# Check if it is possible to form string B from A under the given constraint in Python

Suppose we have two strings s and t, and two values p and q. We have to check whether it is possible to get t from s such that s is separated into groups of p number of characters except the last group which will have characters ≤ p and we can pick at most q number of characters from each group, and also order of characters in t must be same as s.

So, if the input is like s = "mnonnopeqrst", t = "moprst", p = 5, q = 2, then the output will be True as we can make divisions like "mnonn", "opeqr", "st", Now by taking 2 character substrings "mo" and "pr" from "mnonn" and "opeqr" then "st" is already there so using these two length substrings we can generate t by simple concatenation.

To solve this, we will follow these steps −

• temp := an empty map containing empty list for all keys
• l := a list of size same as length of s and fill with 0
• for i in range 0 to size of s, do
• insert i at the end of temp['a']
• low := 0
• for i in range 0 to size of t - 1, do
• indices := temp['a']
• it := index to insert low in the indices list to maintain sorted order
• if it is same as size of indices - 1, then
• return False
• count := quotient of (indices[it] / p)
• l[count] := l[count] + 1
• if l[count] >= q, then
• count := count + 1
• low := count * p
• otherwise,
• low := indices[it] + 1
• return True

## Example

Let us see the following implementation to get better understanding −

Live Demo

from bisect import bisect_left
from collections import defaultdict
def solve(s, t, b, m) :
temp = defaultdict(list)
l = [0] * len(s)
for i in range(len(s)) :
temp['a'].append(i)
low = 0
for i in range(len(t)) :
indices = temp['a']
it = bisect_left(indices, low)
if it == len(indices) :
return False
count = indices[it] // b
l[count] = l[count] + 1
if l[count] >= m :
count += 1
low = count * b
else :
low = indices[it] + 1
return True
s = "mnonnopeqrst"
t = "moprst"
p = 5
q = 2
print (solve(s, t, p, q))

## Input

"mnonnopeqrst", "moprst", 5, 2

## Output

True