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Home/ Questions/Q 7983371
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T10:56:22+00:00 2026-06-04T10:56:22+00:00

Is there a solution for Towers of Hanoi whose running time is less than

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Is there a solution for Towers of Hanoi whose running time is less than O(2n) where n is the number of disks to move? My solution takes O(2n) time.

Also, the below solution is with recursion. Can we use Dynamic Programming with the concept of memoization to solve this in a lesser time?

public void towersOfHanoi(
        int num, 
        MyStack<Integer> from,
        MyStack<Integer> to, 
        MyStack<Integer> spare
) {
    if (num == 1) {
        int i = from.pop();
        to.push(i);
        System.out.println("Move "+i+" from "+from.getName()+" to " + to.getName());
        return;
    }
    towersOfHanoi(num - 1, from, spare, to);
    towersOfHanoi(1, from, to, spare);
    towersOfHanoi(num - 1, spare, to, from);
}

MyStack is an extended version of Stack class in Java that adds a name field and accessor.

Also, are there any variations of the same problem?

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  1. Editorial Team
    Editorial Team
    2026-06-04T10:56:23+00:00Added an answer on June 4, 2026 at 10:56 am

    Given that solving Towers of Hanoi always takes 2^n – 1 steps…no, you’re not going to find a faster algorithm, because it takes O(2^n) just to print out the steps, much less compute them.

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