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# Longest substring with K unique characters using Binary Search

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:
“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 ``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

 `// 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 mp = ``new` `HashMap();``        ``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 uni = ``new` `HashSet();``        ``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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`7`

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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