Largest sub-string where all the characters appear at least K times
Last Updated :
18 Aug, 2021
Given a string str and an integer K, the task is to find the length of the longest sub-string S’ such that every character in S’ appears at least K times.
Input: s = “xyxyyz”, k = 2
Output: 5
“xyxyy” is the longest sub-string where
every character appears at least twice.
Input: s = “geeksforgeeks”, k = 2
Output: 2
Approach: Consider all the possible sub-strings and for every sub-string, calculate the frequency of each of its character and check whether all the characters appear at least K times. For all such valid sub-strings, find the largest length possible.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define MAX 26
bool atLeastK( int freq[], int k)
{
for ( int i = 0; i < MAX; i++) {
if (freq[i] != 0 && freq[i] < k)
return false ;
}
return true ;
}
void setZero( int freq[])
{
for ( int i = 0; i < MAX; i++)
freq[i] = 0;
}
int findlength(string str, int n, int k)
{
int maxLen = 0;
int freq[MAX];
for ( int i = 0; i < n; i++) {
setZero(freq);
for ( int j = i; j < n; j++) {
freq[str[j] - 'a' ]++;
if (atLeastK(freq, k)) {
maxLen = max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
int main()
{
string str = "xyxyyz" ;
int n = str.length();
int k = 2;
cout << findlength(str, n, k);
return 0;
}
|
Java
class GFG
{
static final int MAX = 26 ;
static boolean atLeastK( int freq[], int k)
{
for ( int i = 0 ; i < MAX; i++)
{
if (freq[i] != 0 && freq[i] < k)
return false ;
}
return true ;
}
static void setZero( int freq[])
{
for ( int i = 0 ; i < MAX; i++)
freq[i] = 0 ;
}
static int findlength(String str, int n, int k)
{
int maxLen = 0 ;
int freq[] = new int [MAX];
for ( int i = 0 ; i < n; i++)
{
setZero(freq);
for ( int j = i; j < n; j++)
{
freq[str.charAt(j) - 'a' ]++;
if (atLeastK(freq, k))
{
maxLen = Math.max(maxLen, j - i + 1 );
}
}
}
return maxLen;
}
public static void main(String args[])
{
String str = "xyxyyz" ;
int n = str.length();
int k = 2 ;
System.out.println(findlength(str, n, k));
}
}
|
Python3
MAX = 26
def atLeastK(freq, k) :
for i in range ( MAX ) :
if (freq[i] ! = 0 and freq[i] < k) :
return False ;
return True ;
def setZero(freq) :
for i in range ( MAX ) :
freq[i] = 0 ;
def findlength(string, n, k) :
maxLen = 0 ;
freq = [ 0 ] * MAX ;
for i in range (n) :
setZero(freq);
for j in range (i,n) :
freq[ ord (string[j]) - ord ( 'a' )] + = 1 ;
if (atLeastK(freq, k)) :
maxLen = max (maxLen, j - i + 1 );
return maxLen;
if __name__ = = "__main__" :
string = "xyxyyz" ;
n = len (string);
k = 2 ;
print (findlength(string, n, k));
|
C#
using System;
class GFG
{
static int MAX = 26;
static Boolean atLeastK( int []freq, int k)
{
for ( int i = 0; i < MAX; i++)
{
if (freq[i] != 0 && freq[i] < k)
return false ;
}
return true ;
}
static void setZero( int []freq)
{
for ( int i = 0; i < MAX; i++)
freq[i] = 0;
}
static int findlength(String str, int n, int k)
{
int maxLen = 0;
int []freq = new int [MAX];
for ( int i = 0; i < n; i++)
{
setZero(freq);
for ( int j = i; j < n; j++)
{
freq[str[j] - 'a' ]++;
if (atLeastK(freq, k))
{
maxLen = Math.Max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
public static void Main(String []args)
{
String str = "xyxyyz" ;
int n = str.Length;
int k = 2;
Console.WriteLine(findlength(str, n, k));
}
}
|
Javascript
<script>
var MAX = 26;
function atLeastK(freq, k)
{
for ( var i = 0; i < MAX; i++) {
if (freq[i] != 0 && freq[i] < k)
return false ;
}
return true ;
}
function setZero(freq)
{
for ( var i = 0; i < MAX; i++)
freq[i] = 0;
}
function findlength(str, n, k)
{
var maxLen = 0;
var freq = Array(MAX).fill(0);
for ( var i = 0; i < n; i++) {
setZero(freq);
for ( var j = i; j < n; j++) {
freq[str[j].charCodeAt(0) - 'a' .charCodeAt(0)]++;
if (atLeastK(freq, k)) {
maxLen = Math.max(maxLen, j - i + 1);
}
}
}
return maxLen;
}
var str = "xyxyyz" ;
var n = str.length;
var k = 2;
document.write( findlength(str, n, k));
</script>
|
Time Complexity: O(n2)
Auxiliary Space: O(MAX), where MAX is the maximum number of unique characters in the string.
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