Search in 2D Matrix in DSA C++ - Build from Scratch

Search in 2D Matrix

📖 Scenario: You have a grid of numbers arranged in rows and columns, like seats in a theater. You want to find if a specific number is present in this grid.

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

📋 What You'll Learn

Create a 2D matrix with exact values

Create a variable for the target number to search

Use nested loops to search the target in the matrix

Print 'Found' if the target is in the matrix, else print 'Not Found'

💡 Why This Matters

🌍 Real World

Searching in grids or tables is common in games, maps, and spreadsheets.

💼 Career

Understanding how to search in 2D data structures is important for software development and data analysis.

1

Create the 2D matrix

Create a 2D vector called matrix with these exact rows and columns:
{ {1, 3, 5}, {7, 9, 11}, {13, 15, 17} }

DSA C++

// Create the 2D vector matrix with the given values
// Your code here

Hint

Use std::vector> to create a 2D matrix with the exact rows and columns.

2

Set the target number to search

Create an integer variable called target and set it to 9

DSA C++

#include 

std::vector> matrix = {
    {1, 3, 5},
    {7, 9, 11},
    {13, 15, 17}
};

// Create the target variable and set it to 9
// Your code here

Hint

Use int target = 9; to create the target number.

3

Search the target in the matrix

Use nested for loops with variables i and j to iterate over matrix. Create a boolean variable found set to false. If matrix[i][j] equals target, set found to true and break both loops.

DSA C++

#include 

std::vector> matrix = {
    {1, 3, 5},
    {7, 9, 11},
    {13, 15, 17}
};

int target = 9;

// Search for target in matrix using nested loops
// Set found to true if target is found
// Your code here

Hint

Use two for loops to check each element. Use found to remember if target is found and break loops early.

4

Print the search result

Print "Found" if found is true, otherwise print "Not Found"

DSA C++

#include 
#include 

std::vector> matrix = {
    {1, 3, 5},
    {7, 9, 11},
    {13, 15, 17}
};

int target = 9;

bool found = false;
for (int i = 0; i < matrix.size(); i++) {
    for (int j = 0; j < matrix[i].size(); j++) {
        if (matrix[i][j] == target) {
            found = true;
            break;
        }
    }
    if (found) {
        break;
    }
}

// Print "Found" if found is true, else print "Not Found"
// Your code here

Hint

Use if (found) to decide what to print. Use std::cout to print the result.

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