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Find the longest substring with k unique characters in a given string

  • Difficulty Level : Hard
  • Last Updated : 09 Sep, 2021
Geek Week

Given a string you need to print longest possible substring that has exactly M unique characters. If there are more than one substring of longest possible length, then print any one of them.

Examples: 

"aabbcc", k = 1
Max substring can be any one from {"aa" , "bb" , "cc"}.

"aabbcc", k = 2
Max substring can be any one from {"aabb" , "bbcc"}.

"aabbcc", k = 3
There are substrings with exactly 3 unique characters
{"aabbcc" , "abbcc" , "aabbc" , "abbc" }
Max is "aabbcc" with length 6.

"aaabbb", k = 3
There are only two unique characters, thus show error message. 

Source: Google Interview Question.

Method 1 (Brute Force) 
If the length of string is n, then there can be n*(n+1)/2 possible substrings. A simple way is to generate all the substring and check each one whether it has exactly k unique characters or not. If we apply this brute force, it would take O(n2) to generate all substrings and O(n) to do a check on each one. Thus overall it would go O(n3).
We can further improve this solution by creating a hash table and while generating the substrings, check the number of unique characters using that hash table. Thus it would improve up to O(n2).

Method 2 (Linear Time) 
The problem can be solved in O(n). Idea is to maintain a window and add elements to the window till it contains less or equal k, update our result if required while doing so. If unique elements exceeds than required in window, start removing the elements from left side. 
Below are the implementations of above. The implementations assume that the input string alphabet contains only 26 characters (from ‘a’ to ‘z’). The code can be easily extended to 256 characters. 



C++




// C++ program to find the longest substring with k unique
// characters in a given string
#include <iostream>
#include <cstring>
#define MAX_CHARS 26
using namespace std;
 
// This function calculates number of unique characters
// using a associative array count[]. Returns true if
// no. of characters are less than required else returns
// false.
bool isValid(int count[], int k)
{
    int val = 0;
    for (int i=0; i<MAX_CHARS; i++)
        if (count[i] > 0)
            val++;
 
    // Return true if k is greater than or equal to val
    return (k >= val);
}
 
// Finds the maximum substring with exactly k unique chars
void kUniques(string s, int k)
{
    int u = 0; // number of unique characters
    int n = s.length();
 
    // Associative array to store the count of characters
    int count[MAX_CHARS];
    memset(count, 0, sizeof(count));
 
    // Traverse the string, Fills the associative array
    // count[] and count number of unique characters
    for (int i=0; i<n; i++)
    {
        if (count[s[i]-'a']==0)
            u++;
        count[s[i]-'a']++;
    }
 
    // If there are not enough unique characters, show
    // an error message.
    if (u < k)
    {
        cout << "Not enough unique characters";
        return;
    }
 
    // Otherwise take a window with first element in it.
    // start and end variables.
    int curr_start = 0, curr_end = 0;
 
    // Also initialize values for result longest window
    int max_window_size = 1, max_window_start = 0;
 
    // Initialize associative array count[] with zero
    memset(count, 0, sizeof(count));
 
    count[s[0]-'a']++;  // put the first character
 
    // Start from the second character and add
    // characters in window according to above
    // explanation
    for (int i=1; i<n; i++)
    {
        // Add the character 's[i]' to current window
        count[s[i]-'a']++;
        curr_end++;
 
        // If there are more than k unique characters in
        // current window, remove from left side
        while (!isValid(count, k))
        {
            count[s[curr_start]-'a']--;
            curr_start++;
        }
 
        // Update the max window size if required
        if (curr_end-curr_start+1 > max_window_size)
        {
            max_window_size = curr_end-curr_start+1;
            max_window_start = curr_start;
        }
    }
 
    cout << "Max substring is : "
         << s.substr(max_window_start, max_window_size)
         << " with length " << max_window_size << endl;
}
 
// Driver function
int main()
{
    string s = "aabacbebebe";
    int k = 3;
    kUniques(s, k);
    return 0;
}

Java




import java.util.Arrays;
 
// Java program to find the longest substring with k unique
// characters in a given string
class GFG {
 
    final static int MAX_CHARS = 26;
 
   // This function calculates number
   // of unique characters
   // using a associative array
   // count[]. Returns true if
   // no. of characters are less
   // than required else returns
   // false.
    static boolean isValid(int count[],
                                   int k)
    {
        int val = 0;
        for (int i = 0; i < MAX_CHARS; i++)
        {
            if (count[i] > 0)
            {
                val++;
            }
        }
 
        // Return true if k is greater
        // than or equal to val
        return (k >= val);
    }
 
    // Finds the maximum substring
    // with exactly k unique chars
    static void kUniques(String s, int k)
    {
        int u = 0;
        int n = s.length();
 
        // Associative array to store
        // the count of characters
        int count[] = new int[MAX_CHARS];
        Arrays.fill(count, 0);
        // Traverse the string, Fills
        // the associative array
        // count[] and count number
        // of unique characters
        for (int i = 0; i < n; i++)
        {
            if (count[s.charAt(i) - 'a'] == 0)
            {
                u++;
            }
            count[s.charAt(i) - 'a']++;
        }
 
        // If there are not enough
        // unique characters, show
        // an error message.
        if (u < k) {
            System.out.print("Not enough unique characters");
            return;
        }
 
        // Otherwise take a window with
        // first element in it.
        // start and end variables.
        int curr_start = 0, curr_end = 0;
 
        // Also initialize values for
        // result longest window
        int max_window_size = 1;
        int max_window_start = 0;
 
        // Initialize associative
        // array count[] with zero
        Arrays.fill(count, 0);
         
        // put the first character
        count[s.charAt(0) - 'a']++;
 
        // Start from the second character and add
        // characters in window according to above
        // explanation
        for (int i = 1; i < n; i++) {
            // Add the character 's[i]'
            // to current window
            count[s.charAt(i) - 'a']++;
            curr_end++;
 
            // If there are more than k
            // unique characters in
            // current window, remove from left side
            while (!isValid(count, k)) {
                count[s.charAt(curr_start) - 'a']--;
                curr_start++;
            }
 
            // Update the max window size if required
            if (curr_end - curr_start + 1 > max_window_size)
            {
                max_window_size = curr_end - curr_start + 1;
                max_window_start = curr_start;
            }
        }
 
        System.out.println("Max substring is : "
                + s.substring(max_window_start,
                    max_window_start + max_window_size)
                + " with length " + max_window_size);
    }
 
    // Driver Code
    static public void main(String[] args) {
        String s = "aabacbebebe";
        int k = 3;
        kUniques(s, k);
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python program to find the longest substring with k unique
# characters in a given string
MAX_CHARS = 26
 
# This function calculates number of unique characters
# using a associative array count[]. Returns true if
# no. of characters are less than required else returns
# false.
def isValid(count, k):
    val = 0
    for i in range(MAX_CHARS):
        if count[i] > 0:
            val += 1
 
    # Return true if k is greater than or equal to val
    return (k >= val)
 
# Finds the maximum substring with exactly k unique characters
def kUniques(s, k):
    u = 0 # number of unique characters
    n = len(s)
 
    # Associative array to store the count
    count = [0] * MAX_CHARS
 
    # Tranverse the string, fills the associative array
    # count[] and count number of unique characters
    for i in range(n):
        if count[ord(s[i])-ord('a')] == 0:
            u += 1
        count[ord(s[i])-ord('a')] += 1
 
    # If there are not enough unique characters, show
    # an error message.
    if u < k:
        print ("Not enough unique characters")
        return
 
    # Otherwise take a window with first element in it.
    # start and end variables.
    curr_start = 0
    curr_end = 0
 
    # Also initialize values for result longest window
    max_window_size = 1
    max_window_start = 0
 
    # Initialize associative array count[] with zero
    count = [0] * len(count)
 
    count[ord(s[0])-ord('a')] += 1 # put the first character
 
    # Start from the second character and add
    # characters in window according to above
    # explanation
    for i in range(1,n):
 
        # Add the character 's[i]' to current window
        count[ord(s[i])-ord('a')] += 1
        curr_end+=1
 
        # If there are more than k unique characters in
        # current window, remove from left side
        while not isValid(count, k):
            count[ord(s[curr_start])-ord('a')] -= 1
            curr_start += 1
 
        # Update the max window size if required
        if curr_end-curr_start+1 > max_window_size:
            max_window_size = curr_end-curr_start+1
            max_window_start = curr_start
 
    print ("Max substring is : " + s[max_window_start:max_window_start  + max_window_size]
    + " with length " + str(max_window_size))
 
# Driver function
s = "aabacbebebe"
k = 3
kUniques(s, k)
 
# This code is contributed by BHAVYA JAIN

C#




// C# program to find the longest substring with k unique 
// characters in a given string 
using System;
public class GFG
{
 
  static int MAX_CHARS = 26;
 
  // This function calculates number 
  // of unique characters 
  // using a associative array 
  // count[]. Returns true if 
  // no. of characters are less 
  // than required else returns 
  // false. 
  static bool isValid(int[] count, 
                      int k) 
  
    int val = 0; 
    for (int i = 0; i < MAX_CHARS; i++) 
    
      if (count[i] > 0) 
      
        val++; 
      
    
 
    // Return true if k is greater
    // than or equal to val 
    return (k >= val); 
  
 
  // Finds the maximum substring 
  // with exactly k unique chars 
  static void kUniques(string s, int k) 
  
    int u = 0; 
    int n = s.Length; 
 
    // Associative array to store 
    // the count of characters 
    int[] count = new int[MAX_CHARS]; 
    Array.Fill(count, 0);
 
    // Traverse the string, Fills 
    // the associative array 
    // count[] and count number 
    // of unique characters 
    for (int i = 0; i < n; i++) 
    
      if (count[s[i] - 'a'] == 0) 
      
        u++; 
      
      count[s[i] - 'a']++; 
    
 
    // If there are not enough 
    // unique characters, show 
    // an error message. 
    if (u < k) { 
      Console.Write("Not enough unique characters"); 
      return
    
 
    // Otherwise take a window with
    // first element in it. 
    // start and end variables. 
    int curr_start = 0, curr_end = 0; 
 
    // Also initialize values for
    // result longest window 
    int max_window_size = 1;
    int max_window_start = 0; 
 
    // Initialize associative 
    // array count[] with zero 
    Array.Fill(count, 0); 
 
    // put the first character 
    count[s[0] - 'a']++; 
 
    // Start from the second character and add 
    // characters in window according to above 
    // explanation 
    for (int i = 1; i < n; i++)
    {
       
      // Add the character 's[i]' 
      // to current window 
      count[s[i] - 'a']++; 
      curr_end++; 
 
      // If there are more than k 
      // unique characters in 
      // current window, remove from left side 
      while (!isValid(count, k)) { 
        count[s[curr_start] - 'a']--; 
        curr_start++; 
      
 
      // Update the max window size if required 
      if (curr_end - curr_start + 1 > max_window_size) 
      
        max_window_size = curr_end - curr_start + 1; 
        max_window_start = curr_start; 
      
    
 
    Console.WriteLine("Max substring is : "+
                      s.Substring(max_window_start, max_window_size) +
                      " with length " + max_window_size); 
  
 
  // Driver code
  static public void Main (){
    string s = "aabacbebebe"
    int k = 3; 
    kUniques(s, k); 
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript




<script>
 
// Javascript program to find the longest
// substring with k unique characters in
// a given string
let MAX_CHARS = 26;
 
// This function calculates number of
// unique characters using a associative
// array count[]. Returns true if no. of
// characters are less than required else
// returns false.
function isValid(count, k)
{
    let val = 0;
    for(let i = 0; i < MAX_CHARS; i++)
    {
        if (count[i] > 0)
        {
            val++;
        }
    }
 
    // Return true if k is greater
    // than or equal to val
    return (k >= val);
}
 
// Finds the maximum substring
// with exactly k unique chars
function kUniques(s,k)
{
     
    // Number of unique characters
    let u = 0;
    let n = s.length;
    let count = new Array(MAX_CHARS);
     
    for(let i = 0; i < MAX_CHARS; i++)
    {
        count[i] = 0;
    }
     
    // Traverse the string, Fills
    // the associative array
    // count[] and count number
    // of unique characters
    for(let i = 0; i < n; i++)
    {
        if (count[s[i].charCodeAt(0) -
                   'a'.charCodeAt(0)] == 0)
        {
            u++;
        }
        count[s[i].charCodeAt(0) -
              'a'.charCodeAt(0)]++;
    }
 
    // If there are not enough
    // unique characters, show
    // an error message.
    if (u < k)
    {
        document.write("Not enough unique characters");
        return;
    }
 
    // Otherwise take a window with
    // first element in it.
    // start and end variables.
    let curr_start = 0, curr_end = 0;
 
    // Also initialize values for
    // result longest window
    let max_window_size = 1;
    let max_window_start = 0;
 
    // Initialize associative
    // array count[] with zero
    for(let i = 0; i < MAX_CHARS; i++)
    {
        count[i] = 0;
    }
     
    // put the first character
    count[s[0].charCodeAt(0) -
           'a'.charCodeAt(0)]++;
 
    // Start from the second character and add
    // characters in window according to above
    // explanation
    for(let i = 1; i < n; i++)
    {
         
        // Add the character 's[i]'
        // to current window
        count[s[i].charCodeAt(0) -
               'a'.charCodeAt(0)]++;
        curr_end++;
 
        // If there are more than k
        // unique characters in
        // current window, remove from left side
        while (!isValid(count, k))
        {
            count[s[curr_start].charCodeAt(0) -
                            'a'.charCodeAt(0)]--;
            curr_start++;
        }
 
        // Update the max window size if required
        if (curr_end - curr_start + 1 > max_window_size)
        {
            max_window_size = curr_end - curr_start + 1;
            max_window_start = curr_start;
        }
    }
 
    document.write("Max substring is : " +
                   s.substring(max_window_start,
                   max_window_start +
                   max_window_size + 1) +
                   " with length " + max_window_size);
}
 
// Driver Code
let s = "aabacbebebe";
let k = 3;
 
kUniques(s, k);
 
// This code is contributed by rag2127
 
</script>

Output: 

Max substring is : cbebebe with length 7

Time Complexity: Considering function “isValid()” takes constant time, time complexity of above solution is O(n).
This article is contributed by Gaurav Sharma. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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