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".

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

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 the 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 the implementation of this approach:

C++

 // C++ program to Queries on substring palindrome // formation. #include 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 = { 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; }

Java

 // Java program to Queries on substring // palindrome formation.    class GFG {        // Query type 1: update string     // position i with character x.     static void qType1(int l, int x, char str[])     {         str[l - 1] = (char)x;     }        // Print "Yes" if range [L..R] can form     // palindrome, else print "No".     static void qType2(int l, int r, char str[])     {         int freq[] = new int;            // 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 != 0) {                 count++;             }         }         if (count <= 1) {             System.out.println("Yes");         }         else {             System.out.println("No");         }     }        // Driven code     public static void main(String[] args)     {         char str[] = "geeksforgeeks".toCharArray();         int n = str.length;            qType1(4, 'g', str);         qType2(1, 4, str);         qType2(2, 3, str);         qType1(10, 't', str);         qType2(10, 11, str);     } }    // This code is contributed by 29AjayKumar

Python3

 # Python3 program to Queries on substring  # palindrome formation.    # Query type 1: update str1ing position  # i with character x. def qType1(l, x, str1):     str1[l - 1] = x    # Pr"Yes" if range [L..R] can form palindrome, # else pr"No". def qType2(l, r, str1):        freq = [0 for i in range(27)]         # Find the frequency of      # each character in S[L...R].     for i in range(l - 1, r):         freq[ord(str1[i]) - ord('a')] += 1        # Checking if more than one character      # have frequency greater than 1.     count = 0     for j in range(26):         if (freq[j] % 2):             count += 1     if count <= 1:         print("Yes")     else:         print("No")    # Driver Code str1 = "geeksforgeeks" str2 = [i for i in str1] n = len(str2)    qType1(4, 'g', str2) qType2(1, 4, str2) qType2(2, 3, str2) qType1(10, 't', str2) qType2(10, 11, str2)    # This code is contributed by mohit kumar

C#

 // C# program to Queries on substring // palindrome formation. using System;    class GFG {        // Query type 1: update string     // position i with character x.     static void qType1(int l, int x, char[] str)     {         str[l - 1] = (char)x;     }        // Print "Yes" if range [L..R] can form     // palindrome, else print "No".     static void qType2(int l, int r, char[] str)     {         int[] freq = new int;            // 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 != 0) {                 count++;             }         }         if (count <= 1) {             Console.WriteLine("Yes");         }         else {             Console.WriteLine("No");         }     }        // Driver code     public static void Main(String[] args)     {         char[] str = "geeksforgeeks".ToCharArray();         int n = str.Length;            qType1(4, 'g', str);         qType2(1, 4, str);         qType2(2, 3, str);         qType1(10, 't', str);         qType2(10, 11, str);     } }    // This code contributed by Rajput-Ji

PHP



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 the implementation of this approach:

C++

 // C++ program to Queries on substring palindrome // formation. #include #define max 1000 using namespace std;    // Return the frequency of the character in the // i-th prefix. int getFrequency(int tree[max], 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], 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], 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 character in range // L...R can form palindrome void qType2(int tree[max], int l, int r, char str[]) {     int count = 0;        for (int i = 1; i <= 26; i++) {         // Checking on the first character 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], 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];     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; }

Java

 // Java program to Queries on substring palindrome // formation.    import java.util.*;    class GFG {        static int max = 1000;        // Return the frequency of the character in the     // i-th prefix.     static int getFrequency(int tree[][], int idx, int i)     {         int sum = 0;            while (idx > 0) {             sum += tree[idx][i];             idx -= (idx & -idx);         }            return sum;     }        // Updating the BIT     static void update(int tree[][], int idx,                        int val, int i)     {         while (idx <= max) {             tree[idx][i] += val;             idx += (idx & -idx);         }     }        // Query to update the character in the string.     static void qType1(int tree[][], 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] = (char)x;            // Adding +1 at R position         update(tree, l, 1, str[l - 1] - 97 + 1);     }        // Query to find if rearrangement of character in range     // L...R can form palindrome     static void qType2(int tree[][], int l, int r, char str[])     {         int count = 0;            for (int i = 1; i <= 26; i++) {             // Checking on the first character 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++;             }         }            if (count <= 1)             System.out.println("Yes");         else             System.out.println("No");     }        // Creating the Binary Index Tree of all aphabet     static void buildBIT(int tree[][], char str[], int n)     {            for (int i = 0; i < n; i++)             update(tree, i + 1, 1, str[i] - 97 + 1);     }        // Driver code     public static void main(String[] args)     {         char str[] = "geeksforgeeks".toCharArray();         int n = str.length;            int tree[][] = new int[max];         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);     } }    /* This code contributed by PrinciRaj1992 */

Python3

 # Python3 program to Queries on substr1ing palindrome # formation.    max = 1000;    # Return the frequency of the character in the # i-th prefix. def getFrequency(tree, idx, i):     sum = 0;        while (idx > 0):         sum += tree[idx][i];         idx -= (idx & -idx);        return sum;    # Updating the BIT def update(tree, idx, val, i):     while (idx <= max):         tree[idx][i] += val;         idx += (idx & -idx);    # Query to update the character in the str1ing. def qType1(tree, l, x, str1):        # Adding -1 at L position     update(tree, l, -1, ord(str1[l - 1]) - 97 + 1);        # Updating the character     list1 = list(str1)     list1[l - 1] = x;     str1 = ''.join(list1);        # Adding +1 at R position     update(tree, l, 1, ord(str1[l - 1]) - 97 + 1);    # Query to find if rearrangement of character in range # L...R can form palindrome def qType2(tree, l, r, str1):     count = 0;        for i in range(1, 27):                    # Checking on the first character of the str1ing S.         if (l == 1):             if (getFrequency(tree, r, i) % 2 == 1):                 count+=1;         else:             # Checking if frequency of character is even or odd.             if ((getFrequency(tree, r, i) -                  getFrequency(tree, l - 1, i)) % 2 == 1):                 count += 1;        print("Yes") if(count <= 1) else print("No");    # Creating the Binary Index Tree of all aphabet def buildBIT(tree,str1, n):        for i in range(n):         update(tree, i + 1, 1, ord(str1[i]) - 97 + 1);    # Driver code    str1 = "geeksforgeeks"; n = len(str1);    tree = [[0 for x in range(27)] for y in range(max)]; buildBIT(tree, str1, n);    qType1(tree, 4, 'g', str1); qType2(tree, 1, 4, str1); qType2(tree, 2, 3, str1); qType1(tree, 10, 't', str1); qType2(tree, 10, 11, str1);       # This code is contributed by mits

C#

 // C# program to Queries on substring palindrome // formation. using System;    class GFG {        static int max = 1000;        // Return the frequency of the character in the     // i-th prefix.     static int getFrequency(int [,]tree, int idx, int i)     {         int sum = 0;            while (idx > 0)         {             sum += tree[idx,i];             idx -= (idx & -idx);         }            return sum;     }        // Updating the BIT     static void update(int [,]tree, int idx,                     int val, int i)     {         while (idx <= max)          {             tree[idx,i] += val;             idx += (idx & -idx);         }     }        // Query to update the character in the string.     static void qType1(int [,]tree, 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] = (char)x;            // Adding +1 at R position         update(tree, l, 1, str[l - 1] - 97 + 1);     }        // Query to find if rearrangement of character in range     // L...R can form palindrome     static void qType2(int [,]tree, int l, int r, char []str)     {         int count = 0;            for (int i = 1; i <= 26; i++)          {             // Checking on the first character 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++;             }         }            if (count <= 1)             Console.WriteLine("Yes");         else             Console.WriteLine("No");     }        // Creating the Binary Index Tree of all aphabet     static void buildBIT(int [,]tree, char []str, int n)     {            for (int i = 0; i < n; i++)             update(tree, i + 1, 1, str[i] - 97 + 1);     }        // Driver code     static void Main()     {         char []str = "geeksforgeeks".ToCharArray();         int n = str.Length;            int[,] tree = new int[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);     } }    // This code contributed by mits

Output:

Yes
Yes
No

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