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DSA Cprogramming~5 mins

Tower of Hanoi Problem in DSA C - Time & Space Complexity

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Time Complexity: Tower of Hanoi Problem
O(2^n)
Understanding Time Complexity

We want to understand how the time needed to solve the Tower of Hanoi grows as we add more disks.

How does the number of moves change when the number of disks increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
    if (n == 1) {
        printf("Move disk 1 from rod %c to rod %c\n", from_rod, to_rod);
        return;
    }
    towerOfHanoi(n - 1, from_rod, aux_rod, to_rod);
    printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
    towerOfHanoi(n - 1, aux_rod, to_rod, from_rod);
}
    

This code moves n disks from one rod to another using recursion and an auxiliary rod.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Recursive calls that move disks one by one.
  • How many times: Each call splits into two smaller calls with n-1 disks, plus one move.
How Execution Grows With Input

Each time we add a disk, the number of moves roughly doubles and adds one more move.

Input Size (n)Approx. Operations (Moves)
11
23
37
415
531

Pattern observation: The moves grow exponentially, doubling each time plus one extra move.

Final Time Complexity

Time Complexity: O(2^n)

This means the time needed doubles with each additional disk, growing very fast as n increases.

Common Mistake

[X] Wrong: "The time grows linearly with the number of disks because we move each disk once."

[OK] Correct: Actually, disks are moved multiple times in a recursive pattern, causing exponential growth, not just once each.

Interview Connect

Understanding this problem helps you think about recursion and exponential growth, which are common in many coding challenges.

Self-Check

"What if we changed the problem to move disks iteratively without recursion? How would the time complexity change?"

Practice

(1/5)
1. What is the minimum number of moves required to solve the Tower of Hanoi problem with 3 disks?
easy
A. 6
B. 5
C. 9
D. 7

Solution

  1. Step 1: Understand the formula for minimum moves

    The minimum moves required for n disks is 2^n - 1.
  2. Step 2: Calculate for 3 disks

    For 3 disks, minimum moves = 2^3 - 1 = 8 - 1 = 7.
  3. Final Answer:

    7 -> Option D
  4. Quick Check:

    2^3 - 1 = 7 [OK]
Hint: Use formula 2^n - 1 for minimum moves [OK]
Common Mistakes:
  • Using n instead of 2^n - 1
  • Calculating 2*n - 1 instead of 2^n - 1
  • Confusing number of disks with moves
2. Which of the following is the correct recursive function signature in C for solving Tower of Hanoi?
easy
A. void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod);
B. int towerOfHanoi(int n, char from_rod, char to_rod);
C. void towerOfHanoi(int n, int from_rod, int to_rod);
D. void towerOfHanoi(int n, char from_rod, char to_rod);

Solution

  1. Step 1: Identify parameters needed for Tower of Hanoi

    We need number of disks and names of three rods: source, destination, auxiliary.
  2. Step 2: Check function signature matching these parameters

    void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod); has int n and three char rods, matching the problem requirements.
  3. Final Answer:

    void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod); -> Option A
  4. Quick Check:

    Function needs 3 rods and disk count [OK]
Hint: Tower of Hanoi needs 3 rods as parameters [OK]
Common Mistakes:
  • Missing auxiliary rod parameter
  • Using int instead of char for rods
  • Returning int instead of void
3. What will be the output of the following C code snippet for Tower of Hanoi with 2 disks?
void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
  if (n == 1) {
    printf("Move disk 1 from rod %c to rod %c\n", from_rod, to_rod);
    return;
  }
  towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
  printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
  towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}

int main() {
  towerOfHanoi(2, 'A', 'C', 'B');
  return 0;
}
medium
A. Move disk 2 from rod A to rod C Move disk 1 from rod A to rod B Move disk 2 from rod B to rod C
B. Move disk 1 from rod A to rod C Move disk 2 from rod A to rod B Move disk 1 from rod C to rod B
C. Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C
D. Move disk 1 from rod B to rod C Move disk 2 from rod A to rod C Move disk 1 from rod A to rod B

Solution

  1. Step 1: Trace recursive calls for n=2

    First move disk 1 from A to B, then disk 2 from A to C, then disk 1 from B to C.
  2. Step 2: Match output with options

    Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C matches the exact sequence of moves printed by the code.
  3. Final Answer:

    Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C -> Option C
  4. Quick Check:

    Recursive moves for 2 disks = Move disk 1 from rod A to rod B Move disk 2 from rod A to rod C Move disk 1 from rod B to rod C [OK]
Hint: For 2 disks, move smaller disk to aux first [OK]
Common Mistakes:
  • Swapping auxiliary and destination rods
  • Printing moves in wrong order
  • Confusing disk numbers
4. Identify the error in the following Tower of Hanoi C function and select the correct fix:
void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
  if (n == 0) {
    return;
  }
  towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
  printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
  towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}
medium
A. Change base case condition to if (n == 1) instead of if (n == 0)
B. Add else before recursive calls
C. Change printf to print n-1 instead of n
D. Swap aux_rod and to_rod in recursive calls

Solution

  1. Step 1: Analyze base case condition

    Base case should stop recursion when n == 1, not 0, to print the last move.
  2. Step 2: Understand effect of current base case

    With n == 0, no moves are printed for the smallest disk, causing no output.
  3. Final Answer:

    Change base case condition to if (n == 1) instead of if (n == 0) -> Option A
  4. Quick Check:

    Base case must be n == 1 to print moves [OK]
Hint: Base case must be n == 1 to print last move [OK]
Common Mistakes:
  • Using n == 0 as base case
  • Changing print statement unnecessarily
  • Swapping rods incorrectly
5. You want to solve Tower of Hanoi for 4 disks but print only the moves where disk 1 is moved. Which modification to the recursive function is correct?
void towerOfHanoi(int n, char from_rod, char to_rod, char aux_rod) {
  if (n == 1) {
    printf("Move disk 1 from rod %c to rod %c\n", from_rod, to_rod);
    return;
  }
  towerOfHanoi(n-1, from_rod, aux_rod, to_rod);
  printf("Move disk %d from rod %c to rod %c\n", n, from_rod, to_rod);
  towerOfHanoi(n-1, aux_rod, to_rod, from_rod);
}
hard
A. Remove all recursive calls except for n == 1 case
B. Add condition to print only if n == 1 inside the function
C. Print all moves but filter disk 1 moves outside the function
D. Change base case to n == 0 and print only disk 1 moves

Solution

  1. Step 1: Understand requirement to print only disk 1 moves

    We must print moves only when disk number is 1 inside the recursive function.
  2. Step 2: Modify print statements accordingly

    Keep recursive calls intact but print only when n == 1, skipping other prints.
  3. Final Answer:

    Add condition to print only if n == 1 inside the function -> Option B
  4. Quick Check:

    Print condition inside recursion for disk 1 only [OK]
Hint: Print moves only when disk number is 1 inside recursion [OK]
Common Mistakes:
  • Removing recursive calls breaks solution
  • Filtering outside loses recursion context
  • Changing base case incorrectly
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10 questions~5 mineasy hard