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Program to find out the substrings of given strings at given positions in a set of all possible substrings in python
Suppose we are provided n number of strings; str1, str2, str3,.....,strn. Now, let's suppose that substri is a set that contains all the substrings of stri. The union of all the substr sets is substr_union. We now are given q number of queries, and we have to find the q-th element of the set substr_union. The set substr_union is lexicographically sorted and the indexes start from 1.
So, if the input is like list of strings are = ['pqr', 'pqt'], queries are = [4, 7, 9], then the output will be ['pqt', 'qt', 't']
The substrings from the first string are subs_str_1 = {p, pq, pqr, q, qr, r }, sub_str_2 = {p, pq, pqt, q, qt, t}.
The union of these two sets, or substr_union is {p, pq, pqr, pqt, q, qr, qt, r, t}.
So the items at index 4, 7, and 9 are 'pqt', qt', and 't' respectively.
To solve this, we will follow these steps −
- Define a function lng_i() . This will take suff, lng, i
- d := a new tuple containing (suff, lng)
- lo := 0
- hi := 0
- for each tuple (suf, lng) in d, do
- if lng is similar as null, then
- lng := 0
- hi := hi + size of suf - lng
- if hi - 1 is same as i, then
- return suf
- otherwise when hi - 1 > i, then
- for index p and item q in list of values from lng to size of suf, do
- if lo + p is same as i, then
- return suf[from index 0 to j+1]
- if lo + p is same as i, then
- for index p and item q in list of values from lng to size of suf, do
- lo := hi
- if lng is similar as null, then
- return False
- Define a function hlp_ii() . This will take str1,str2
- ub := minimum of size of str1 , size of str2
- cnt := 0
- for i in range 0 to ub, do
- if str1[i] is same as str2[i], then
- cnt := cnt + 1
- otherwise,
- return cnt
- return cnt
- if str1[i] is same as str2[i], then
- t_dict := a new map
- suff := a new list
- lng := a new list
- for each str in strings, do
for i in range 0 to size of str, do
- value := str[from index i to end]
- if value is not present in t_dict, then
- t_dict[value] := 1
- insert value at the end of suff
- sort the list suff
- suff_len := size of suff
- for i in range 0 to size of suff_len, do
- if i is same as 0, then
- insert null at the end of lng
- otherwise,
- insert hlp_ii(suff[i-1], suff[i]) at the end of lng
- if i is same as 0, then
- res := a new list
- for each q in q_list, do
- insert (lng_i(suff, lng, q-1)) at the end of res
- return res
Example
Let us see the following implementation to get better understanding −
def lng_i(suff, lng, i):
d = zip(suff,lng)
lo = hi = 0
for suf, lng in d:
if lng is None:
lng = 0
hi += len(suf) - lng
if hi - 1 == i:
return suf
elif hi - 1 > i:
for p, q in enumerate(list(range(lng, len(suf)))):
if lo + p == i:
return suf[:q+1]
lo = hi
return False
def hlp_ii(str1,str2):
ub = min(len(str1), len(str2))
cnt = 0
for i in range(ub):
if str1[i] == str2[i]:
cnt += 1
else:
return cnt
return cnt
def solve(strings,q_list):
t_dict = {}
suff = []
lng = []
for str in strings:
for i in range(len(str)):
value = str[i:]
if value not in t_dict:
t_dict[value] = 1
suff.append(value)
suff.sort()
suff_len = len(suff)
for i in range(suff_len):
if i == 0:
lng.append(None)
else:
lng.append(hlp_ii(suff[i-1], suff[i]))
res = []
for q in q_list:
(res.append(lng_i(suff, lng, q-1)))
return res
print(solve(['pqr', 'pqt'], [4, 7, 9]))Input
['pqr', 'pqt'], [4, 7, 9]
Output
['pqt', 'qt', 't']
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