Auto comment: topic has been updated by weily (previous revision, new revision, compare).

Auto comment: topic has been updated by weily (previous revision, new revision, compare).

Edit for completeness: In the original post, I mistakenly assumed that the calls are $$$O(\log n)$$$ instead of $$$O(\log \log n)$$$.

Assume that we traverse the branch $$$\mathrm{solve}(c)$$$ first when given an input $$$n$$$. Let $$$k$$$ be the maximum integer such that $$$k^3 = n$$$. The main branch will go through all perfect cubes and resulting in calls of $$$\mathrm{solve}(n - k^3)$$$ and $$$\mathrm{solve}((i+1)^3 - 1 - i^3) = \mathrm{solve}(3i^2 + 3i)$$$ for $$$i < k$$$ (the lower bound depends on your preprocessing).

This is a very rough sketch, but notice that the calls to $$$\mathrm{solve}(3i^2 + 3i)$$$ will always be present in any call of $$$\mathrm{solve}$$$. Call these important calls. There are $$$O(\sqrt[3]{n})$$$ of them.

Assume that we have precomputed all of them. How many calls will $$$\mathrm{solve}(n - k^3)$$$ perform? Well, note that $$$n - k^3 \le (k+1)^3 - k^3 = O(k^2)$$$. Therefore, ignoring the important calls, we see that $$$n$$$ becomes $$$O(n^{2/3})$$$ in one call, thus we will have $$$O(\log \log n)$$$ non-important calls for a value of $$$n$$$.

Now, what do we do with the important calls? We can also assume an upper bound that all the important calls also will produce this amount of calls in precomputing them.

Thus, the number of states visited by $$$\mathrm{solve}$$$ are at most $$$O(\sqrt[3]{n} \log \log n)$$$.

Let's call the original value of n as S.

At each step, we can see clearly that (n-c) will always be less than cbrt(S).

Furthermore, it's also pretty obvious that, for everytime solve(n) is called where n is larger than cbrt(S), n will be a cube (except for the first call).

Since n is always a cube, we can write it as d^3. Since c is also a cube, we can precisely write it as (d-1)^3 as well.

Note that the difference between n and c is O(d^2)=O(n^(2/3)). So, in each step solve(n), we will need to call a number that's O(n^(2/3)) less than n. The number of calls where n>cbrt(S) very roughly works out to be O(n/(n^(2/3)))=O(n^(1/3)).

For all steps where n<cbrt(S), we only need to calculate all of them at most O(cbrt(S)) times. For all steps otherwise, it doesnt really matter since there are only O(cbrt(S)) of them.

So the total complexity is indeed O(cbrt(S)). Of course I made some algebraic gymnastics but I guess an intuition is enough.

EDIT:

I made a mistake when calculating cnt_ops.

I did cnt_ops[i] = cnt_ops[c] + cnt_ops[i - c];

when I should've done cnt_ops[i] = cnt_ops[c] + (c != i - c ? cnt_ops[i - c] : 0);

ORIGINAL COMMENT:

I am not completely sure about the complexity of the function but I would think that it runs in $$$O(n)$$$ time (assuming that the cuberoot function runs in $$$O(1)$$$ time.) This is because when solving for $$$n$$$ you make a call to $$$c = \lfloor \sqrt[3]{n - 1} \rfloor^3$$$. It is easy to prove that $$$c < n$$$ always but it is also true $$$c ~ n$$$. I wrote some code to count the number of operations done when solve(i) is called. I found on printing the results for small values of $$$i$$$ that the number of operations is exactly equal to $$$i$$$ except when $$$i = 0$$$ (base case, one operation). To check for larger values of $$$i$$$ I added an assert to the code.

I ran the code the below for $$$n = 10^5$$$ and it did not terminate with the error message implying that that for $$$n \le 10^5$$$ the runtime of the recursive function is $$$O(n)$$$.

#include "bits/stdc++.h"
using namespace std;

int64_t cube_root(int64_t n) { // rounded down
    if(n == 0) return 0;
    int64_t lo = 1, hi = n;
    while(lo + 1 < hi) {
        int64_t mid = (lo + hi) / 2;
        (mid * mid * mid <= n ? lo : hi) = mid;
    }
    return lo;
}

void CHECK(bool ch, int64_t x, int64_t y) {
    if(ch) return;
    cout << "Fails for " << x << " which takes " << y << " operations" << endl;
    abort(); // terminates program
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int64_t n; cin >> n;
    int64_t ump[n + 1], cnt_ops[n + 1];
    for(int64_t i = 0; i <= n; ++i) {
        // base case
        if(i < 2) {
            ump[i] = i;
            cnt_ops[i] = 1;
        }
        else {
            int64_t p = cube_root(i - 1);
            int64_t c = p * p * p;
            ump[i] = ump[c] + ump[i - c] + (i - c);
            cnt_ops[i] = cnt_ops[c] + cnt_ops[i - c];
        }
        CHECK(i == 0 || i == cnt_ops[i], i, cnt_ops[i]);
    }
}