- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

Suppose we want to make a target string of lowercase letters. At first, we have the sequence as n '?' marks (n is the length of target string). We also have a stamp of lowercase letters. On each turn, we can place the stamp over the sequence, and replace every letter in the with the corresponding letter from that stamp. You can make up to 10 * n turns.

As an example consider the initial sequence is "?????", and the stamp is "abc", then we may make strings like "abc??", "?abc?", "??abc" in the first turn. If the sequence is possible to stamp, then return an array of the index with the left-most letter being stamped at each turn. If that is not possible then return an empty array. So when the sequence is "ababc", and the stamp is "abc", then the answer can be like [0, 2], Because we can form like "?????" -> "abc??" - > "ababc".

So, if the input is like s = "abcd" t = "abcdbcd", then the output will be [3,0]

To solve this, we will follow these steps −

if size of s is same as 1, then

return a list from 0 to t when all characters in t are same and they are s[0], otherwise a new blank list

ans := a new list

while t is not same as size of t number of "?" marks, do

tmp := t

for i in range 0 to size of s, do

for j in size of s down to i+1:

search := i number of "?" concatenate substring of s[from index i to j-1] concatenate (size of s - j)number of "?"

while search is in t, do

insert where search is present in t at the end of ans

t := replace search with size of s number of "?" only once

if t is same as size of t number of "?", then

come out from loop

if t is same as size of t number of "?", then

come out from loop

if tmp is same as t, then

come out from loop

return reverse of ans.

Let us see the following implementation to get better understanding

def solve(s, t): if len(s) == 1: return [i for i in range(len(t))] if all(t==s[0] for t in t)else [] ans = [] while t != "?" * len(t): tmp = t for i in range(len(s)): for j in reversed(range(i+1, len(s)+1)): search = "?" * i + s[i:j] + "?" * (len(s)-j) while t.find(search) != -1: ans.append(t.find(search)) t = t.replace(search, "?"*len(s), 1) if t == "?" * len(t): break if t == "?" * len(t): break if tmp == t: return [] return ans[::-1] s = "abcd" t = "abcdbcd" print(solve(s, t))

"abcd", "abcdbcd"

[3,0]

- Related Questions & Answers
- Number of Squareful Arrays in C++
- Program to find number of sub-arrays with odd sum using Python
- Program to find equal sum arrays with minimum number of operations in Python
- Program to find number of good pairs in Python
- Program to find number of good triplets in Python
- Program to find number of distinct subsequences in Python
- Python program to find common elements in three sorted arrays?
- Python program to find better divisor of a number
- Python program to find factorial of a large number
- Program to find maximum number of eaten apples in Python
- Program to find number of different subsequences GCDs in Python
- Program to find super digit of a number in Python
- Program to find winner of number reducing game in Python
- Program to find number of bit 1 in the given number in Python
- Program to find minimum number of people to teach in Python

Advertisements