# Check for Palindrome after every character replacement Query

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: “YES”
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 “seses” which is
palindrome so output “YES”

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”

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 C++ implementation of above idea

 `// C++ program to find if string becomes palindrome ` `// after every query. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if string is Palindrome or Not ` `bool` `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 program ` `int` `main() ` `{ ` `    ``char` `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) (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

 `// C++/c program check if given string is palindrome ` `// or not after every query ` `#include ` `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> 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 "``` `: cout << ````"NO "````; ` `    ``} ` `} ` ` `  `// Driver program ` `int` `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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

## tags:

Hash Strings Hash Strings

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