Longest substring where all the characters appear at least K times | Set 3
Last Updated :
03 Jun, 2022
Given a string str and an integer K, the task is to find the length of the longest substring S such that every character in S appears at least K times.
Examples:
Input: str = “aabbba”, K = 3
Output: 6
Explanation: In substring "aabbba", each character repeats at least k times and its length is 6.
Input: str = “ababacb”, K = 3
Output: 0
Explanation: There is no substring where each character repeats at least k times.
Naive Approach: The simplest approach to solve the given problem is discussed in Set 1.
Time Complexity: O(N2)
Auxiliary Space: O(26)
Divide and Conquer Approach: The divide and conquer approach for the given problem is discussed in the Set 2.
Time Complexity: O(N*log N)
Auxiliary Space: O(26)
Efficient Approach: The above two approaches can be optimized further by using Sliding Window technique. Follow the steps below to solve the problem:
- Store the number of unique characters in the string str in a variable, say unique.
- Initialize an array freq[] of size 26 with {0} and store the frequency of each character in this array.
- Iterate over the range [1, unique] using the variable curr_unique. In each iteration, curr_unique is the maximum number of unique characters that must be present in the window.
- Reinitialize the array freq[] with {0} to store the frequency of each character in this window.
- Initialize start and end as 0, to store the starting and the ending point of the window respectively.
- Use two variables cnt, for storing the number of unique characters and countK, for storing the number of characters with at least K repeating characters in the current window.
- Now, iterate a loop while end < N, and perform the following:
- If the value of cnt is less than or equals to curr_unique, then expand the window from the right by adding a character to the end of the window. And increment its frequency by 1 in freq[].
- Otherwise, reduce the window from the left by removing a character from start and decrementing its frequency by 1 in freq[].
- At every step, update the values of cnt and countK.
- If the value of cnt is same as curr_unique and each character occurs at least K times, then update the overall maximum length and store it in ans.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the length of
// the longest substring
int longestSubstring(string s, int k)
{
// Store the required answer
int ans = 0;
// Create a frequency map of the
// characters of the string
int freq[26] = { 0 };
// Store the length of the string
int n = s.size();
// Traverse the string, s
for (int i = 0; i < n; i++)
// Increment the frequency of
// the current character by 1
freq[s[i] - 'a']++;
// Stores count of unique characters
int unique = 0;
// Find the number of unique
// characters in string
for (int i = 0; i < 26; i++)
if (freq[i] != 0)
unique++;
// Iterate in range [1, unique]
for (int curr_unique = 1;
curr_unique <= unique;
curr_unique++) {
// Initialize frequency of all
// characters as 0
memset(freq, 0, sizeof(freq));
// Stores the start and the
// end of the window
int start = 0, end = 0;
// Stores the current number of
// unique characters and characters
// occurring atleast K times
int cnt = 0, count_k = 0;
while (end < n) {
if (cnt <= curr_unique) {
int ind = s[end] - 'a';
// New unique character
if (freq[ind] == 0)
cnt++;
freq[ind]++;
// New character which
// occurs atleast k times
if (freq[ind] == k)
count_k++;
// Expand window by
// incrementing end by 1
end++;
}
else {
int ind = s[start] - 'a';
// Check if this character
// is present atleast k times
if (freq[ind] == k)
count_k--;
freq[ind]--;
// Check if this character
// is unique
if (freq[ind] == 0)
cnt--;
// Shrink the window by
// incrementing start by 1
start++;
}
// If there are curr_unique
// characters and each character
// is atleast k times
if (cnt == curr_unique
&& count_k == curr_unique)
// Update the overall
// maximum length
ans = max(ans, end - start);
}
}
// return the answer
return ans;
}
// Driver Code
int main()
{
string S = "aabbba";
int K = 3;
cout << longestSubstring(S, K) << endl;
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the length of
// the longest subString
static void longestSubString(char[] s, int k)
{
// Store the required answer
int ans = 0;
// Create a frequency map of the
// characters of the String
int freq[] = new int[26];
// Store the length of the String
int n = s.length;
// Traverse the String, s
for (int i = 0; i < n; i++)
// Increment the frequency of
// the current character by 1
freq[s[i] - 'a']++;
// Stores count of unique characters
int unique = 0;
// Find the number of unique
// characters in String
for (int i = 0; i < 26; i++)
if (freq[i] != 0)
unique++;
// Iterate in range [1, unique]
for (int curr_unique = 1;
curr_unique <= unique;
curr_unique++)
{
// Initialize frequency of all
// characters as 0
Arrays.fill(freq, 0);
// Stores the start and the
// end of the window
int start = 0, end = 0;
// Stores the current number of
// unique characters and characters
// occurring atleast K times
int cnt = 0, count_k = 0;
while (end < n)
{
if (cnt <= curr_unique)
{
int ind = s[end] - 'a';
// New unique character
if (freq[ind] == 0)
cnt++;
freq[ind]++;
// New character which
// occurs atleast k times
if (freq[ind] == k)
count_k++;
// Expand window by
// incrementing end by 1
end++;
}
else
{
int ind = s[start] - 'a';
// Check if this character
// is present atleast k times
if (freq[ind] == k)
count_k--;
freq[ind]--;
// Check if this character
// is unique
if (freq[ind] == 0)
cnt--;
// Shrink the window by
// incrementing start by 1
start++;
}
// If there are curr_unique
// characters and each character
// is atleast k times
if (cnt == curr_unique
&& count_k == curr_unique)
// Update the overall
// maximum length
ans = Math.max(ans, end - start);
}
}
// Print the answer
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
String S = "aabbba";
int K = 3;
longestSubString(S.toCharArray(), K);
}
}
// This code is contributed by 29AjayKumar
# Python3 program for the above approach
# Function to find the length of
# the longest substring
def longestSubstring(s, k) :
# Store the required answer
ans = 0
# Create a frequency map of the
# characters of the string
freq = [0]*26
# Store the length of the string
n = len(s)
# Traverse the string, s
for i in range(n) :
# Increment the frequency of
# the current character by 1
freq[ord(s[i]) - ord('a')] += 1
# Stores count of unique characters
unique = 0
# Find the number of unique
# characters in string
for i in range(26) :
if (freq[i] != 0) :
unique += 1
# Iterate in range [1, unique]
for curr_unique in range(1, unique + 1) :
# Initialize frequency of all
# characters as 0
Freq = [0]*26
# Stores the start and the
# end of the window
start, end = 0, 0
# Stores the current number of
# unique characters and characters
# occurring atleast K times
cnt, count_k = 0, 0
while (end < n) :
if (cnt <= curr_unique) :
ind = ord(s[end]) - ord('a')
# New unique character
if (Freq[ind] == 0) :
cnt += 1
Freq[ind] += 1
# New character which
# occurs atleast k times
if (Freq[ind] == k) :
count_k += 1
# Expand window by
# incrementing end by 1
end += 1
else :
ind = ord(s[start]) - ord('a')
# Check if this character
# is present atleast k times
if (Freq[ind] == k) :
count_k -= 1
Freq[ind] -= 1
# Check if this character
# is unique
if (Freq[ind] == 0) :
cnt -= 1
# Shrink the window by
# incrementing start by 1
start += 1
# If there are curr_unique
# characters and each character
# is atleast k times
if ((cnt == curr_unique) and (count_k == curr_unique)) :
# Update the overall
# maximum length
ans = max(ans, end - start)
# Print the answer
print(ans)
S = "aabbba"
K = 3
longestSubstring(S, K)
# This code is contributed by divyesh072019.
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to find the length of
// the longest substring
static void longestSubstring(string s, int k)
{
// Store the required answer
int ans = 0;
// Create a frequency map of the
// characters of the string
int[] freq = new int[26];
// Store the length of the string
int n = s.Length;
// Traverse the string, s
for (int i = 0; i < n; i++)
// Increment the frequency of
// the current character by 1
freq[s[i] - 'a']++;
// Stores count of unique characters
int unique = 0;
// Find the number of unique
// characters in string
for (int i = 0; i < 26; i++)
if (freq[i] != 0)
unique++;
// Iterate in range [1, unique]
for (int curr_unique = 1;
curr_unique <= unique;
curr_unique++)
{
// Initialize frequency of all
// characters as 0
for (int i = 0; i < freq.Length; i++)
{
freq[i] = 0;
}
// Stores the start and the
// end of the window
int start = 0, end = 0;
// Stores the current number of
// unique characters and characters
// occurring atleast K times
int cnt = 0, count_k = 0;
while (end < n)
{
if (cnt <= curr_unique)
{
int ind = s[end] - 'a';
// New unique character
if (freq[ind] == 0)
cnt++;
freq[ind]++;
// New character which
// occurs atleast k times
if (freq[ind] == k)
count_k++;
// Expand window by
// incrementing end by 1
end++;
}
else
{
int ind = s[start] - 'a';
// Check if this character
// is present atleast k times
if (freq[ind] == k)
count_k--;
freq[ind]--;
// Check if this character
// is unique
if (freq[ind] == 0)
cnt--;
// Shrink the window by
// incrementing start by 1
start++;
}
// If there are curr_unique
// characters and each character
// is atleast k times
if (cnt == curr_unique
&& count_k == curr_unique)
// Update the overall
// maximum length
ans = Math.Max(ans, end - start);
}
}
// Print the answer
Console.Write(ans);
}
// Driver Code
public static void Main()
{
string S = "aabbba";
int K = 3;
longestSubstring(S, K);
}
}
// This code is contributed by splevel62.
<script>
// Javascript program to implement
// the above approach
// Function to find the length of
// the longest substring
function longestSubstring(s, k)
{
// Store the required answer
let ans = 0;
// Create a frequency map of the
// characters of the string
let freq = new Array(26);
freq.fill(0);
// Store the length of the string
let n = s.length;
// Traverse the string, s
for (let i = 0; i < n; i++)
// Increment the frequency of
// the current character by 1
freq[s[i].charCodeAt() -
'a'.charCodeAt()]++;
// Stores count of unique characters
let unique = 0;
// Find the number of unique
// characters in string
for (let i = 0; i < 26; i++)
if (freq[i] != 0)
unique++;
// Iterate in range [1, unique]
for (let curr_unique = 1;
curr_unique <= unique;
curr_unique++)
{
// Initialize frequency of all
// characters as 0
for (let i = 0; i < freq.length; i++)
{
freq[i] = 0;
}
// Stores the start and the
// end of the window
let start = 0, end = 0;
// Stores the current number of
// unique characters and characters
// occurring atleast K times
let cnt = 0, count_k = 0;
while (end < n)
{
if (cnt <= curr_unique)
{
let ind = s[end].charCodeAt() -
'a'.charCodeAt();
// New unique character
if (freq[ind] == 0)
cnt++;
freq[ind]++;
// New character which
// occurs atleast k times
if (freq[ind] == k)
count_k++;
// Expand window by
// incrementing end by 1
end++;
}
else
{
let ind = s[start].charCodeAt() -
'a'.charCodeAt();
// Check if this character
// is present atleast k times
if (freq[ind] == k)
count_k--;
freq[ind]--;
// Check if this character
// is unique
if (freq[ind] == 0)
cnt--;
// Shrink the window by
// incrementing start by 1
start++;
}
// If there are curr_unique
// characters and each character
// is atleast k times
if (cnt == curr_unique
&& count_k == curr_unique)
// Update the overall
// maximum length
ans = Math.max(ans, end - start);
}
}
// Print the answer
document.write(ans);
}
let S = "aabbba";
let K = 3;
longestSubstring(S, K);
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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