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DSA C++programming~10 mins

Number of Islands BFS and DFS in DSA C++ - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to start BFS from the current cell.

DSA C++
queue<pair<int, int>> q;
q.push([1]);
Drag options to blanks, or click blank then click option'
Aj, i
Bi, j
C0, 0
Di + 1, j + 1
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping i and j coordinates.
Pushing incorrect or fixed coordinates like (0,0).
2fill in blank
medium

Complete the code to mark the current cell as visited.

DSA C++
grid[[1]][j] = '0';
Drag options to blanks, or click blank then click option'
Aj
Bvisited
C0
Di
Attempts:
3 left
💡 Hint
Common Mistakes
Using column index instead of row index.
Using a variable that does not exist.
3fill in blank
hard

Fix the error in the DFS recursive call to explore neighbors.

DSA C++
if (row >= 0 && row < m && col >= 0 && col < n && grid[row][col] == '1') {
    [1](grid, row, col);
}
Drag options to blanks, or click blank then click option'
Adfs
Bbfs
Cexplore
Dsearch
Attempts:
3 left
💡 Hint
Common Mistakes
Calling BFS inside DFS.
Using undefined function names.
4fill in blank
hard

Fill both blanks to check all four directions in BFS.

DSA C++
int directions[4][2] = {{-1, 0}, {1, 0}, [1], [2];
Drag options to blanks, or click blank then click option'
A0, -1
B0, 1
C1, 1
D-1, -1
Attempts:
3 left
💡 Hint
Common Mistakes
Using diagonal directions instead of left/right.
Swapping row and column changes.
5fill in blank
hard

Fill all three blanks to complete the DFS function signature and base case.

DSA C++
void [1](vector<vector<char>>& grid, int [2], int [3]) {
    int m = grid.size();
    int n = grid[0].size();
    if ([2] < 0 || [2] >= m || [3] < 0 || [3] >= n || grid[[2]][[3]] == '0') return;
    grid[[2]][[3]] = '0';
    // explore neighbors
}
Drag options to blanks, or click blank then click option'
Adfs
Brow
Ccol
Dsearch
Attempts:
3 left
💡 Hint
Common Mistakes
Using wrong function names.
Using incorrect parameter names.

Practice

(1/5)
1. What does the "Number of Islands" problem ask you to find in a 2D grid of '1's (land) and '0's (water)?
easy
A. Count total number of '1's in the grid
B. Count how many separate groups of connected '1's exist
C. Find the largest island size
D. Find the shortest path between two islands

Solution

  1. Step 1: Understand the problem definition

    The problem asks to count distinct islands, where an island is a group of connected '1's horizontally or vertically.
  2. Step 2: Clarify what is being counted

    We count how many separate clusters of '1's exist, not the total number of '1's or sizes.
  3. Final Answer:

    Count how many separate groups of connected '1's exist -> Option B
  4. Quick Check:

    Number of islands = count of connected '1' groups [OK]
Hint: Islands = groups of connected '1's, count groups not cells [OK]
Common Mistakes:
  • Counting total '1's instead of groups
  • Including diagonal connections as connected
  • Confusing island size with island count
2. Which of the following C++ code snippets correctly initializes a 2D vector grid of chars with given rows and columns filled with '0'?
easy
A. std::vector grid(rows, std::vector<char>(cols, '0'));
B. std::vector<char> grid(rows, std::vector<char>(cols, '0'));
C. std::vector grid(cols, std::vector<char>(rows, '0'));
D. std::vector grid(rows, cols, '0');

Solution

  1. Step 1: Understand 2D vector initialization syntax

    To create a 2D vector of chars, use vector of vectors: outer vector size = rows, inner vector size = cols.
  2. Step 2: Check each option

    std::vector grid(rows, std::vector<char>(cols, '0')); correctly creates 'rows' vectors each with 'cols' chars initialized to '0'. Others have wrong types or parameters.
  3. Final Answer:

    std::vector grid(rows, std::vector<char>(cols, '0')); -> Option A
  4. Quick Check:

    2D vector init = vector of vectors with sizes [OK]
Hint: Use vector(rows, vector<char>(cols, '0')) [OK]
Common Mistakes:
  • Swapping rows and cols sizes
  • Using wrong vector types
  • Trying to initialize 2D vector with three parameters
3. Given the following grid and BFS-based code snippet to count islands, what is the output?
std::vector grid = {
  {'1','1','0','0'},
  {'1','0','0','1'},
  {'0','0','1','1'},
  {'0','0','0','0'}
};
int countIslands(const std::vector& grid) {
  int count = 0;
  int rows = grid.size();
  int cols = grid[0].size();
  std::vector visited(rows, std::vector<bool>(cols, false));
  std::queue> q;
  int directions[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
  for(int i=0; i=0 && nr=0 && nc
medium
A. 3
B. 4
C. 2
D. 1

Solution

  1. Step 1: Identify islands in the grid

    Islands are groups of connected '1's. Here, islands are: top-left cluster (3 cells), single '1' at (1,3), and bottom cluster (2 cells at (2,2),(2,3)). Total 3 islands.
  2. Step 2: Understand BFS traversal

    The BFS marks all connected '1's visited per island, so count increments once per island found.
  3. Final Answer:

    3 -> Option A
  4. Quick Check:

    Count islands = 3 [OK]
Hint: Count connected groups by BFS marking visited [OK]
Common Mistakes:
  • Counting diagonal connections as islands
  • Missing isolated '1's as separate islands
  • Not marking visited correctly causing overcount
4. The following DFS code for counting islands has a bug. What is the bug?
void dfs(int r, int c, std::vector& grid) {
  int rows = grid.size();
  int cols = grid[0].size();
  if(r < 0 || c < 0 || r >= rows || c >= cols || grid[r][c] == '0')
    return;
  grid[r][c] = '0';
  dfs(r+1, c, grid);
  dfs(r-1, c, grid);
  dfs(r, c+1, grid);
  dfs(r, c-1, grid);
}

int numIslands(std::vector& grid) {
  int count = 0;
  for(int i=0; i
medium
A. The base case does not check for visited cells
B. The DFS does not check boundaries correctly
C. The DFS does not mark visited cells, causing infinite recursion
D. The grid is modified during DFS, which is incorrect

Solution

  1. Step 1: Analyze DFS grid modification

    The DFS marks visited cells by changing '1' to '0' in the original grid, modifying input.
  2. Step 2: Identify why this is a bug

    Modifying input grid may be unintended side effect; better to use a separate visited array to avoid altering input data.
  3. Final Answer:

    The grid is modified during DFS, which is incorrect -> Option D
  4. Quick Check:

    Modifying input grid = bug [OK]
Hint: Avoid changing input grid; use visited array instead [OK]
Common Mistakes:
  • Ignoring that grid is changed in-place
  • Assuming no visited tracking needed
  • Misreading boundary checks as incorrect
5. You want to count islands in a grid where diagonal connections also count as connected (8 directions). How should you modify the BFS or DFS approach?
hard
A. Count islands by summing all '1's without traversal
B. Only check horizontal neighbors, ignore vertical and diagonal
C. Add diagonal directions (like (1,1), (1,-1), (-1,1), (-1,-1)) to the directions array
D. Use BFS for horizontal and DFS for vertical connections separately

Solution

  1. Step 1: Understand connectivity change

    Diagonal connections mean neighbors include 8 directions: up, down, left, right, and 4 diagonals.
  2. Step 2: Modify directions array

    Add diagonal moves (1,1), (1,-1), (-1,1), (-1,-1) to directions used in BFS or DFS to explore all connected neighbors.
  3. Final Answer:

    Add diagonal directions (like (1,1), (1,-1), (-1,1), (-1,-1)) to the directions array -> Option C
  4. Quick Check:

    Include all 8 neighbors for diagonal connectivity [OK]
Hint: Add diagonal moves to directions for 8-way connectivity [OK]
Common Mistakes:
  • Ignoring diagonal neighbors
  • Counting only horizontal or vertical neighbors
  • Trying to count without traversal
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