'Getting permutation index and permutation at index faster than the provided solution

Thanks to this answer, here is how I'm getting permutation index and permutation at an index:

import time


def get_Cl(distinct):
    Cl = []
    for i in range(1, distinct + 1):  # i is distincct
        c = [0] * i + [1, 0]
        C = [c]
        for l in range(2, distinct + 1):
            c = [
                    c[d] * d + c[d + 1] * (distinct - d)
                    for d in range(i + 1)
                ] + [0]
            C.append(c)
        Cl.append(C)
    return Cl


def item_index(item, distinct, n_symbols, Cl):
    length = len(item)
    offset = 0
    seen = set()
    for i, di in enumerate(item):
        for d in range(n_symbols):
            if d == di:
                break
            if d in seen:
                # test = Cl[distinct][length - 1 - i][len(seen)]
                offset += Cl[distinct][length - 1 - i][len(seen)]
            else:
                offset += Cl[distinct][length - 1 - i][len(seen) + 1]
        seen.add(di)
    return offset


def item_at(idx, length, distinct, n_symbols, Cl):
    seen = [0] * n_symbols
    prefix = [0] * length
    used = 0
    for i in range(length):
        for d in range(n_symbols):
            if seen[d] != 0:
                branch_count = Cl[distinct][length - 1 - i][used]
            else:
                branch_count = Cl[distinct][length - 1 - i][used + 1]
            if branch_count <= idx:
                idx -= branch_count
            else:
                prefix[i] = d
                if seen[d] == 0:
                    used += 1
                seen[d] = 1
                break
    return prefix


if __name__ == "__main__":
    start_time = time.time()
    Cl = get_Cl(512)
    end_time = time.time()
    print(f'{(end_time - start_time)} seconds for Cl')
    start_time = time.time()
    item = item_at(idx=432, length=512, distinct=350, n_symbols=512, Cl=Cl)
    end_time = time.time()
    print(f'{(end_time - start_time)} seconds for item_at')
    print(item)
    start_time = time.time()
    print(item_index(item=item, distinct=350, n_symbols=512, Cl=Cl))
    end_time = time.time()
    print(f'{(end_time - start_time)} seconds for item_index')
356.3069865703583 seconds for Cl
2.5428783893585205 seconds for item_at  
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432
0.025868892669677734 seconds for item_index

It works fine unless numbers get bigger, then it gets very slow. Wondered if it is possible to improve this code like this answer that is improved version of the same slow function to calculating all the permutations?

The reason I get Cl in a separate line is that for a fixed distinct there will be thousands of calls on item_at and item_index, so the Cl is the same if distinct is the same thus no need for call it for each item_at or item_index.

Update: Test result from answer

0.008994340896606445 seconds for item_at
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432
0.006995677947998047 seconds for item_index


Sources

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Source: Stack Overflow

Solution Source