Sudoku Solver Using Backtracking in DSA Typescript - Time & Space Complexity

Time Complexity: Sudoku Solver Using Backtracking

O(9^n)

Understanding Time Complexity

We want to understand how the time needed to solve a Sudoku puzzle changes as the puzzle size or difficulty changes.

Specifically, how the backtracking method explores possibilities and how that affects the time it takes.

Scenario Under Consideration

Analyze the time complexity of the following Sudoku solver using backtracking.

function solveSudoku(board: number[][]): boolean {
  for (let row = 0; row < 9; row++) {
    for (let col = 0; col < 9; col++) {
      if (board[row][col] === 0) {
        for (let num = 1; num <= 9; num++) {
          if (isValid(board, row, col, num)) {
            board[row][col] = num;
            if (solveSudoku(board)) return true;
            board[row][col] = 0;
          }
        }
        return false;
      }
    }
  }
  return true;
}

This code tries numbers 1 to 9 in empty cells and uses recursion to fill the board step-by-step.

Identify Repeating Operations

Look at what repeats most in this code.

• Primary operation: Trying numbers 1 to 9 in each empty cell and checking if the number fits.
• How many times: Potentially for every empty cell, and for each number, the function calls itself recursively to try the next cell.

How Execution Grows With Input

Imagine the puzzle has n empty cells.

Input Size (empty cells n)	Approx. Operations
10	Up to 9^10 tries (about 3.5 billion)
20	Up to 9^20 tries (very large, trillions)
40	Up to 9^40 tries (astronomically large)

Pattern observation: The number of tries grows exponentially as the number of empty cells increases.

Final Time Complexity

Time Complexity: O(9^n)

This means the time needed can grow very fast, roughly nine times more work for each extra empty cell.

Common Mistake

[X] Wrong: "The solver always tries all 9 numbers for every empty cell."

[OK] Correct: The solver stops trying numbers as soon as it finds a valid one that leads to a solution, so it often tries fewer than 9 numbers per cell.

Interview Connect

Understanding this exponential growth helps you explain why backtracking can be slow and why pruning or smarter checks matter.

It shows your grasp of how recursive search explores many possibilities and how complexity grows with input size.

Self-Check

"What if we used a heuristic to pick the next empty cell with the fewest valid options? How would the time complexity change?"

Practice

1. What is the main idea behind using backtracking to solve a Sudoku puzzle?

easy

A. Use a sorting algorithm to arrange numbers in each row

B. Try placing numbers one by one and backtrack if a conflict occurs

C. Fill all empty cells randomly and check if the puzzle is solved

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 undoing choices

   If a number causes a conflict later, backtracking removes it and tries the next number.

3. Final Answer:

   Try placing numbers one by one and backtrack if a conflict occurs -> Option B

4. Quick Check:

   Backtracking = Try and undo choices [OK]

Hint: Backtracking means try, check, undo if needed [OK]

Common Mistakes:

• Thinking backtracking fills cells randomly
• Assuming sorting solves Sudoku
• Replacing zeros without checks

2. Which of the following TypeScript function signatures correctly represents a function to check if placing a number is valid in Sudoku?

easy

A. function isValid(board: number[][], row: number, col: number, num: number): boolean

B. function isValid(board: number[], row: number, col: number): number

C. function isValid(board: string[][], num: string): void

D. function isValid(board: number[][], num: number): string

Solution

1. Step 1: Analyze function parameters

   The function needs the board (2D array), row, column, and number to check validity.

2. Step 2: Check return type

   It should return a boolean indicating if placing the number is valid or not.

3. Final Answer:

   function isValid(board: number[][], row: number, col: number, num: number): boolean -> Option A

4. Quick Check:

   Correct parameters and boolean return [OK]

Hint: Check function inputs and boolean output for validity [OK]

Common Mistakes:

• Using wrong parameter types
• Missing row or column parameters
• Returning wrong data type

3. Given the following partial Sudoku board and the backtracking function call solveSudoku(board), what will be the value at board[0][2] after the first successful placement?

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]
];

medium

A. 4

B. 2

C. 1

D. 7

Solution

1. Step 1: Identify empty cell and possible numbers

   At position (0,2), possible numbers are those not in row 0, column 2, or 3x3 box. Candidates: 1, 2, 4.

2. Step 2: Check numbers 1 to 9 for validity

   1, 2, 4 are initially valid. Backtracking tries 1 first, recurses but backtracks due to later conflicts; tries 2, same; 4 succeeds and solves the puzzle.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   First successful number placed = 4 [OK]

Hint: Check row, column, box for smallest valid number [OK]

Common Mistakes:

• Choosing a number already in the same column
• Ignoring 3x3 box constraints
• Assuming first empty cell is filled with 1 always

4. Identify the bug in this TypeScript snippet for checking if a number can be placed in a Sudoku cell:

function isValid(board: number[][], row: number, col: number, num: number): boolean {
  for (let i = 0; i < 9; i++) {
    if (board[row][i] === num) return false;
    if (board[i][col] === num) return false;
  }
  const startRow = row - row % 3;
  const startCol = col - col % 3;
  for (let i = startRow; i < startRow + 3; i++) {
    for (let j = startCol; j < startCol + 3; j++) {
      if (board[i][j] === num) return false;
    }
  }
  return true;
}

medium

A. The function does not check the entire row correctly

B. The calculation of startRow and startCol is incorrect

C. The function checks the row and column twice causing errors

D. No bug; the function correctly checks validity

Solution

1. Step 1: Review row and column checks

   The loop checks all cells in the given row and column for the number.

2. Step 2: Review 3x3 box check

   startRow and startCol correctly find the top-left cell of the 3x3 box; nested loops check all 9 cells.

3. Final Answer:

   No bug; the function correctly checks validity -> Option D

4. Quick Check:

   All constraints checked properly [OK]

Hint: Check row, column, and box loops carefully [OK]

Common Mistakes:

• Miscomputing startRow or startCol
• Forgetting to check 3x3 box
• Returning false too early

5. You want to optimize the Sudoku solver by stopping early when the puzzle is solved. Which modification to the backtracking function is best to achieve this?

hard

A. Only check rows and columns, skip 3x3 box checks

B. Continue trying all numbers in all cells even after solution found

C. Return true immediately when a valid number is placed in the last empty cell

D. Remove the base case and let recursion run fully

Solution

1. Step 1: Understand base case in backtracking

   The base case returns true when all cells are filled correctly, stopping recursion.

2. Step 2: Modify to return true immediately on last cell

   Returning true immediately when the last empty cell is filled avoids unnecessary further checks.

3. Final Answer:

   Return true immediately when a valid number is placed in the last empty cell -> Option C

4. Quick Check:

   Stop recursion early by returning true [OK]

Hint: Return true at last cell to stop recursion early [OK]

Common Mistakes:

• Ignoring base case and continuing recursion
• Skipping box checks causing wrong solutions
• Trying all numbers even after solution