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”
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<bits/stdc++.h> 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<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
" : 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.
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