# Program to Find Out the Smallest Substring Containing a Specific String in Python

Suppose we have two strings s and t. We have to find the smallest substring in s, where t is also a subsequence of the substring. If that type of substring does not exist, we will return a blank string, and if there are multiple smallest substrings, we will take the leftmost one.

So, if the input is like s = "abcbfbghfb", t = "fg", then the output will be fbg

To solve this, we will follow these steps −

• N := size of S

• dp := a new list of size N initialized with infinity

• for i in range 0 to N − 1, do

• if S[i] is same as T, then

• dp[i] := 1

• for j in range 1 to size of T − 1, do

• last := a new map

• dp2 := a new list of size N initialized with infinity

• for i in range 0 to N, do

• if S[i] is same as T[j], then

• prev_i := return the value of T[j − 1] from last

• if prev_i is not null, then

• dp2[i] := dp[prev_i] + (i − prev_i)

• last[S[i]] := i

• dp := dp2

• m := minimum of dp

• i := return the index containing m in dp

• if m is infinity, then

• return blank string

• return S[from index i − dp[i] + 1 to i]

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution:
def solve(self, S, T):
INF = float("inf")
N = len(S)
dp = [INF] * N
for i in range(N):
if S[i] == T:
dp[i] = 1
for j in range(1, len(T)):
last = {}
dp2 = [INF] * N
for i in range(N):
if S[i] == T[j]:
prev_i = last.get(T[j − 1], None)
if prev_i is not None:
dp2[i] = dp[prev_i] + (i − prev_i)
last[S[i]] = i
dp = dp2
m = min(dp)
i = dp.index(m)
if m == INF:
return ""
return S[i − dp[i] + 1 : i + 1]
ob = Solution()
print(ob.solve("abcbfbghfb","fg"))

## Input

"abcbfbghfb","fg"

## Output

fbg