# Minimum operations to make strings equal

Contributed by
Ekansh Saxena
Last Updated: 23 Feb, 2023
Medium
0/80
Avg time to solve 10 mins
Success Rate 80 %
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## Problem Statement

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#### You have been given two strings A and B consisting of lower case English letters. The task is to count the minimum number of pre-processing moves on the string A required to make it equal to string B after applying below operations:

``````1. Choose any index i (0 <= i < n) and swap characters a[i]  and b[i].
2. Choose any index i (0 <= i < n) and swap characters a[i]  and a[n-i-1] .
3. Choose any index i (0 <= i < n) and swap characters b[i]  and b[n-i-1] .
``````

#### In one preprocess move, you can replace a character in A with any other character of the English alphabet.

##### Note:
``````1. The number of changes you make after the preprocess move does not matter.
2. You cannot apply to preprocess moves to the String B or make any preprocess moves after the first change is made.
``````
Detailed explanation ( Input/output format, Notes, Images )
##### Constraints:
``````1 <= T <= 100
1 <= N <= 5000

Where ‘N’ is the length of the strings.

Time Limit: 1 sec
``````
##### Sample Input 1:
``````2
abacaba
bacabaa
zcabd
dbacz
``````
##### Sample Output 1:
``````4
0
``````
##### Explanation of Sample Input 1:
``````For the first test case, preprocess moves are as follows: A[0] = ‘b’, A[2] = ‘c’, A[3] = ‘a’ and A[4] = ‘b’. Afterwards, A = ‘“bbcabba”. Then we can obtain equal strings by the following sequence of changes: swap(A[2], B[2]) and swap(A[2], A[6]). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make strings equal, so the answer in this test case is 4.

For the second test case, no preprocess moves are required. We can use the following sequence of changes to make A and B equal: swap(B[1], B[5]), swap(A[2], A[4]).
``````
##### Sample Input 2:
``````3
zxyyxzx
zyzyxyy
jfhjfl
jhkkjs
abcd
abcde
``````
##### Sample Output 2:
``````3
4
-1
``````
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