# Program to count number of operations needed to make string as concatenation of same string twice in Python

Suppose we have a lowercase string s. Now consider an operation, where we can delete, insert or update any character in s. We have to count minimum number of operations required to make s = (t concatenate t) for any string t.

So, if the input is like s = "pqrxqsr", then the output will be 2, because, we can update the "x" with "p" and delete "s", then s is "pqrpqr", this is s = t concatenate t, for t = "pqr".

To solve this, we will follow these steps −

• Define a function edit_distance(). This will take s1, s2
• m := size of s1
• n := size of s2
• cur := a new list from range 0 to n
• for i in range 0 to m - 1, do
• prev := cur
• cur := a list with (i + 1) and then n number of 0s
• for j in range 0 to n - 1, do
• cur[j + 1] := prev[j] if s1[i] and s2[j] are same otherwise (minimum of cur[j], prev[j], prev[j + 1]) + 1
• return cur[n]
• From the main method, do the following −
• res := size of s
• for i in range 0 to size of s - 1, do
• res := minimum of edit_distance(substring of s from index 0 to i - 1, substring of s from index i to end) and res
• return res

## Example

Let us see the following implementation to get better understanding −

def solve(s):
def edit_distance(s1, s2):
m, n = len(s1), len(s2)
cur = list(range(n + 1))
for i in range(m):
prev, cur = cur, [i + 1] + [0] * n
for j in range(n):
cur[j + 1] = (prev[j]
if s1[i] == s2[j] else min(cur[j], prev[j], prev[j + 1]) + 1)
return cur[n]

res = len(s)
for i in range(len(s)):
res = min(edit_distance(s[:i], s[i:]), res)
return res

s = "pqrxqsr"
print(solve(s))

## Input

"pqrxqsr"

## Output

None