Find the kth eleven-non-free number

Question

Let\'s define eleven-non-free numbers:

If we consider a number as a string, then if any substring inside is a (non-zero) power of 11, then this

Answer 1

This looks like an O(klog^2(k)) algorithm:

def add(vals, new_vals, m, val):
    if val < m and val not in vals and val not in new_vals:
        new_vals.append(val)

def kth11(k):
    vals = [ 11**i for i in range(1, k+1) ]
    new_vals = [ ]
    while True:
        vals = sorted(vals + new_vals)
        vals = vals[:k]
        m = vals[-1]
        new_vals = [ ] 
        for v in vals:
            for s in range(1, 10):
                val = int(str(s) + str(v))
                add(vals, new_vals, m, val)
            for s in range(0, 10):
                val = int(str(v) + str(s))
                add(vals, new_vals, m, val)
        if len(new_vals) == 0: break
    return vals[-1]

print kth11(130)