Python 3:

def ashtonString(s, k): 
    # # Write your code here. 
    kv = k - 1 
    N = len(s) 
    sr = [[0 for _ in range(N)] for __ in range(17)] 
    sr[0] = [ord(c)-97 for c in s]
    L = [0]*N
    cnt = 1
    kf = lambda x: x[0]*(N+1) + x[1]
    for i in range(1, 17):
        for j in range(N):
            L[j] = (sr[i-1][j], sr[i-1][j+cnt] if j+cnt < N else -1, j)
        L.sort(key=kf)

        sr[i][L[0][2]] = 0
        cr = 0
        for j in range(1, N):
            if L[j-1][0] != L[j][0] or L[j-1][1] != L[j][1]:
                cr += 1
            sr[i][L[j][2]] = cr
        cnt *= 2
        if cnt >= N:
            break

    sa = [l[2] for l in L]
    rank = [0]*N
    lcp = [0]*N

    for i in range(N):
        rank[sa[i]] = i

    k = 0
    for i in range(N):
        if rank[i] == N-1:
            k = 0
            continue
        j = sa[rank[i] + 1]
        while j + k < N and i + k < N and s[i+k] == s[j+k]:
            k += 1
        lcp[rank[i]] = k
        if k > 0:
            k -= 1

    numprev = 0
    tri = lambda x: ((x+1)*x)>>1
    print(sa)
    print(lcp)
    for i in range(N):
        mylen = N - sa[i]
        tt = tri(mylen) - tri(numprev)
        if kv < tt:
            for j in range(1 + numprev, 1 + mylen):
                if kv < j:
                    return s[sa[i]+kv]
                kv -= j
        kv -= tt
        numprev = lcp[i]

    return ''

Here is my solution in java, javascript, python, C, C++, Csharp HackerRank Ashton and String Problem Solution

solution in python it passes all test cases

def ashtonString(s, k): kv = k - 1 N = len(s) sr = [[0 for _ in range(N)] for __ in range(17)] sr[0] = [ord(c)-97 for c in s]

L = [0]*N
cnt = 1
kf = lambda x: x[0]*(N+1) + x[1]
for i in range(1, 17):
    for j in range(N):
        L[j] = (sr[i-1][j], sr[i-1][j+cnt] if j+cnt < N else -1, j)
    L.sort(key=kf)

    sr[i][L[0][2]] = 0
    cr = 0
    for j in range(1, N):
        if L[j-1][0] != L[j][0] or L[j-1][1] != L[j][1]:
            cr += 1
        sr[i][L[j][2]] = cr
    cnt *= 2
    if cnt >= N:
        break

sa = [l[2] for l in L]
rank = [0]*N
lcp = [0]*N

for i in range(N):
    rank[sa[i]] = i

k = 0
for i in range(N):
    if rank[i] == N-1:
        k = 0
        continue
    j = sa[rank[i] + 1]
    while j + k < N and i + k < N and s[i+k] == s[j+k]:
        k += 1
    lcp[rank[i]] = k
    if k > 0:
        k -= 1

numprev = 0
tri = lambda x: ((x+1)*x)>>1
print(sa)
print(lcp)
for i in range(N):
    mylen = N - sa[i]
    tt = tri(mylen) - tri(numprev)
    if kv < tt:
        for j in range(1 + numprev, 1 + mylen):
            if kv < j:
                return s[sa[i]+kv]
            kv -= j
    kv -= tt
    numprev = lcp[i]

return ''

solution in python it passes all test cases

def ashtonString(s, k): # # Write your code here. # kv = k - 1 N = len(s) sr = [[0 for _ in range(N)] for __ in range(17)] sr[0] = [ord(c)-97 for c in s]

L = [0]*N
cnt = 1
kf = lambda x: x[0]*(N+1) + x[1]
for i in range(1, 17):
    for j in range(N):
        L[j] = (sr[i-1][j], sr[i-1][j+cnt] if j+cnt < N else -1, j)
    L.sort(key=kf)

    sr[i][L[0][2]] = 0
    cr = 0
    for j in range(1, N):
        if L[j-1][0] != L[j][0] or L[j-1][1] != L[j][1]:
            cr += 1
        sr[i][L[j][2]] = cr
    cnt *= 2
    if cnt >= N:
        break

sa = [l[2] for l in L]
rank = [0]*N
lcp = [0]*N

for i in range(N):
    rank[sa[i]] = i

k = 0
for i in range(N):
    if rank[i] == N-1:
        k = 0
        continue
    j = sa[rank[i] + 1]
    while j + k < N and i + k < N and s[i+k] == s[j+k]:
        k += 1
    lcp[rank[i]] = k
    if k > 0:
        k -= 1

numprev = 0
tri = lambda x: ((x+1)*x)>>1
print(sa)
print(lcp)
for i in range(N):
    mylen = N - sa[i]
    tt = tri(mylen) - tri(numprev)
    if kv < tt:
        for j in range(1 + numprev, 1 + mylen):
            if kv < j:
                return s[sa[i]+kv]
            kv -= j
    kv -= tt
    numprev = lcp[i]

return ''

Here is the solution of Ashton and String Click Here