Queries on substring palindrome formation

Given a string S, and two type of queries.

Type 1: 1 L x, Indicates update Lth index 
               of string S by x character.
Type 2: 2 L R, Find if characters between position L and R 
               of string S can form a palindrome string. 
               If palindrome can be formed print "Yes", 
               else print "No".
1 <= L, R <= |S| 

Examples:

Input : S = "geeksforgeeks"
Query 1: 1 4 g
Query 2: 2 1 4
Query 3: 2 2 3
Query 4: 1 10 t
Query 5: 2 10 11
Output :
Yes
Yes
No

Query 1: update index 3 (position 4) of string S by 
character 'g'. So new string S = "geegsforgeeks".

Query 2: find if rearrangement between index 0 and 3
can form a palindrome. "geegs" is palindrome, print "Yes".

Query 3: find if rearrangement between index 1 and 2 
can form a palindrome. "ee" is palindrome, print "Yes".

Query 4: update index 9 (position 10) of string S by 
character 't'. So new string S = "geegsforgteks".

Query 3: find if rearrangement between index 9 and 10 
can form a palindrome. "te" is not palindrome, print "No".

Substring S[L…R] form a palindrome only if frequencies of all the character in S[L…R] are even, with one except allowed.

For query of type 1, simply update string 
S[L] by character x.

For each query of type 2, calculate the 
frequency of character and check if 
frequencies of all characters is even (with)
one exception allowed.

Following are two different methods to find frequency of each character in S[L…R]:

Method 1: Use a frequency array to find the frequency of each element in S[L…R].

Below is C++ implementation of this approach:

// C++ program to Queries on substring palindrome
// formation.
#include<bits/stdc++.h>
using namespace std;

// Query type 1: update string position i with
// character x.
void qType1(int l, int x, char str[])
{
    str[l-1] = x;
}

// Print "Yes" if range [L..R] can form palindrome,
// else print "No".
void qType2(int l, int r, char str[])
{
    int freq[27] = { 0 };

    // Find the frequency of each character in
    // S[L...R].
    for (int i = l-1; i<=r-1; i++)
        freq[str[i] - 'a']++;

    // Checking if more than one character have
    // frequency greater than 1.
    int count = 0;
    for (int j = 0; j < 26; j++)
        if (freq[j] % 2)
            count++;

    (count<=1)? (cout << "Yes" << endl):
                 (cout << "No" << endl);
}

// Driven Program
int main()
{
    char str[] = "geeksforgeeks";
    int n = strlen(str);

    qType1(4, 'g', str);
    qType2(1, 4, str);
    qType2(2, 3, str);
    qType1(10, 't', str);
    qType2(10, 11, str);

    return 0;
}

Output:

Yes
Yes
No

 

Method 2 : Use Binary Indexed Tree

The efficient approach can be maintain 26 Binary Index Tree for each alphabet.
Define a function getFrequency(i,u) which returns the frequency of ‘u’ in the ith prefix. Frequency of character ‘u’ in range L…R can be find by getFrequency(R, u) – getFrequency(L-1, u).
Whenever update(Query 1) comes to change S[i] from character ‘u’ to ‘v’. BIT[u] is updated with -1 at index i and BIT[v] is updated with +1 at index i.

Below is C++ implementation of this approach:

// C++ program to Queries on substring palindrome
// formation.
#include <bits/stdc++.h>
#define max 1000
using namespace std;

// Return the frequency of the character in the
// i-th prefix.
int getFrequency(int tree[max][27], int idx, int i)
{
    int sum = 0;

    while (idx > 0)
    {
        sum += tree[idx][i];
        idx -= (idx & -idx);
    }

    return sum;
}

// Updating the BIT
void update(int tree[max][27], int idx, int val, int i)
{
    while (idx <= max)
    {
        tree[idx][i] += val;
        idx += (idx & -idx);
    }
}

// Query to update the character in the string.
void qType1(int tree[max][27], int l, int x, char str[])
{
    // Adding -1 at L position
    update(tree, l, -1, str[l-1]-97+1);

    // Updating the character
    str[l-1] = x;

    // Adding +1 at R position
    update(tree, l, 1, str[l-1]-97+1);
}

// Query to find if rearrangement of charcater in range
// L...R can form palindrome
void qType2(int tree[max][27], int l, int r, char str[])
{
    int count = 0;

    for (int i = 1; i <= 26; i++)
    {
        // Checking on the first charcter of the string S.
        if (l == 1)
        {
            if (getFrequency(tree, r, i)%2 == 1)
                count++;
        }
        else
        {
            // Checking if frequency of character is even or odd.
            if ((getFrequency(tree, r, i) -
                 getFrequency(tree, l-1, i))%2 == 1)
                count++;
        }
    }

    (count<=1)?(cout << "Yes" << endl):(cout << "No" << endl);
}

// Creating the Binary Index Tree of all aphabet
void buildBIT(int tree[max][27], char str[], int n)
{
    memset(tree,0,sizeof(tree));

    for (int i = 0; i < n; i++)
        update(tree, i+1, 1, str[i]-97+1);
}

// Driven Program
int main()
{
    char str[] = "geeksforgeeks";
    int n = strlen(str);

    int tree[max][27];
    buildBIT(tree, str, n);

    qType1(tree, 4, 'g', str);
    qType2(tree, 1, 4, str);
    qType2(tree, 2, 3, str);
    qType1(tree, 10, 't', str);
    qType2(tree, 10, 11, str);

    return 0;
}

Output:

Yes
Yes
No

This article is contributed by Anuj Chauhan. 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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