Minimize replacements or swapping of same indexed characters required to make two given strings palindromic
- Replace any character of the strings to any other character([a – z]).
- Swap any two characters present at the same index in both the strings.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
Input: str1 = “abbd”, str2 = “dbca”
Swapping (str1, str2) modifies strings str1 to “dbbd” and str2 to “abca”
Replacing str2 to ‘c’ modifies string str2 to “acca”.
Therefore, after above 2 operations, strings str1 and str2 become palindromic.
Input: str1 = “geeksforgeeks”, str2 = “geeksforgeeks”
Approach: Follow the steps below to solve the problem:
- Initialize two variables, say i = 0 and j = 0 to store the index of left pointer and right pointer of both the strings respectively.
- Initialize a variable, say cntOp to store the count of minimum operations required to make both the strings palindromic string.
- Traverse both the string and check the following conditions:
- If str1[i] == str1[j] and str2[i] != str2[j] then replace the value of str2[i] to str2[j] and increment the value of cntOp by 1.
- If str1[i] != str1[j] and str2[i] == str2[j] then replace the value of str1[i] to str1[j] and increment the value of cntOp by 1.
- If str1[i] != str1[j] and str2[i] != str2[j] then check the if the value of (str1[i] == str2[j] and str2[i] == str1[j]) equal to true or not. If found to be true then swap(str1[i], str2[j]) and increment the value of cntOp by 1.
- Otherwise, replace str1[i] to str1[j], str2[i] to str2[j] and increment the value of cntOp by 2.
- Finally, print the value of cntOp.
Below is the implementation of the above approach:
Time Complexity: O(N)
Auxiliary Space: O(1)