Check for Palindrome after every character replacement Query - GeeksforGeeks

10 Jul, 2021
Given a string str and Q queries. Each query contains a pair of integers (i1, i2) and a character ‘ch’. We need to replace characters at indexes i1 and i2 with new character ‘ch’ and then tell if string str is palindrome or not. (0 <= i1, i2 < string_length)Examples:  
Input : str = "geeks"  Q = 2
        query 1: i1 = 3 ,i2 = 0, ch = 'e'
        query 2: i1 = 0 ,i2 = 2, ch = 's'
Output : query 1: "NO"
         query 2: "NO"
Explanation :
        In query 1 : i1 = 3 , i2 = 0 ch = 'e'
                    After replacing char at index i1, i2
                    str[3] = 'e', str[0] = 'e'
                    string become "eeees" which is not
                    palindrome so output "NO"
        In query 2 : i1 = 0 i2 = 2  ch = 's'
                    After replacing char at index i1 , i2
                     str[0] = 's', str[2] = 's'
                    string become "sesks" which is
                    palindrome so output "NO"

Input : str = "jasonamat"  Q = 3
        query 1: i1 = 3, i2 = 8 ch = 'j'
        query 2: i1 = 2, i2 = 6 ch = 'n'
        query 3: i1 = 3, i2 = 7 ch = 'a'
Output :
       query 1: "NO"
       query 2: "NO"
       query 3: "YES"
 
A Simple solution is that for each query , we replace character at indexes (i1 & i2) with a new character ‘ch’ and then check if string is palindrome or not.Below is implementation of above idea  
C++
Python3
// C++ program to find if string becomes palindrome// after every query.#include<bits/stdc++.h>using namespace std; // Function to check if string is Palindrome or Notbool IsPalindrome(string &str){    int n = strlen(str);    for (int i = 0; i < n/2 ; i++)        if (str[i] != str[n-1-i])            return false;    return true;} // Takes two inputs for Q queries. For every query, it// prints Yes if string becomes palindrome and No if not.void Query(string &str, int Q){    int i1, i2;    char ch;     // Process all queries one by one    for (int q = 1 ; q <= Q ; q++ )    {        cin >> i1 >> i2 >> ch;         // query 1: i1 = 3 ,i2 = 0, ch = 'e'        // query 2: i1 = 0 ,i2 = 2 , ch = 's'        // replace character at index i1 & i2 with new 'ch'        str[i1] = str[i2] = ch;         // check string is palindrome or not        (isPalindrome(str)== true) ? cout << "YES" << endl :                                     cout << "NO" << endl;    }} // Driver programint main(){    char str[] = "geeks";    int Q = 2;    Query(str, Q);    return 0;}
# Python3 program to find if# string becomes palindrome# after every query. # Function to check if string# is Palindrome or Notdef isPalindrome(string: list) -> bool:    n = len(string)    for i in range(n // 2):        if string[i] != string[n - 1 - i]:            return False    return True # Takes two inputs for Q queries.# For every query, it prints Yes# if string becomes palindrome# and No if not.def Query(string: list, Q: int) -> None:     # Process all queries one by one    for i in range(Q):         # To get space separated        # input from user        inp = list(input().split())         # parsing user inputs as integers        # and strings/char        i1 = int(inp[0])        i2 = int(inp[1])        ch = inp[2]         # query 1: i1 = 3 ,i2 = 0, ch = 'e'        # query 2: i1 = 0 ,i2 = 2 , ch = 's'        # replace character at index        # i1 & i2 with new 'ch'        string[i1] = string[i2] = ch         # check string is palindrome or not        if isPalindrome(string):            print("Yes")        else:            print("No") # Driver Codeif __name__ == "__main__":    string = list("geeks")    Q = 2    Query(string, Q) # This code is contributed by# sanjeev2552
Input:  
3 0 e
0 2 s
Output:  
"NO"
"YES"
Time complexity O(Q*n) (n is length of string )An efficient solution is to use hashing. We create an empty hash set that stores indexes that are unequal in palindrome (Note: ” we have to store indexes only first half of string that are unequal “).  
Given string "str" and length 'n'.
Create an empty set S and store unequal indexes in first half.
Do following for each query :
   1. First replace character at indexes i1 & i2 with 
      new char "ch"

   2. If i1 and/or i2 are/is greater than n/2 then convert 
      into first half index(es)

   3. In this step we make sure that S contains maintains 
      unequal indexes of first half.
      a) If str[i1] == str [n - 1 - i1] means i1 becomes 
         equal after replacement, remove it from S (if present)
         Else add i1 to S 
      b) Repeat step a) for i2 (replace i1 with i2)  

   4. If S is empty then string is palindrome else NOT
Below is C++ implementation of above idea  
CPP
// C++/c program check if given string is palindrome// or not after every query#include<bits/stdc++.h>using namespace std; // This function makes sure that set S contains// unequal characters from first half. This is called// for every character.void addRemoveUnequal(string &str, int index, int n,                              unordered_set<int> &S){    // If character becomes equal after query    if (str[index] == str[n-1-index])    {        // Remove the current index from set if it        // is present        auto it = S.find(index);        if (it != S.end())            S.erase(it) ;    }     // If not equal after query, insert it into set    else        S.insert(index);} // Takes two inputs for Q queries. For every query, it// prints Yes if string becomes palindrome and No if not.void Query(string &str, int Q){    int n = str.length();     // create an empty set that store indexes of    // unequal location in palindrome    unordered_set<int> S;     // we store indexes that are unequal in palindrome    // traverse only first half of string    for (int i=0; i<n/2; i++)        if (str[i] != str[n-1-i])            S.insert(i);     // traversal the query    for (int q=1; q<=Q; q++)    {        // query 1: i1 = 3, i2 = 0, ch = 'e'        // query 2: i1 = 0, i2 = 2, ch = 's'        int i1, i2;        char ch;        cin >> i1 >> i2 >> ch;         // Replace characters at indexes i1 & i2 with        // new char 'ch'        str[i1] = str [i2] = ch;         // If i1 and/or i2 greater than n/2        // then convert into first half index        if (i1 > n/2)            i1 = n- 1 -i1;        if (i2 > n/2)            i2 = n -1 - i2;         // call addRemoveUnequal function to insert and remove        // unequal indexes        addRemoveUnequal(str, i1 , n, S );        addRemoveUnequal(str, i2 , n, S );         // if set is not empty then string is not palindrome        S.empty()? cout << "YES\n" : cout << "NO\n";    }} // Driver programint main(){    string str = "geeks";    int Q = 2 ;    Query(str, Q);    return 0;}
Input:
3 0 e
0 2 s
Output:  
"NO"
"YES"
Time Complexity : O(Q + n) under the assumption that set insert, delete and find operations take O(1) time.This article is contributed by Nishant_Singh(Pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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