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++ 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 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 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# 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

<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>

32629

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