Maximum count of sub-strings of length K consisting of same characters
Given a string str and an integer k. The task is to count the occurrences of sub-strings of length k that consist of the same characters. There can be multiple such sub-strings possible of length k, choose the count of the one which appears the maximum number of times as the sub-string (non-overlapping) of str.
Examples:
Input: str = “aaacaabbaa”, k = 2
Output: 3
“aa” and “bb” are the only sub-strings of length 2 that consist of the same characters.
“bb” appears only once as a sub-string of str whereas “aa” appears thrice (which is the answer)
Input: str = “abab”, k = 2
Output: 0
Approach: Iterate over all the characters from ‘a’ to ‘z’ and count the number of times a string of length k consisting only of the current character appears as a sub-string of str. Print the maximum of these counts in the end.
Steps to solve the problem:
- Initialize maxSubStr to 0 and n to the size of the string s.
- Iterate over all characters of the English alphabet using a loop from 0 to 25.
- For each character, store it in a variable ch.
- Initialize a variable curr to 0.
- Use another loop to iterate over all possible substrings of length k in the string s.
- If the current character of the string s is not the same as the character ch, continue with the next iteration of the loop.
- If the current character is the same as ch, count the number of consecutive characters in the substring that are the same as ch, using a variable cnt.
- If cnt is not equal to k, continue with the next iteration of the loop.
- If cnt is equal to k, increment the variable curr.
- After the inner loop, update maxSubStr to the maximum of its current value and curr.
- After the outer loop, return maxSubStr as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maxSubStrings(string s, int k)
{
int maxSubStr = 0, n = s.size();
for ( int c = 0; c < 26; c++) {
char ch = 'a' + c;
int curr = 0;
for ( int i = 0; i <= n - k; i++) {
if (s[i] != ch)
continue ;
int cnt = 0;
while (i < n && s[i] == ch && cnt != k) {
i++;
cnt++;
}
i--;
if (cnt == k)
curr++;
}
maxSubStr = max(maxSubStr, curr);
}
return maxSubStr;
}
int main()
{
string s = "aaacaabbaa" ;
int k = 2;
cout << maxSubStrings(s, k);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
static int maxSubStrings(String s, int k)
{
int maxSubStr = 0 , n = s.length();
for ( int c = 0 ; c < 26 ; c++)
{
char ch = ( char )(( int ) 'a' + c);
int curr = 0 ;
for ( int i = 0 ; i <= n - k; i++)
{
if (s.charAt(i) != ch)
continue ;
int cnt = 0 ;
while (i < n && s.charAt(i) == ch &&
cnt != k)
{
i++;
cnt++;
}
i--;
if (cnt == k)
curr++;
}
maxSubStr = Math.max(maxSubStr, curr);
}
return maxSubStr;
}
public static void main(String []args)
{
String s = "aaacaabbaa" ;
int k = 2 ;
System.out.println(maxSubStrings(s, k));
}
}
|
Python3
def maxSubStrings(s, k):
maxSubStr = 0
n = len (s)
for c in range ( 27 ):
ch = chr ( ord ( 'a' ) + c)
curr = 0
for i in range (n - k):
if (s[i] ! = ch):
continue
cnt = 0
while (i < n and s[i] = = ch and
cnt ! = k):
i + = 1
cnt + = 1
i - = 1
if (cnt = = k):
curr + = 1
maxSubStr = max (maxSubStr, curr)
return maxSubStr
if __name__ = = '__main__' :
s = "aaacaabbaa"
k = 2
print (maxSubStrings(s, k))
|
C#
using System;
class GFG
{
static int maxSubStrings(String s, int k)
{
int maxSubStr = 0, n = s.Length;
for ( int c = 0; c < 26; c++)
{
char ch = ( char )(( int ) 'a' + c);
int curr = 0;
for ( int i = 0; i <= n - k; i++)
{
if (s[i] != ch)
continue ;
int cnt = 0;
while (i < n && s[i] == ch &&
cnt != k)
{
i++;
cnt++;
}
i--;
if (cnt == k)
curr++;
}
maxSubStr = Math.Max(maxSubStr, curr);
}
return maxSubStr;
}
public static void Main()
{
string s = "aaacaabbaa" ;
int k = 2;
Console.WriteLine(maxSubStrings(s, k));
}
}
|
Javascript
<script>
function maxSubStrings(s, k) {
var maxSubStr = 0,
n = s.length;
for ( var c = 0; c < 26; c++) {
var ch = String.fromCharCode( "a" .charCodeAt(0) + c);
var curr = 0;
for ( var i = 0; i <= n - k; i++) {
if (s[i] !== ch) continue ;
var cnt = 0;
while (i < n && s[i] === ch && cnt !== k) {
i++;
cnt++;
}
i--;
if (cnt === k) curr++;
}
maxSubStr = Math.max(maxSubStr, curr);
}
return maxSubStr;
}
var s = "aaacaabbaa" ;
var k = 2;
document.write(maxSubStrings(s, k));
</script>
|
Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(1).
Last Updated :
01 Mar, 2023
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