Longest substring with K unique characters using Binary Search
Last Updated :
25 Apr, 2023
Given a string str and an integer K, the task is to print the length of the longest possible substring that has exactly K unique characters. If there is more than one substring of the longest possible length, then print any one of them or print -1 if there is no such substring possible.
Examples:
Input: str = "aabacbebebe", K = 3
Output: 7
"cbebebe" is the required substring.
Input: str = "aabc", K = 4
Output: -1
Approach: An approach to solve this problem has been discussed in this article. In this article, a binary search based approach will be discussed. Binary search will be applied to the length of the substring which has at least K unique characters. Let's say we try for length len and check whether a substring of size len is there which is having at least k unique characters. If it is possible, then try to maximize the size by searching for this length to the maximum possible length, i.e. the size of the input string. If it is not possible, then search for a lower size len.
To check that the length given by binary search will have k unique characters, a set can be used to insert all the characters, and then if the size of the set is less than k then the answer is not possible, else the answer given by the binary search is the max answer.
Binary search is applicable here because it is known if for some len the answer is possible and we want to maximize the len so the search domain changes and we search from this len to n.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if there
// is a substring of length len
// with <=k unique characters
bool isValidLen(string s, int len, int k)
{
// Size of the string
int n = s.size();
// Map to store the characters
// and their frequency
unordered_map<char, int> mp;
int right = 0;
// Update the map for the
// first substring
while (right < len) {
mp[s[right]]++;
right++;
}
if (mp.size() <= k)
return true;
// Check for the rest of the substrings
while (right < n) {
// Add the new character
mp[s[right]]++;
// Remove the first character
// of the previous window
mp[s[right - len]]--;
// Update the map
if (mp[s[right - len]] == 0)
mp.erase(s[right - len]);
if (mp.size() <= k)
return true;
right++;
}
return mp.size() <= k;
}
// Function to return the length of the
// longest substring which has K
// unique characters
int maxLenSubStr(string s, int k)
{
// Check if the complete string
// contains K unique characters
set<char> uni;
for (auto x : s)
uni.insert(x);
if (uni.size() < k)
return -1;
// Size of the string
int n = s.size();
// Apply binary search
int lo = -1, hi = n + 1;
while (hi - lo > 1) {
int mid = lo + hi >> 1;
if (isValidLen(s, mid, k))
lo = mid;
else
hi = mid;
}
return lo;
}
// Driver code
int main()
{
string s = "aabacbebebe";
int k = 3;
cout << maxLenSubStr(s, k);
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true if there
// is a subString of length len
// with <=k unique characters
static boolean isValidLen(String s,
int len, int k)
{
// Size of the String
int n = s.length();
// Map to store the characters
// and their frequency
Map<Character,
Integer> mp = new HashMap<Character,
Integer>();
int right = 0;
// Update the map for the
// first subString
while (right < len)
{
if (mp.containsKey(s.charAt(right)))
{
mp.put(s.charAt(right),
mp.get(s.charAt(right)) + 1);
}
else
{
mp.put(s.charAt(right), 1);
}
right++;
}
if (mp.size() <= k)
return true;
// Check for the rest of the subStrings
while (right < n)
{
// Add the new character
if (mp.containsKey(s.charAt(right)))
{
mp.put(s.charAt(right),
mp.get(s.charAt(right)) + 1);
}
else
{
mp.put(s.charAt(right), 1);
}
// Remove the first character
// of the previous window
if (mp.containsKey(s.charAt(right - len)))
{
mp.put(s.charAt(right - len),
mp.get(s.charAt(right - len)) - 1);
}
// Update the map
if (mp.get(s.charAt(right - len)) == 0)
mp.remove(s.charAt(right - len));
if (mp.size() <= k)
return true;
right++;
}
return mp.size() <= k;
}
// Function to return the length of the
// longest subString which has K
// unique characters
static int maxLenSubStr(String s, int k)
{
// Check if the complete String
// contains K unique characters
Set<Character> uni = new HashSet<Character>();
for (Character x : s.toCharArray())
uni.add(x);
if (uni.size() < k)
return -1;
// Size of the String
int n = s.length();
// Apply binary search
int lo = -1, hi = n + 1;
while (hi - lo > 1)
{
int mid = lo + hi >> 1;
if (isValidLen(s, mid, k))
lo = mid;
else
hi = mid;
}
return lo;
}
// Driver code
public static void main(String[] args)
{
String s = "aabacbebebe";
int k = 3;
System.out.print(maxLenSubStr(s, k));
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation of the approach
# Function that returns True if there
# is a sub of length len
# with <=k unique characters
def isValidLen(s, lenn, k):
# Size of the
n = len(s)
# Map to store the characters
# and their frequency
mp = dict()
right = 0
# Update the map for the
# first sub
while (right < lenn):
mp[s[right]] = mp.get(s[right], 0) + 1
right += 1
if (len(mp) <= k):
return True
# Check for the rest of the subs
while (right < n):
# Add the new character
mp[s[right]] = mp.get(s[right], 0) + 1
# Remove the first character
# of the previous window
mp[s[right - lenn]] -= 1
# Update the map
if (mp[s[right - lenn]] == 0):
del mp[s[right - lenn]]
if (len(mp) <= k):
return True
right += 1
return len(mp)<= k
# Function to return the length of the
# longest sub which has K
# unique characters
def maxLenSubStr(s, k):
# Check if the complete
# contains K unique characters
uni = dict()
for x in s:
uni[x] = 1
if (len(uni) < k):
return -1
# Size of the
n = len(s)
# Apply binary search
lo = -1
hi = n + 1
while (hi - lo > 1):
mid = lo + hi >> 1
if (isValidLen(s, mid, k)):
lo = mid
else:
hi = mid
return lo
# Driver code
s = "aabacbebebe"
k = 3
print(maxLenSubStr(s, k))
# This code is contributed by Mohit Kumar
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function that returns true if there
// is a subString of length len
// with <=k unique characters
static bool isValidLen(String s,
int len, int k)
{
// Size of the String
int n = s.Length;
// Map to store the characters
// and their frequency
Dictionary<char,
int> mp = new Dictionary<char,
int>();
int right = 0;
// Update the map for the
// first subString
while (right < len)
{
if (mp.ContainsKey(s[right]))
{
mp[s[right]] = mp[s[right]] + 1;
}
else
{
mp.Add(s[right], 1);
}
right++;
}
if (mp.Count <= k)
return true;
// Check for the rest of the subStrings
while (right < n)
{
// Add the new character
if (mp.ContainsKey(s[right]))
{
mp[s[right]] = mp[s[right]] + 1;
}
else
{
mp.Add(s[right], 1);
}
// Remove the first character
// of the previous window
if (mp.ContainsKey(s[right - len]))
{
mp[s[right - len]] = mp[s[right - len]] - 1;
}
// Update the map
if (mp[s[right - len]] == 0)
mp.Remove(s[right - len]);
if (mp.Count <= k)
return true;
right++;
}
return mp.Count <= k;
}
// Function to return the length of the
// longest subString which has K
// unique characters
static int maxLenSubStr(String s, int k)
{
// Check if the complete String
// contains K unique characters
HashSet<char> uni = new HashSet<char>();
foreach (char x in s.ToCharArray())
uni.Add(x);
if (uni.Count < k)
return -1;
// Size of the String
int n = s.Length;
// Apply binary search
int lo = -1, hi = n + 1;
while (hi - lo > 1)
{
int mid = lo + hi >> 1;
if (isValidLen(s, mid, k))
lo = mid;
else
hi = mid;
}
return lo;
}
// Driver code
public static void Main(String[] args)
{
String s = "aabacbebebe";
int k = 3;
Console.Write(maxLenSubStr(s, k));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript implementation of the approach
// Function that returns true if there
// is a substring of length len
// with <=k unique characters
function isValidLen(s, len, k)
{
// Size of the string
var n = s.length;
// Map to store the characters
// and their frequency
var mp = new Map();
var right = 0;
// Update the map for the
// first substring
while (right < len) {
if(mp.has(s[right]))
mp.set(s[right],mp.get(s[right])+1)
else
mp.set(s[right], 1)
right++;
}
if (mp.size <= k)
return true;
// Check for the rest of the substrings
while (right < n) {
// Add the new character
if(mp.has(s[right]))
mp.set(s[right],mp.get(s[right])+1)
else
mp.set(s[right], 1)
// Remove the first character
// of the previous window
if(mp.has(s[right - len]))
mp.set(s[right - len], mp.get(s[right - len])-1)
// Update the map
if(mp.has(s[right - len]) && mp.get(s[right - len])==0)
mp.delete(s[right - len]);
if (mp.size <= k)
return true;
right++;
}
return mp.size <= k;
}
// Function to return the length of the
// longest substring which has K
// unique characters
function maxLenSubStr(s, k)
{
// Check if the complete string
// contains K unique characters
var uni = new Set();
s.split('').forEach(x => {
uni.add(x);
});
if (uni.size < k)
return -1;
// Size of the string
var n = s.length;
// Apply binary search
var lo = -1, hi = n + 1;
while (hi - lo > 1) {
var mid = lo + hi >> 1;
if (isValidLen(s, mid, k))
lo = mid;
else
hi = mid;
}
return lo;
}
// Driver code
var s = "aabacbebebe";
var k = 3;
document.write( maxLenSubStr(s, k));
</script>
The time complexity of the given program is O(N*logN), where N is the length of the input string. This is because the program uses binary search to find the maximum length of a substring with K unique characters, and the isValidLen function checks whether a substring of a given length has at most K unique characters.
The space complexity of the given program is O(N), where N is the length of the input string. This is because the program uses an unordered_map to keep track of the frequency of each character in the current substring, and the size of this map can be at most the number of unique characters in the input string, which is O(N) in the worst case. Additionally, the program uses a set to check whether the input string contains at least K unique characters, and the size of this set can also be at most O(N).
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