Search in 2D Matrix in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the target is found in the matrix.

DSA C

if(matrix[i][j] == [1]) {
    return 1;
}

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

Amatrix

Bi

Cj

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with the wrong variable like i or j instead of target.

Using matrix instead of target for comparison.

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

2fill in blank

medium

Complete the code to iterate over rows of the matrix.

DSA C

for(int i = 0; i < [1]; i++) {
    // inner loop
}

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Arows

Bcols

Ctarget

Dmatrix

Attempts:

3 left

💡 Hint

Common Mistakes

Using cols instead of rows for the outer loop.

Using target or matrix as loop limits.

The outer loop should run over the number of rows in the matrix.

3fill in blank

hard

Fix the error in the inner loop to iterate over columns.

DSA C

for(int j = 0; j < [1]; j++) {
    // check element
}

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

Arows

Bcols

Ctarget

Dmatrix

Attempts:

3 left

💡 Hint

Common Mistakes

Using rows instead of cols for the inner loop.

Using target or matrix as loop limits.

The inner loop should run over the number of columns in the matrix.

4fill in blank

hard

Fill both blanks to complete the function signature and return type.

DSA C

[1] searchMatrix(int** matrix, int rows, int cols, int target) {
    for(int i = 0; i < rows; i++) {
        for(int j = 0; j < cols; j++) {
            if(matrix[i][j] == target) {
                return [2];
            }
        }
    }
    return 0;
}

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

Aint

B1

Cvoid

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Using void as return type but returning a value.

Returning true instead of 1 in C.

The function returns an int indicating if the target is found (1) or not (0).

5fill in blank

hard

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

DSA C

int searchMatrix(int** matrix, int rows, int cols, int target) {
    int row = 0, col = [1] - 1;
    while(row < rows && col >= 0) {
        if(matrix[row][col] == target) {
            return [2];
        } else if(matrix[row][col] > target) {
            col[3];
        } else {
            row++;
        }
    }
    return 0;
}

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

Acols

B1

C--

D++

Attempts:

3 left

💡 Hint

Common Mistakes

Using rows instead of cols for column index initialization.

Returning 0 instead of 1 when target is found.

Incrementing column instead of decrementing when current > target.

Start from top-right corner (col = cols - 1), return 1 if found, move left (col--) if current > target.

Practice

1. What is the main idea behind searching a target value in a 2D matrix where each row is sorted and the first integer of each row is greater than the last integer of the previous row?

easy

A. Start from bottom-left corner and move up or right

B. Search each row linearly until the target is found

C. Use a hash map to store all elements for quick lookup

D. Treat the 2D matrix as a sorted 1D array and apply binary search

Solution

1. Step 1: Understand matrix sorting property

   Each row is sorted and the first element of a row is greater than the last element of the previous row, so the entire matrix can be seen as a sorted list.

2. Step 2: Apply binary search on the matrix

   By treating the matrix as a single sorted array, binary search can be used efficiently to find the target.

3. Final Answer:

   Treat the 2D matrix as a sorted 1D array and apply binary search -> Option D

4. Quick Check:

   Matrix sorted row-wise = binary search [OK]

Hint: Think of the matrix as one sorted list for binary search [OK]

Common Mistakes:

• Searching each row linearly wastes time
• Using hash map wastes extra space
• Starting from bottom-left only works if columns are sorted

2. Which of the following C code snippets correctly calculates the middle element index when performing binary search on a 2D matrix with rows and columns?

easy

A. int mid = (low + high) / 2; int row = mid % rows; int col = mid / cols;

B. int mid = (low + high) / 2; int row = mid / cols; int col = mid % cols;

C. int mid = (low + high) >> 1; int row = mid % cols; int col = mid / rows;

D. int mid = (low + high) / 2; int row = mid / rows; int col = mid % rows;

Solution

1. Step 1: Calculate mid index in 1D

   mid = (low + high) / 2 correctly finds the middle index between low and high.

2. Step 2: Convert mid to 2D indices

   Row is mid divided by number of columns (mid / cols), column is mid modulo number of columns (mid % cols).

3. Final Answer:

   int mid = (low + high) / 2; int row = mid / cols; int col = mid % cols; -> Option B

4. Quick Check:

   Row = mid/cols, Col = mid%cols [OK]

Hint: Divide mid by cols for row, modulo cols for column [OK]

Common Mistakes:

• Swapping rows and columns in calculation
• Using rows instead of cols for division/modulo
• Using bit shift incorrectly for mid

3. What will be the output of the following C code snippet searching for target 5 in the matrix?

int matrix[3][4] = {
  {1, 3, 5, 7},
  {10, 11, 16, 20},
  {23, 30, 34, 60}
};
int target = 5;
int rows = 3, cols = 4;
int low = 0, high = rows * cols - 1;
int found = 0;
while (low <= high) {
  int mid = (low + high) / 2;
  int r = mid / cols;
  int c = mid % cols;
  if (matrix[r][c] == target) {
    found = 1;
    break;
  } else if (matrix[r][c] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}
printf("%d\n", found);

medium

A. 1

B. 0

C. 5

D. Compilation error

Solution

1. Step 1: Perform binary search on matrix

   The code treats the matrix as a sorted 1D array and searches for target 5 using binary search.

2. Step 2: Check if target 5 exists

   5 is present at matrix[0][2], so found will be set to 1 and loop breaks.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Target found = 1 [OK]

Hint: Binary search returns 1 if target found, else 0 [OK]

Common Mistakes:

• Confusing indices calculation for row and column
• Not updating low and high correctly
• Printing target value instead of found flag

4. Identify the bug in the following C code snippet that searches for a target in a 2D matrix and suggest the fix:

int matrix[2][3] = {{1, 2, 3}, {4, 5, 6}};
int target = 4;
int rows = 2, cols = 3;
int low = 0, high = rows * cols - 1;
int found = 0;
while (low < high) {
  int mid = (low + high) / 2;
  int r = mid / cols;
  int c = mid % cols;
  if (matrix[r][c] == target) {
    found = 1;
    break;
  } else if (matrix[r][c] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}
printf("%d\n", found);

medium

A. Change mid calculation to (low + high + 1) / 2

B. Swap row and column calculation

C. Change while condition to low <= high

D. Initialize found to -1 instead of 0

Solution

1. Step 1: Analyze loop condition

   The loop uses while (low < high), which misses the case when low == high, possibly skipping the last element.

2. Step 2: Fix loop condition

   Changing to while (low <= high) ensures all elements are checked, including when low equals high.

3. Final Answer:

   Change while condition to low <= high -> Option C

4. Quick Check:

   Loop must check low == high [OK]

Hint: Use while(low <= high) for complete binary search [OK]

Common Mistakes:

• Using < instead of <= in loop condition
• Incorrect mid calculation causing infinite loop
• Swapping row and column indices

5. Given a 2D matrix where each row and each column is sorted in ascending order, but the first integer of each row is NOT necessarily greater than the last integer of the previous row, which approach is best to search for a target value efficiently?

hard

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

B. Search each row using binary search separately

C. Use binary search treating matrix as 1D sorted array

D. Use linear search over all elements

Solution

1. Step 1: Understand matrix property

   Rows are sorted but no guarantee about relation between rows, so matrix is not fully sorted as 1D array.

2. Step 2: Choose search method

   Starting from top-right corner, move left if current > target, move down if current < target, efficiently narrows search.

3. Final Answer:

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

4. Quick Check:

   Partial row sorting = stair-step search [OK]

Hint: Use stair-step search from top-right for partially sorted matrix [OK]

Common Mistakes:

• Applying full binary search on unsorted matrix
• Searching each row linearly wasting time
• Using hash map wasting extra space