Check for Palindrome after every character replacement Query

3.5

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\n" : cout << "NO\n";
    }
}

// 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.

This article is contributed by Nishant_Singh(Pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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