Search in 2D Matrix in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return true if the target is found in the matrix.

DSA C++

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    for (int i = 0; i < matrix.size(); i++) {
        for (int j = 0; j < matrix[0].size(); j++) {
            if (matrix[i][j] == [1]) {
                return true;
            }
        }
    }
    return false;
}

Drag options to blanks, or click blank then click option'

Atarget

Bi

Cj

Dmatrix

Attempts:

3 left

💡 Hint

Common Mistakes

Using loop indices instead of the target value for comparison.

Comparing with the entire matrix instead of a single element.

We compare each element matrix[i][j] with the target value to find a match.

2fill in blank

medium

Complete the code to perform binary search on a sorted row to find the target.

DSA C++

bool binarySearchRow(vector<int>& row, int target) {
    int left = 0, right = row.size() - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (row[mid] == [1]) {
            return true;
        } else if (row[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}

Drag options to blanks, or click blank then click option'

Atarget

Bleft

Cmid

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with indices instead of the target value.

Using wrong variable names for comparison.

We compare the middle element row[mid] with the target value to check for equality.

3fill in blank

hard

Fix the error in the code to correctly search the matrix using binary search on rows.

DSA C++

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int top = 0, bottom = matrix.size() - 1;
    while (top <= bottom) {
        int mid = top + (bottom - top) / 2;
        if (matrix[mid][0] == [1]) {
            return true;
        } else if (matrix[mid][0] < target) {
            top = mid + 1;
        } else {
            bottom = mid - 1;
        }
    }
    return false;
}

Drag options to blanks, or click blank then click option'

Amid

Btarget

Ctop

Dbottom

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with mid index instead of target value.

Using wrong variable for comparison.

We compare the first element of the middle row with the target to decide which half to search next.

4fill in blank

hard

Fill both blanks to complete the code that finds the correct row and then searches it.

DSA C++

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int top = 0, bottom = matrix.size() - 1;
    while (top <= bottom) {
        int mid = top + (bottom - top) / 2;
        if (matrix[mid][0] == target) {
            return true;
        } else if (matrix[mid][0] [1] target) {
            top = mid + 1;
        } else {
            bottom = mid - 1;
        }
    }
    int row = bottom;
    if (row < 0) return false;
    return binarySearchRow(matrix[row], [2]);
}

Drag options to blanks, or click blank then click option'

A<

B>

Ctarget

Drow

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operator.

Passing wrong argument to binary search.

We check if the first element of mid row is less than target to move top pointer. Then we search the found row for the target.

5fill in blank

hard

Fill all three blanks to complete the optimized search in 2D matrix.

DSA C++

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    if (matrix.empty() || matrix[0].empty()) return false;
    int rows = matrix.size(), cols = matrix[0].size();
    int left = 0, right = rows * cols - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        int mid_value = matrix[mid / [1]][mid % [2]];
        if (mid_value == [3]) {
            return true;
        } else if (mid_value < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}

Drag options to blanks, or click blank then click option'

Acols

Brows

Ctarget

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using rows instead of cols for index calculations.

Comparing with wrong variable.

We use 'cols' to convert 1D mid index to 2D indices. We compare mid_value with target to decide search direction.

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