There are some perfect squares with a particular property.
For example the number ```n = 256``` is a perfect square, its square root is ```16```. If we change the position of the digits of n, we may obtain another perfect square``` 625``` (square root = 25).
With these three digits ```2```,```5``` and ```6``` we can get two perfect squares: ```[256,625]```

The number ```1354896``` may generate another ```4``` perfect squares, having with the number itself, a total of five perfect squares: ```[1354896, 3594816, 3481956, 5391684, 6395841]```, being the last one in the list, ```6395841```, the highest value of the set.

Your task is to find the first perfect square above the given lower_limit, that can generate the given k number of perfect squares, and it doesn't contain the digit 0. Then return the maximum perfect square that can be obtained from its digits.

Example with the cases seen above:
```
lower_limit = 200
k = 2 (amount of perfect squares)
result = 625

lower_limit = 3550000
k = 5 (amount of perfect squares)
result  = 6395841
```
Features of the random tests:
```
100 <= lower_limit <= 1e6
2 <= k <= 5
number of tests = 45
```
Have a good time!

Difficulty: introductory

Starter code:

def next_perfectsq_perm(lower_limit, k):
	

Solutions:
from itertools import count, permutations

def next_perfectsq_perm(limit_below, k):
    for n in count(int(limit_below**.5)+1):
        s = str(n**2)
        if '0' not in s:
            sq_set = {x for x in (int(''.join(p)) for p in permutations(s)) if (x**.5).is_integer()}
            if len(sq_set) == k:
                return max(sq_set)

from itertools import permutations
from math import sqrt

def is_per(n):
    return str(sqrt(int(n)))[-1] == '0'
def next_perfectsq_perm(lower_limit, k): 
    perfect = [i**2 for i in range(2000) if '0' not in str(i**2) if i**2 > lower_limit]
    for i in perfect: 
        if i > lower_limit:
            num = set()
            per = set(permutations(str(i)))
            tem = [int(''.join(j)) for j in per if is_per(''.join(j))]
            if len(tem) == k:
                num.add(i)
                ma = max(tem)
                break
    return max(tem)

from itertools import permutations
def next_perfectsq_perm(lower_limit, k): 
    while 1:
        lower_limit += 1
        if '0' not in str(lower_limit) and int(lower_limit ** 0.5 + 0.5) ** 2 == lower_limit:
            perms = list([''.join(x) for x in set(permutations(str(lower_limit)))])
            perms = list(map(int, perms))
            perfects = 0
            perfect_list = []
            for i in perms:
                if int(i ** 0.5 + 0.5) ** 2 == i:
                    perfects += 1
                    perfect_list.append(i)
            if perfects == k:
                return max(perfect_list)   


candidates = ((sq, str(sq)) for sq in (i*i for i in range(12, 3000)))
candidates = ((sq, s) for sq, s in candidates if '0' not in s)

d = {}  # 144: [144, 441], 256: [256, 625], ...
sqs = {}  # 144: [144, 441], 256: [256, 625], 441: [144, 441], 625: [256, 625], ...
for sq, s in candidates:
    sqs[sq] = d.setdefault(int(''.join(sorted(s))), [])
    sqs[sq].append(sq)

def next_perfectsq_perm(lower_limit, k): 
    return sqs[min(x for x in sqs if x > lower_limit and len(sqs[x]) == k)][-1]

from itertools import permutations

def is_perfect(square):
    return round(square**0.5)**2 == square

def permuted(n):
    return (int("".join(p)) for p in set(permutations(str(n))))

def perfect_permutations(n):
    return tuple(p for p in permuted(n) if is_perfect(p))
    
def next_perfectsq_perm(lower_limit, k): 
    root = int(lower_limit**0.5) + 1
    while True:
        square = root**2
        if not "0" in str(square):
            p = perfect_permutations(square)
            if len(p) == k:
                return max(p)
        root += 1


from collections import *

NUM_DIG = 8
palindromes = OrderedDict()
for n in range(int(10 ** (NUM_DIG / 2)) + 1):
    key = Counter(str(n ** 2))
    if '0' not in key:
        palindromes.setdefault(
            frozenset(key.most_common()), []).append(n ** 2)

lasts, lengths = {}, defaultdict(list)
for seq in palindromes.values():
    for n in seq:
        lengths[len(seq)].append(n)
        lasts[n] = seq[-1]
for k in lengths: lengths[k].sort()

from bisect import bisect
def next_perfectsq_perm(lower_limit, k):
    return lasts[lengths[k][bisect(lengths[k], lower_limit)]]

K_PERMS = {
    2: [144, 256, 441, 625, 1369, 1764, 1936, 4761, 11236, 12769, 14884, 21316, 24649, 24964,
        27556, 29584, 34596, 36864, 43264, 45369, 46656, 48841, 51984, 54289, 66564, 74529,
        75625, 79524, 86436, 95481, 96721, 99856, 112225, 113569, 118336, 133956, 139876,
        142884, 147456, 148225, 148996, 154449, 155236, 166464, 167281, 169744, 173889, 174724,
        175561, 194481, 198916, 212521, 216225, 224676, 225625, 227529, 233289, 237169, 238144,
        264196, 265225, 267289, 272484, 287296, 293764, 314721, 316969, 323761, 328329, 329476,
        344569, 345744, 355216, 368449, 369664, 374544, 383161, 391876, 398161, 427716, 432964,
        434281, 447561, 456976, 459684, 463761, 467856, 474721, 498436, 521284, 522729, 546121,
        591361, 594441, 611524, 619369, 622521, 627264, 646416, 649636, 654481, 659344, 667489,
        669124, 675684, 678976, 712336, 729316, 736164, 741321, 748225, 755161, 758641, 769129,
        772641, 779689, 781456, 786769, 793881, 795664, 799236, 817216, 835396, 842724, 844561,
        848241, 874225, 877969, 891136, 894916, 919681, 927369, 933156, 948676, 956484, 964324,
        972196, 976144, 9114361],
    3: [961, 9216, 11664, 12544, 16641, 18225, 23716, 25281, 32761, 41616, 42849, 44521, 49284,
        52441, 72361, 81225, 82944, 131769, 141376, 149769, 164836, 165649, 197136, 319225,
        349281, 364816, 378225, 381924, 417316, 471969, 481636, 541696, 837225, 931225, 974169,
        7112889],
    4: [81796, 112896, 132496, 168921, 195364, 214369, 269361, 298116, 346921, 395641, 436921,
        543169, 962361, 4511376],
    5: [43681, 6395841],
    6: [7241481],
    7: [4532641] }


def next_perfectsq_perm(lower_limit, k, answer = 0):
    
    for number in K_PERMS[k]:
        if not answer and number > lower_limit:
            answer = number
        elif answer and sorted(str(number)) == sorted(str(answer)):
            answer = number
    
    return answer

from itertools import permutations, count

def next_perfectsq_perm(lower_limit, k):
    for i in count(int(lower_limit ** 0.5) + 1):
        j = i ** 2
        if '0' in str(j):
            continue
        k_count = 0
        matches = [j]
        perms = set(permutations(str(j))) - set(matches)
        for p in perms:
            num = int(''.join(p))
            if int(num ** 0.5) ** 2 == num:
                matches.append(num)
                k_count += 1
        if k_count == k:
            return sorted(matches)[-1]

M,U = {},[]
def H(Q) :
    R = 0
    while Q :
        if Q % 10 < 1 : return 0
        R += 1 << (Q % 10 * 3)
        Q //= 10
    return R
for F in range(1,10000) :
    F = F * F
    R = H(F)
    if R :
        U.append(F)
        M.setdefault(R,[]).append(F)
def next_perfectsq_perm(Q,S) :
    for V in U :
        if Q < V :
            T = M.get(H(V))
            if T and S == len(T) : return T[-1]

import itertools
def next_perfectsq_perm(l, k):
    perf = lambda n: n**0.5%1 == 0
    while True:
        l += 1
        if perf(l) and '0' not in str(l):
            r = [x for x in set([int(''.join(x)) for x in list(itertools.permutations(str(l)))]) if perf(x)]
            if len(r) == k: return max(r)