CODEWITHSUNDEEP
algorithmdynamic-programmingmemoization
ddx
ddx

Reputation: 520

Time and space complexity of dynamic programming algorithm

This algorithm is from Cracking the Coding Interview, 5th Edition, found here: https://www.geekbooks.me/book/view/cracking-the-coding-interview-5th-edition

A child is running up a staircase with n steps, and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs. Algorithm:

 public static int countWaysDP(int n, int[] map) {
    if (n < 0) {
        return 0;
    } else if (n == 0) {
        return 1;
    } else if (map[n] > -1) {
        return map[n];
    } else {
        map[n] = countWaysDP(n - 1, map) + 
                 countWaysDP(n - 2, map) + 
                 countWaysDP(n - 3, map);
        return map[n];
        }
}

What is the time and space complexity of this algorithm? I think since memoization is used, the results are stored so values don't get calculated multiple times like in the pure recursive method. Since there are three calls to countWaysDP the time complexity is O(3n) which is an element of O(n). The space complexity would be O(n+n) one n for the size of map and one n for the recursive call stack, which is also an element of O(n). Is my reasoning correct?

Thanks

Upvotes: 4

Views: 12419

Answers (3)

User_Targaryen
User_Targaryen

Reputation: 4225

A simple solution using memoization (O(N) time complexity and space complexity):

public static int countWaysDP(int n, int[] map) {
    map[1] = 1; // when n = 1, answer is 1
    map[2] = 2; // when n = 2, answer is 2, (1+1) and (2)
    map[3] = 4; // (1+1+1), (1+2), (2+1), (3)

    for(int i = 4;i <= n; i++)
    {
        map[i] = map[i-1] + map[i-2] + map[i-3]; 
    }

    return map[n];
}

Hope this helps!!!

Upvotes: 0

Shasha99
Shasha99

Reputation: 1916

Let us execute the code:

Note the notation of recursion stack. 1.2.3. means third recursion countWaysDP(n-5) of second recursion countWaysDP(n-2) of countWaysDP(n).

Consider n=6

1. countWaysDP(6)
1.1. countWaysDP(5)
1.1.1. countWaysDP(4)
1.1.1.1. countWaysDP(3)
1.1.1.1.1. countWaysDP(2)
1.1.1.1.1.1. countWaysDP(1)

1.1.1.1.1.1.1. countWaysDP(0) //returns 1
1.1.1.1.1.1.2. countWaysDP(-1) //returns 0
1.1.1.1.1.1.3. countWaysDP(-2) //returns 0                        -> map[1] = 1 <-

1.1.1.1.1.2. countWaysDP(0) //return 1
1.1.1.1.1.3. countWaysDP(-1) //return 0                           -> map[2] = 2 <-

1.1.1.1.2. countWaysDP(1) //already claculated, returns map[1]
1.1.1.1.3. countWaysDP(0) //returns 1                             ->map[3] = 4 <-

1.1.1.2. countWaysDP(2) //already present, returns map[2]
1.1.1.3. countWaysDP(1) //already present, returns map[1]         -> map[4] = 7 <-

1.1.2. countWaysDP(3) //already present, returns map[3]
1.1.3. countWaysDP(2) //already present, returns map[2]           -> map[5] = 13 <-

1.2. countWaysDP(4) //already present, returns map[4]
1.3. countWaysDP(3) //already present, returns map[3]             -> map[6] = 24 <-

Now assume that involking a method is O(1) operation. The total time taken for this example would be:

6 + 3 + (2 + 2 + 2 + 2 + 2) = 19

So yes, you are correct about the TIME. Its 3n as the leftmost recursion path is taking O(n) and then all other calls are O(2n).

The recursion stack would take O(n) as the maximum stack depth is n + 3 and your map will take O(n) space. So yes again, the SPACE is O(n + n) = O(n).

Upvotes: 3

Erik
Erik

Reputation: 3048

Regarding your algorithm: Although you store results in your map, your algorithms does 3 recursive call each time and only afterwards inserts the solution into the map. This means you don't reuse intermediate results.

In Order to reuse you would need to start with n=1 and then iterate until you reach your desired number of steps. This way you can be sure to reuse the result for all smaller steps:

for (int i = 1; i <= n; i++) {
  int current = map[i-1]; // you can make 1 step
  if (i > 1) 
    current += map[i-2]; // you can make 2 steps
  if (i > 2)
    current += map[i-3]; // you can make 3 steps
  map[i] = current;
}
return map[n];

Now for the complexity:

We use one map that at the end has exactly n entries. Therefore the space complexity is O(n).

We iterate once from 1 to n to make the calculation. Therefore the time complexity is also O(n).

Upvotes: 0

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