Maximum length subsequence possible of the form R^N K^N

Given a string containing only two characters i.e. R and K (like RRKRRKKKKK). The task is to find the maximum value of N for a subsequence possible of the form R—N times and then K—N times (i.e. of the form R^N K^N).

Note: String of k should be started after the string of R i.e. first k that would be considered for ‘K’ string must occur after the last R of the ‘R’ string in the given string. Also, the length of the resulting subsequence will be 2*N.

Examples: 

Input: RRRKRRKKRRKKK 
Output: 5 
If we take R’s at indexes 0, 1, 2, 4, 5 and K’s at indexes 6, 7, 10, 11, 12 
then we get a maximum subsequence of the form R^N K^N, where N = 5.

Input: KKKKRRRRK 
Output: 1 
If we take R at index 4( or 5 or 6 or 7) and K at index 8 
then we get the desired subsequence with N = 1. 

Approach:

1. Calculate the number of R’s before a K .
2. Calculate the number of K’s after a K, including that K.
3. Store them in a table with a number of R’s in table[x][0] and a number of K’s in table[x][1].
4. Minimum of the two gives the value of n for each K and we will the return the maximum n.

Below is the implementation of the above approach: 

C++

// C++ program to find the maximum
// length of a substring of form R^nK^n
#include<bits/stdc++.h>
using namespace std;

    // function to calculate the maximum
    // length of substring of the form R^nK^n
    int find(string s)
    {
        int max = 0, i, j = 0, countk = 0, countr = 0;
        int table[s.length()][2];

        // Count no. Of R's before a K
        for (i = 0; i < s.length(); i++) {
            if (s[i] == 'R')
                countr++;
            else
                table[j++][0] = countr;
        }
        j--;

        // Count no. Of K's after a K
        for (i = s.length() - 1; i >= 0; i--) {
            if (s[i] == 'K') {
                countk++;
                table[j--][1] = countk;
            }

            // Update maximum length
            if (min(table[j + 1][0], table[j + 1][1]) > max)
                max = min(table[j + 1][0], table[j + 1][1]);
        }

        return max;
    }

    // Driver code
    int main()
    {
        string s = "RKRRRKKRRKKKKRR";
        int n = find(s);
        cout<<(n);
    }
// This code is contributed by
// Surendra_Gangwar

Java

// Java program to find the maximum
// length of a substring of form R^nK^n
public class gfg {

    // function to calculate the maximum
    // length of substring of the form R^nK^n
    int find(String s)
    {
        int max = 0, i, j = 0, countk = 0, countr = 0;
        int table[][] = new int[s.length()][2];

        // Count no. Of R's before a K
        for (i = 0; i < s.length(); i++) {
            if (s.charAt(i) == 'R')
                countr++;
            else
                table[j++][0] = countr;
        }
        j--;

        // Count no. Of K's after a K
        for (i = s.length() - 1; i >= 0; i--) {
            if (s.charAt(i) == 'K') {
                countk++;
                table[j--][1] = countk;
            }

            // Update maximum length
            if (Math.min(table[j + 1][0], table[j + 1][1]) > max)
                max = Math.min(table[j + 1][0], table[j + 1][1]);
        }

        return max;
    }

    // Driver code
    public static void main(String srgs[])
    {
        String s = "RKRRRKKRRKKKKRR";
        gfg ob = new gfg();
        int n = ob.find(s);
        System.out.println(n);
    }
}

Python3

# Python3 program to find the maximum
# length of a substring of form R^nK^n

# Function to calculate the maximum
# length of substring of the form R^nK^n
def find(s):

    Max = j = countk = countr = 0
    table = [[0, 0] for i in range(len(s))]

    # Count no. Of R's before a K
    for i in range(0, len(s)):
        if s[i] == 'R':
            countr += 1
        else:
            table[j][0] = countr
            j += 1

    j -= 1

    # Count no. Of K's after a K
    for i in range(len(s) - 1, -1, -1):
        if s[i] == 'K':
            countk += 1
            table[j][1] = countk
            j -= 1

        # Update maximum length
        if min(table[j + 1][0], table[j + 1][1]) > Max:
            Max = min(table[j + 1][0], table[j + 1][1])

    return Max

# Driver code
if __name__ == "__main__":

    s = "RKRRRKKRRKKKKRR"
    print(find(s))

# This code is contributed by Rituraj Jain

C#

// C# program to find the maximum
// length of a substring of
// form R^nK^n
using System;

class GFG
{

// function to calculate the
// maximum length of substring
// of the form R^nK^n
static int find(String s)
{
    int max = 0, i, j = 0,
        countk = 0, countr = 0;
    int [,]table= new int[s.Length, 2];

    // Count no. Of R's before a K
    for (i = 0; i < s.Length; i++)
    {
        if (s[i] == 'R')
            countr++;
        else
            table[(j++),0] = countr;
    }
    j--;

    // Count no. Of K's after a K
    for (i = s.Length - 1; i >= 0; i--)
    {
        if (s[i] == 'K')
        {
            countk++;
            table[j--, 1] = countk;
        }

        // Update maximum length
        if (Math.Min(table[j + 1, 0],
                     table[j + 1, 1]) > max)
            max = Math.Min(table[j + 1, 0],
                           table[j + 1, 1]);
    }

    return max;
}

// Driver code
static public void Main(String []srgs)
{
    String s = "RKRRRKKRRKKKKRR";
    int n = find(s);
    Console.WriteLine(n);
}
}

// This code is contributed
// by Arnab Kundu

Javascript

<script>

// JavaScript program to find the maximum
// length of a substring of form R^nK^n

// function to calculate the maximum
    // length of substring of the form R^nK^n
function find(s)
{
    let max = 0, i, j = 0, countk = 0, countr = 0;
        let table = new Array(s.length);
        for(let i=0;i<s.length;i++)
        {
            table[i]=new Array(2);
        }

        // Count no. Of R's before a K
        for (i = 0; i < s.length; i++) {
            if (s[i] == 'R')
                countr++;
            else
                table[j++][0] = countr;
        }
        j--;

        // Count no. Of K's after a K
        for (i = s.length - 1; i >= 0; i--) {
            if (s[i] == 'K') {
                countk++;
                table[j--][1] = countk;
            }

            // Update maximum length
            if (Math.min(table[j + 1][0], table[j + 1][1]) > max)
                max = Math.min(table[j + 1][0], table[j + 1][1]);
        }

        return max;
}

// Driver code
let s = "RKRRRKKRRKKKKRR";
let n=find(s);
document.write(n);


// This code is contributed by avanitrachhadiya2155

</script>

Output

32629

Time Complexity: O(n) where n is the length of the given string.
Auxiliary Space: O(n)