如何计算DFS算法的时间复杂度？

Question

Like the function below, how to calculate its time complexity? I think it's should be O(m*n)...

int uniquePaths(int m, int n) {
    if (m < 1 || n < 1) return 0;     
    if (m == 1 && n == 1) return 1;      
    return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}

Answer 1

You can model the time complexity with the recursive function T(m,n) = T(m-1, n) + T(m, n-1) , where T(1,1)=1 and T(m,n)=0 whenever min(m,n)<1 .

It looks like T(m,n)=(m+n choose m) . To see this note that (m+n choose n) = (m+n-1 choose m) + (m+n-1 choose m-1) by the recurrence for the binomial coefficients , and that T(m,n) = T(m-1, n) + T(m,n-1) is exactly the same recurrence.

Thus T(m,n) = O((m+n)^n / n!) . (There are many other bounds .)

Answer 2

The Recursion Tree:

To represent a problem with m = M and n = N, let's write it as <M, N> . If we try to draw the recursion tree of this problem:

• At root, we have one problem: <M, N>
• Then, root breaks to two problems: <M-1, N> and <M, N-1>
• And, if we continue down this recursion tree, we will reach to leaves, which will be <0, 0> .

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Tree's Depth:

But, what is the maximum depth possible for this tree? It's ¹ (M + N). Further, each node <M, N> can break to at most 2 paths, viz. <M-1, N> and <M, N-1> .

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Inferring Complexity From Tree:

So, what is the maximum number of leaves possible? (Hint: 2 ^{(M + N)} )

Well, since every node breaks to two nodes, at every level, the number of leaves will be multiplied by 2, starting from 1 at the root.

• So, starting from root, total leaves possible = 2 ^{(M + N - 1)} .
• Multiplying and dividing by 2 gives us, (1 / 2) * 2 ^{(M + N)}
• Getting rid of the constant value (1 / 2) gives us the complexity = 2 ^{(M + N)} !

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Hence, the complexity of your algorithm can be upper bounded to O(2 ^{(m + n)} ).

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^{1 The longest path will start from <M, N> and first go to <0, N> . Then from there it will go to <0, 0> . So, longest possible path length = M + N.}