols, val)) [OK]

Common Mistakes:

• Using wrong vector type for inner vector
• Swapping rows and columns sizes
• Trying to pass three arguments to vector constructor

3. What will be the output of the following C++ code snippet?

std::vector matrix = {
  {1, 4, 7, 11},
  {2, 5, 8, 12},
  {3, 6, 9, 16},
  {10, 13, 14, 17}
};
int target = 5;
int row = 0, col = 3;
bool found = false;
while (row < 4 && col >= 0) {
  if (matrix[row][col] == target) {
    found = true;
    break;
  } else if (matrix[row][col] > target) {
    col--;
  } else {
    row++;
  }
}
std::cout << (found ? "Found" : "Not Found") << std::endl;

medium

A. Found

B. Not Found

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Trace the search starting at matrix[0][3]

   matrix[0][3] = 11, target=5, 11 > 5 so col-- to 2.

2. Step 2: Continue search

   matrix[0][2] = 7, 7 > 5 so col-- to 1; matrix[0][1] = 4, 4 < 5 so row++ to 1; matrix[1][1] = 5 equals target, found = true.

3. Final Answer:

   Found -> Option A

4. Quick Check:

   Search finds 5 at (1,1) = Found [OK]

Hint: Start top-right, move left if bigger, down if smaller [OK]

Common Mistakes:

• Not updating row and col correctly
• Confusing comparison operators
• Assuming target not found without full search

4. Identify the bug in the following C++ code that searches for a target in a sorted 2D matrix:

bool searchMatrix(std::vector& matrix, int target) {
  int row = 0, col = matrix[0].size() - 1;
  while (row <= matrix.size() && col >= 0) {
    if (matrix[row][col] == target) return true;
    else if (matrix[row][col] > target) col--;
    else row++;
  }
  return false;
}

medium

A. The while loop condition should use < instead of <= for row

B. The initial column index is incorrect

C. The function should return false inside the loop

D. The matrix should be passed by value, not reference

Solution

1. Step 1: Analyze loop boundary for row index

   matrix.size() gives number of rows; valid row indices are 0 to size()-1, so row < size() is correct.

2. Step 2: Identify off-by-one error

   Using row <= matrix.size() allows row to equal size(), causing out-of-bounds access.

3. Final Answer:

   The while loop condition should use < instead of <= for row -> Option A

4. Quick Check:

   Row index must be less than size() [OK]

Hint: Use < size() for indices, not <= size() [OK]

Common Mistakes:

• Using <= causing out-of-bounds errors
• Incorrect initial column index
• Returning false too early inside loop

5. Given a 2D matrix where each row and column is sorted ascending, write a function to count how many elements are less than or equal to a target value. Which approach is most efficient?

hard

A. Flatten matrix and sort, then count using binary search (O(m*n log(m*n)))

B. Use binary search on each row to find count per row (O(m*log n))

C. Traverse each element and count if <= target (O(m*n))

D. Start from bottom-left and move up or right to count (O(m+n))

Solution

1. Step 1: Understand matrix sorted property

   Rows and columns sorted ascending allows efficient traversal from bottom-left.

2. Step 2: Use bottom-left traversal to count elements

   Start at bottom-left; if element <= target, all elements above in that column are <= target, move right and add count; else move up.

3. Step 3: Compare efficiency

   This approach visits at most m+n elements, better than binary search on each row or full traversal.

4. Final Answer:

   Start from bottom-left and move up or right to count (O(m+n)) -> Option D

5. Quick Check:

   Bottom-left traversal counts in O(m+n) [OK]

Hint: Count using bottom-left traversal for O(m+n) time [OK]

Common Mistakes:

• Using full traversal causing O(m*n) time
• Applying binary search on unsorted columns
• Flattening matrix unnecessarily increasing complexity