Bird
Raised Fist0
DSA Javascriptprogramming~10 mins

Sudoku Solver Using Backtracking in DSA Javascript - Interactive Practice

Choose your learning style10 modes available
Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to check if a number is already in the given row.

DSA Javascript
function isInRow(board, row, num) {
  for (let col = 0; col < 9; col++) {
    if (board[row][col] === [1]) {
      return true;
    }
  }
  return false;
}
Drag options to blanks, or click blank then click option'
Anum
Brow
Cboard
Dcol
Attempts:
3 left
💡 Hint
Common Mistakes
Using the wrong variable instead of the number to check.
2fill in blank
medium

Complete the code to check if a number is already in the given column.

DSA Javascript
function isInCol(board, col, num) {
  for (let row = 0; row < 9; row++) {
    if (board[row][col] === [1]) {
      return true;
    }
  }
  return false;
}
Drag options to blanks, or click blank then click option'
Acol
Brow
Cnum
Dboard
Attempts:
3 left
💡 Hint
Common Mistakes
Using the wrong variable instead of the number to check.
3fill in blank
hard

Fix the error in checking if a number is in the 3x3 box.

DSA Javascript
function isInBox(board, startRow, startCol, num) {
  for (let row = 0; row < 3; row++) {
    for (let col = 0; col < 3; col++) {
      if (board[startRow + row][startCol + col] === [1]) {
        return true;
      }
    }
  }
  return false;
}
Drag options to blanks, or click blank then click option'
Anum
BstartRow
CstartCol
Drow
Attempts:
3 left
💡 Hint
Common Mistakes
Comparing with wrong variables like startRow or startCol.
4fill in blank
hard

Fill both blanks to check if placing a number is valid in the Sudoku board.

DSA Javascript
function isValid(board, row, col, num) {
  return !isInRow(board, row, num) && !isInCol(board, col, num) && !isInBox(board, row - row % 3, col - col % 3, [1]) && board[row][col] === [2];
}
Drag options to blanks, or click blank then click option'
Anum
B0
Cnull
Dundefined
Attempts:
3 left
💡 Hint
Common Mistakes
Using null or undefined instead of 0 for empty cells.
5fill in blank
hard

Fill all three blanks to complete the backtracking solver function.

DSA Javascript
function solveSudoku(board) {
  for (let row = 0; row < 9; row++) {
    for (let col = 0; col < 9; col++) {
      if (board[row][col] === [1]) {
        for (let num = 1; num <= 9; num++) {
          if (isValid(board, row, col, num)) {
            board[row][col] = [2];
            if (solveSudoku(board)) {
              return true;
            } else {
              board[row][col] = [3];
            }
          }
        }
        return false;
      }
    }
  }
  return true;
}
Drag options to blanks, or click blank then click option'
A0
Bnum
Dnull
Attempts:
3 left
💡 Hint
Common Mistakes
Using null instead of 0 for empty cells.
Not resetting the cell on backtrack.

Practice

(1/5)
1. What is the main idea behind using backtracking to solve a Sudoku puzzle?
easy
A. Try placing numbers one by one and backtrack if a conflict occurs
B. Fill all empty cells randomly and check if the puzzle is solved
C. Use a sorting algorithm to arrange numbers in order
D. Replace all zeros with the number 9 directly

Solution

  1. Step 1: Understand backtracking approach

    Backtracking tries numbers in empty cells one by one and checks if they fit the Sudoku rules.
  2. Step 2: Handle conflicts by backtracking

    If a number causes a conflict, the algorithm removes it and tries the next number, going back if needed.
  3. Final Answer:

    Try placing numbers one by one and backtrack if a conflict occurs -> Option A
  4. Quick Check:

    Backtracking = Try and undo [OK]
Hint: Backtracking means try, check, undo if wrong [OK]
Common Mistakes:
  • Thinking backtracking fills cells randomly
  • Assuming sorting solves Sudoku
  • Replacing zeros without checks
2. Which of the following is the correct way to check if a number can be placed in a Sudoku cell in JavaScript?
easy
A. if isSafe(board, row, col, num) { placeNumber(); }
B. if isSafe(board, row, col, num) then placeNumber();
C. if (isSafe board, row, col, num) { placeNumber(); }
D. if (isSafe(board, row, col, num)) { placeNumber(); }

Solution

  1. Step 1: Understand JavaScript if syntax

    JavaScript requires parentheses around the condition in an if statement.
  2. Step 2: Check function call syntax

    Function calls need parentheses with arguments inside, like isSafe(board, row, col, num).
  3. Final Answer:

    if (isSafe(board, row, col, num)) { placeNumber(); } -> Option D
  4. Quick Check:

    if(condition) with parentheses [OK]
Hint: Use parentheses around conditions in if statements [OK]
Common Mistakes:
  • Missing parentheses in if condition
  • Using 'then' keyword in JavaScript
  • Incorrect function call syntax
3. What will be the output of the following code snippet when solving a Sudoku board?
const board = [
  [5,3,0,0,7,0,0,0,0],
  [6,0,0,1,9,5,0,0,0],
  [0,9,8,0,0,0,0,6,0],
  [8,0,0,0,6,0,0,0,3],
  [4,0,0,8,0,3,0,0,1],
  [7,0,0,0,2,0,0,0,6],
  [0,6,0,0,0,0,2,8,0],
  [0,0,0,4,1,9,0,0,5],
  [0,0,0,0,8,0,0,7,9]
];
solveSudoku(board);
console.log(board);
medium
A. An error because solveSudoku is not defined
B. The original board unchanged with zeros intact
C. A fully solved Sudoku board with no zeros
D. A partially solved board with some zeros replaced

Solution

  1. Step 1: Understand solveSudoku function purpose

    solveSudoku uses backtracking to fill all empty cells correctly, solving the puzzle.
  2. Step 2: Check console.log output after solving

    After solveSudoku runs, the board is fully solved with no zeros remaining.
  3. Final Answer:

    A fully solved Sudoku board with no zeros -> Option C
  4. Quick Check:

    Backtracking solves all cells [OK]
Hint: Backtracking fills all empty cells correctly [OK]
Common Mistakes:
  • Expecting original board unchanged
  • Thinking partial fill is final
  • Assuming function is undefined
4. Consider this buggy code snippet for checking if a number can be placed in a Sudoku cell:
function isSafe(board, row, col, num) {
  for (let x = 0; x < 9; x++) {
    if (board[row][x] === num || board[x][col] === num) {
      return false;
    }
  }
  // Missing 3x3 box check
  return true;
}
What is the main problem here?
medium
A. It returns false too early
B. It does not check the 3x3 subgrid for duplicates
C. It uses wrong loop variable x
D. It modifies the board inside the function

Solution

  1. Step 1: Analyze row and column checks

    The code correctly checks if num exists in the same row or column.
  2. Step 2: Identify missing 3x3 subgrid check

    Sudoku rules require checking the 3x3 box; this code does not do that, causing errors.
  3. Final Answer:

    It does not check the 3x3 subgrid for duplicates -> Option B
  4. Quick Check:

    3x3 box check missing = bug [OK]
Hint: Always check row, column, and 3x3 box [OK]
Common Mistakes:
  • Ignoring 3x3 box validation
  • Thinking loop variable causes error
  • Assuming early return is wrong
5. You want to optimize the Sudoku solver by choosing the next empty cell with the fewest possible valid numbers to try. Which technique does this describe?
hard
A. Heuristic with Minimum Remaining Value (MRV) strategy
B. Random guessing of empty cells
C. Depth-first search without pruning
D. Sorting the board rows before solving

Solution

  1. Step 1: Understand MRV heuristic

    MRV chooses the empty cell with the fewest valid options to reduce branching and speed up solving.
  2. Step 2: Compare with other options

    Random guessing and depth-first search without pruning are less efficient; sorting rows does not help solving.
  3. Final Answer:

    Heuristic with Minimum Remaining Value (MRV) strategy -> Option A
  4. Quick Check:

    MRV = choose cell with fewest options [OK]
Hint: Pick cell with fewest choices first [OK]
Common Mistakes:
  • Confusing random guessing with heuristics
  • Ignoring pruning benefits
  • Thinking sorting helps solving
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard