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:
- Calculate the number of R’s before a K .
- Calculate the number of K’s after a K, including that K.
- 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].
- 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^nusing System;
class GFG
{// function to calculate the // maximum length of substring// of the form R^nK^nstatic 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 codestatic 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)
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