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Program to find smallest value of K for K-Similar Strings in Python
Suppose we have two strings s and t. These two strings are K-similar if we can swap the positions of two letters in s exactly K times so that the resulting string is t. So, we have two anagrams s and t, we have to find the smallest K for which s and t are K-similar.
So, if the input is like s = "abc" t = "bac", then the output will be 1.
To solve this, we will follow these steps −
Define a function neighbors() . This will take new_data
for each index i and value c in new_data, do
if c is not same as t[i], then
come out from loop
for j in range i+1 to size of new_data - 1, do
if u[j] is same as t[i], then
exchange new_data[i] new_data[j]
generate a string by joining all elements from new_data and return, from next call, start again from this area
exchange new_data[i] new_data[j]
From the main method, do the following:
q := make one queue ans insert pair (s, 0)
seen := a new set from s
while q is not empty, do
(u, swap_cnt) := first item of q and delete it from q
if u is same as t, then
return swap_cnt
for each v in neighbors(u as list), do
if v is not in seen, then
mark v as seen
insert (v, swap_cnt+1) at the end of q
return 0
Example
Let us see the following implementation to get better understanding
from collections import deque
def solve(s, t):
def swap(data, i, j):
data[i], data[j] = data[j], data[i]
def neighbors(new_data):
for i, c in enumerate(new_data):
if c != t[i]: break
for j in range(i+1, len(new_data)):
if u[j] == t[i]:
swap(new_data, i, j)
yield "".join(new_data)
swap(new_data, i, j)
q = deque([(s, 0)])
seen = set(s)
while q:
u, swap_cnt = q.popleft()
if u == t:
return swap_cnt
for v in neighbors(list(u)):
if v not in seen:
seen.add(v)
q.append((v, swap_cnt+1))
return 0
s = "abc"
t = "bac"
print(solve(s, t))Input
"abc, bac"
Output
1
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