Maximum length palindrome that can be created with characters in range L and R

Given a string str and Q queries. Each query consists of two numbers L and R. The task is to find the maximum length palindrome that can be created with characters in the range [L, R]. 
Examples:

Input: str = “amim”, Q[] = {{1, 4}, {3, 4} 
Output:
3 
1 
In range [1, 4], only two palindromes “mam” and “mim” can be formed. 
In range [3, 4], only “i” or “m” can be created using the characters in range.
Input: str = “aaaaa”, Q[] = {{1, 5}, {5, 5} 
Output:
5 
1 

Approach: Let prefix[i][j] be an array which denotes the frequency of character char(j+97) in range 1 to i. For any range L to R, count the even frequencies and the odd frequencies. Since odd-1 is even, it can also contribute to the palindromic string. Also keep a mark for an odd frequency character, which can be inserted in the middle. Hence the length of the longest palindrome possible will be the sum of all even and the sum of odd-1 frequencies, adding 1 if there exists an odd frequency character. 
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 4

// Function to return the length of the
// longest palindrome that can be formed
// using the characters in the range [l, r]
int performQueries(int l, int r, int prefix[N][26])
{

    // 0-based indexing
    l--;
    r--;

    // Marks if there is an
    // odd frequency character
    bool flag = false;

    // Length of the longest palindrome
    // possible from characters in range
    int count = 0;

    // Traverse for all characters
    // and count their frequencies
    for (int i = 0; i < 26; i++) {

        // Find the frequency in range 1 - r
        int cnt = prefix[r][i];

        // Exclude the frequencies in range 1 - (l - 1)
        if (l > 0)
            cnt -= prefix[l - 1][i];

        // If frequency is odd, then add 1 less than
        // the original frequency to make it even
        if (cnt % 2 == 1) {
            flag = true;
            count += cnt - 1;
        }
        // Else completely add if even
        else
            count += cnt;
    }

    // If any odd frequency character
    // is present then add 1
    if (flag)
        count += 1;

    return count;
}

// Function to pre-calculate the frequencies
// of the characters to reduce complexity
void preCalculate(string s, int prefix[N][26])
{
    int n = s.size();

    // Iterate and increase the count
    for (int i = 0; i < n; i++) {
        prefix[i][s[i] - 'a']++;
    }

    // Create a prefix type array
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < 26; j++)
            prefix[i][j] += prefix[i - 1][j];
    }
}

// Driver code
int main()
{
    string s = "amim";

    // Pre-calculate prefix array
    int prefix[N][26];
    memset(prefix, 0, sizeof prefix);
    preCalculate(s, prefix);

    int queries[][2] = { { 1, 4 }, { 3, 4 } };
    int q = sizeof(queries) / sizeof(queries[0]);

    // Perform queries
    for (int i = 0; i < q; i++) {
        cout << performQueries(queries[i][0],
                               queries[i][1], prefix)
             << endl;
    }

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

static int N = 4 ;

// Function to return the length of the
// longest palindrome that can be formed
// using the characters in the range [l, r]
static int performQueries(int l, int r, int prefix[][])
{

    // 0-based indexing
    l--;
    r--;

    // Marks if there is an
    // odd frequency character
    boolean flag = false;

    // Length of the longest palindrome
    // possible from characters in range
    int count = 0;

    // Traverse for all characters
    // and count their frequencies
    for (int i = 0; i < 26; i++)
    {

        // Find the frequency in range 1 - r
        int cnt = prefix[r][i];

        // Exclude the frequencies in range 1 - (l - 1)
        if (l > 0)
            cnt -= prefix[l - 1][i];

        // If frequency is odd, then add 1 less than
        // the original frequency to make it even
        if (cnt % 2 == 1)
        {
            flag = true;
            count += cnt - 1;
        }

        // Else completely add if even
        else
            count += cnt;
    }

    // If any odd frequency character
    // is present then add 1
    if (flag)
        count += 1;

    return count;
}

// Function to pre-calculate the frequencies
// of the characters to reduce complexity
static void preCalculate(String s, int prefix[][])
{
    int n = s.length();

    // Iterate and increase the count
    for (int i = 0; i < n; i++)
    {
        prefix[i][s.charAt(i) - 'a']++;
    }

    // Create a prefix type array
    for (int i = 1; i < n; i++)
    {
        for (int j = 0; j < 26; j++)
            prefix[i][j] += prefix[i - 1][j];
    }
}

// Driver code
public static void main(String args[])
{
    String s = "amim";

    // Pre-calculate prefix array
    int prefix[][] = new int[N][26];
    preCalculate(s, prefix);

    int queries[][] = { { 1, 4 }, { 3, 4 } };
    int q = queries.length;

    // Perform queries
    for (int i = 0; i < q; i++)
    {
        System.out.println( performQueries(queries[i][0],
                            queries[i][1], prefix) );
    }
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
N = 4

# Function to return the length of the
# longest palindrome that can be formed
# using the characters in the range [l, r]
def performQueries(l, r, prefix):

    # 0-based indexing
    l -= 1
    r -= 1

    # Marks if there is an
    # odd frequency character
    flag = False

    # Length of the longest palindrome
    # possible from characters in range
    count = 0

    # Traverse for all characters
    # and count their frequencies
    for i in range(26):

        # Find the frequency in range 1 - r
        cnt = prefix[r][i]

        # Exclude the frequencies in range 1 - (l - 1)
        if (l > 0):
            cnt -= prefix[l - 1][i]

        # If frequency is odd, then add 1 less than
        # the original frequency to make it even
        if (cnt % 2 == 1):
            flag = True
            count += cnt - 1

        # Else completely add if even
        else:
            count += cnt

    # If any odd frequency character
    # is present then add 1
    if (flag):
        count += 1

    return count

# Function to pre-calculate the frequencies
# of the characters to reduce complexity
def preCalculate(s, prefix):

    n = len(s)

    # Iterate and increase the count
    for i in range(n):
        prefix[i][ord(s[i]) - ord('a')] += 1


    # Create a prefix type array
    for i in range(1, n):
        for j in range(26):
            prefix[i][j] += prefix[i - 1][j]

# Driver code
s = "amim"

# Pre-calculate prefix array
prefix = [[0 for i in range(26)]
             for i in range(N)]

preCalculate(s, prefix)

queries = [[1, 4] , [3, 4]]
q = len(queries)

# Perform queries
for i in range(q):
    print(performQueries(queries[i][0],
                         queries[i][1],
                         prefix))

# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System;

class GFG
{

static int N = 4 ;

// Function to return the length of the
// longest palindrome that can be formed
// using the characters in the range [l, r]
static int performQueries(int l, int r, int[,] prefix)
{

    // 0-based indexing
    l--;
    r--;

    // Marks if there is an
    // odd frequency character
    bool flag = false;

    // Length of the longest palindrome
    // possible from characters in range
    int count = 0;

    // Traverse for all characters
    // and count their frequencies
    for (int i = 0; i < 26; i++)
    {

        // Find the frequency in range 1 - r
        int cnt = prefix[r, i];

        // Exclude the frequencies in range 1 - (l - 1)
        if (l > 0)
            cnt -= prefix[l - 1, i];

        // If frequency is odd, then add 1 less than
        // the original frequency to make it even
        if (cnt % 2 == 1)
        {
            flag = true;
            count += cnt - 1;
        }

        // Else completely add if even
        else
            count += cnt;
    }

    // If any odd frequency character
    // is present then add 1
    if (flag)
        count += 1;

    return count;
}

// Function to pre-calculate the frequencies
// of the characters to reduce complexity
static void preCalculate(string s, int[,] prefix)
{
    int n = s.Length;

    // Iterate and increase the count
    for (int i = 0; i < n; i++)
    {
        prefix[i, s[i] - 'a']++;
    }

    // Create a prefix type array
    for (int i = 1; i < n; i++)
    {
        for (int j = 0; j < 26; j++)
            prefix[i, j] += prefix[i - 1, j];
    }
}

// Driver code
public static void Main()
{
    string s = "amim";

    // Pre-calculate prefix array
    int[,] prefix = new int[N, 26];
    preCalculate(s, prefix);

    int[,] queries = { { 1, 4 }, { 3, 4 } };
    int q = queries.Length;

    // Perform queries
    for (int i = 0; i < q; i++)
    {
        Console.WriteLine( performQueries(queries[i, 0],
                            queries[i, 1], prefix) );
    }
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation of the approach
let N = 4 ;

// Function to return the length of the
// longest palindrome that can be formed
// using the characters in the range [l, r]
function performQueries(l, r, prefix)
{

    // 0-based indexing
    l--;
    r--;

    // Marks if there is an
    // odd frequency character
    let flag = false;

    // Length of the longest palindrome
    // possible from characters in range
    let count = 0;

    // Traverse for all characters
    // and count their frequencies
    for (let i = 0; i < 26; i++)
    {

        // Find the frequency in range 1 - r
        let cnt = prefix[r][i];

        // Exclude the frequencies in range 1 - (l - 1)
        if (l > 0)
            cnt -= prefix[l - 1][i];

        // If frequency is odd, then add 1 less than
        // the original frequency to make it even
        if (cnt % 2 == 1)
        {
            flag = true;
            count += cnt - 1;
        }

        // Else completely add if even
        else
            count += cnt;
    }

    // If any odd frequency character
    // is present then add 1
    if (flag)
        count += 1;

    return count;
}

// Function to pre-calculate the frequencies
// of the characters to reduce complexity
function preCalculate(s,prefix)
{
    let n = s.length;

    // Iterate and increase the count
    for (let i = 0; i < n; i++)
    {
        prefix[i][s[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }

    // Create a prefix type array
    for (let i = 1; i < n; i++)
    {
        for (let j = 0; j < 26; j++)
            prefix[i][j] += prefix[i - 1][j];
    }
}

// Driver code
let s = "amim";

    // Pre-calculate prefix array
    let prefix = new Array(N);
    for(let i = 0; i < 26; i++)
    {
        prefix[i] = new Array(26);
        for(let j = 0; j < 26; j++)
        {
            prefix[i][j] = 0;
        }
    }
    preCalculate(s, prefix);

    let queries = [[ 1, 4 ], [ 3, 4 ]];
    let q = queries.length;

    // Perform queries
    for (let i = 0; i < q; i++)
    {
        document.write( performQueries(queries[i][0],
                            queries[i][1], prefix) +"<br>");
    }

// This code is contributed by patel2127
</script>

Output:

3
1

Time Complexity: O(26*N), as we are using nested loops to traverse 26*N times. Where N is the length of the string.

Auxiliary Space: O(26*N), as we are using extra space for the prefix matrix. Where N is the length of the string.