Given a string you need to print longest possible substring that has exactly M unique characters. If there is more than one substring of longest possible length, then print any one of them.
Examples:
Input: Str = “aabbcc”, k = 1
Output: 2
Explanation: Max substring can be any one from {“aa” , “bb” , “cc”}.Input: Str = “aabbcc”, k = 2
Output: 4
Explanation: Max substring can be any one from {“aabb” , “bbcc”}.Input: Str = “aabbcc”, k = 3
Output: 6
Explanation:
There are substrings with exactly 3 unique characters
{“aabbcc” , “abbcc” , “aabbc” , “abbc” }
Max is “aabbcc” with length 6.Input: Str = “aaabbb”, k = 3
Output: Not enough unique characters
Explanation: 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).
// C++ program to find the longest substring // with k unique characters in a given string #include <bits/stdc++.h> using namespace std;
// Function to calculate length of // longest substring with k characters void longestKSubstr(string s, int k)
{ int n = s.length();
int answer = -1;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j <= n; j++) {
unordered_set< char > distinct(s.begin() + i,
s.begin() + j);
if (distinct.size() == k) {
answer = max(answer, j - i);
}
}
}
cout << answer;
} // Driver function int main()
{ string s = "aabacbebebe" ;
int k = 3;
// Function Call
longestKSubstr(s, k);
return 0;
} |
// Java program to find the longest substring // with k unique characters in a given string import java.util.*;
public class GFG {
// Function to calculate length of
// longest substring with k characters
static void longestKSubstr(String s, int k)
{
int n = s.length();
int answer = - 1 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = i + 1 ; j <= n; j++) {
HashSet<Character> distinct
= new HashSet<Character>();
for ( int x = i; x < j; x++) {
distinct.add(s.charAt(x));
}
if (distinct.size() == k) {
answer = Math.max(answer, j - i);
}
}
}
System.out.println(answer);
}
public static void main(String[] args)
{
String s = "aabacbebebe" ;
int k = 3 ;
// Function Call
longestKSubstr(s, k);
}
} // This code is contributed by garg28harsh. |
// C# program to find ceil of a given value in BST using System;
using System.Collections.Generic;
public class GFG {
// Function to calculate length of
// longest substring with k characters
static void longestKSubstr( string s, int k)
{
int n = s.Length;
int answer = -1;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j <= n; j++) {
HashSet< char > distinct
= new HashSet< char >();
for ( int x = i; x < j; x++) {
distinct.Add(s[x]);
}
if (distinct.Count == k) {
answer = Math.Max(answer, j - i);
}
}
}
Console.WriteLine(answer);
}
public static void Main( string [] args)
{
String s = "aabacbebebe" ;
int k = 3;
// Function Call
longestKSubstr(s, k);
}
} // This code is contributed by karandeep1234 |
# Python program to find the longest substring # with k unique characters in a given string # Function to calculate length of # longest substring with k characters def longestKSubstr(s, k):
n = len (s)
answer = - 1
for i in range (n):
for j in range (i + 1 , n + 1 ):
distinct = set (s[i:j])
if len (distinct) = = k:
answer = max (answer, j - i)
print (answer)
s = "aabacbebebe"
k = 3
# Function Call longestKSubstr(s, k) |
// JavaScript function to find the longest substring // with k unique characters in a given string function longestKSubstr(s, k) {
let n = s.length;
let answer = -1;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j <= n; j++) {
let distinct = new Set();
for (let x = i; x < j; x++) {
distinct.add(s.charAt(x));
}
if (distinct.size === k) {
answer = Math.max(answer, j - i);
}
}
}
console.log(answer);
} let s = "aabacbebebe" ;
let k = 3; // Function call longestKSubstr(s, k); //This code is contributed by shivamsharma215 |
7
Time Complexity: O(n^2)
Auxiliary Space: O(n)
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++ 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;
} |
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 |
# 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
# Traverse 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# 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 |
<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> |
Max substring is : cbebebe with length 7
Time Complexity: Considering function “isValid()” takes constant time, time complexity of above solution is O(n).
Auxiliary Space: O(MAX_CHARS).
ANOTHER APPROACH:(Linear Time)
INTUITION :
- We declare a Map<Character,Integer> data structure to store frequency of characters.
- Here we use acquire and release property .
- In this we declare two pointers i and j, we traverse from i pointer and acquire characters until map size is greater than K. When map size get equals to k we store the length of i-j in a variable and traverse forward.
- Once map size exceeds K,then we start releasing characters from map with the help of j pointer till map size again equals K, and we store the length and compare with previous length and this way the algorithm continues.
Implementation:
#include <iostream> #include <unordered_map> using namespace std;
int longestKSubstr(string s, int k)
{ int i = 0;
int j = 0;
int ans = -1;
unordered_map< char , int > mp;
while (j < s.size()) {
mp[s[j]]++;
while (mp.size() > k) {
mp[s[i]]--;
if (mp[s[i]] == 0)
mp.erase(s[i]);
i++;
}
if (mp.size() == k) {
ans = max(ans, j - i + 1);
}
j++;
}
return ans;
} int main()
{ string s = "aabacbebebe" ;
int k = 3;
cout << longestKSubstr(s, k) << endl;
return 0;
} |
//Java program to find the longest substring with k unique characters in a given string import java.io.*;
import java.util.*;
class GFG {
public static int longestkSubstr(String S, int k)
{
// code here
Map<Character, Integer> map = new HashMap<>();
int i = - 1 ;
int j = - 1 ;
int ans = - 1 ;
while ( true ) {
boolean flag1 = false ;
boolean flag2 = false ;
while (i < S.length() - 1 ) {
flag1 = true ;
i++;
char ch = S.charAt(i);
map.put(ch, map.getOrDefault(ch, 0 ) + 1 );
if (map.size() < k)
continue ;
else if (map.size() == k) {
int len = i - j;
ans = Math.max(len, ans);
}
else
break ;
}
while (j < i) {
flag2 = true ;
j++;
char ch = S.charAt(j);
if (map.get(ch) == 1 )
map.remove(ch);
else
map.put(ch, map.get(ch) - 1 );
if (map.size() == k) {
int len = i - j;
ans = Math.max(ans, len);
break ;
}
else if (map.size() > k)
continue ;
}
if (flag1 == false && flag2 == false )
break ;
}
return ans;
}
public static void main(String[] args)
{
String s = "aabacbebebe" ;
int k = 3 ;
// Function Call
int ans = longestkSubstr(s, k);
System.out.println(ans);
}
} // This code is contributed by Raunak Singh |
def longestkSubstr(S, k):
# Create an empty dictionary to store character counts
map = {}
# Initialize two pointers i and j to -1
i = - 1
j = - 1
# Initialize ans to -1
ans = - 1
# Loop until break statement is reached
while True :
# Initialize two flags to False
flag1 = False
flag2 = False
# Loop until i reaches end of string S
while i < len (S) - 1 :
# Set flag1 to True
flag1 = True
# Increment i and get character at index i
i + = 1
ch = S[i]
# Add character count to dictionary
map [ch] = map .get(ch, 0 ) + 1
# If number of unique characters is less than k, continue loop
if len ( map ) < k:
continue
# If number of unique characters is equal to k, update ans if necessary and continue loop
elif len ( map ) = = k:
len_ = i - j
ans = max (len_, ans)
# If number of unique characters is greater than k, break loop
else :
break
# Loop until j reaches i
while j < i:
# Set flag2 to True
flag2 = True
# Increment j and get character at index j
j + = 1
ch = S[j]
# If character count is 1, remove character from dictionary
if map [ch] = = 1 :
del map [ch]
# If character count is greater than 1, decrement character
# count in dictionary
else :
map [ch] - = 1
# If number of unique characters is equal to k,
# update ans if necessary and break loop
if len ( map ) = = k:
len_ = i - j
ans = max (ans, len_)
break
# If number of unique characters is greater than k, continue loop
elif len ( map ) > k:
continue
# If both flags are False, break outer loop (while True)
if not flag1 and not flag2:
break
return ans
# Test case inputs and function call s = "aabacbebebe"
k = 3
ans = longestkSubstr(s, k)
print (ans)
|
using System;
using System.Collections.Generic;
class MainClass {
public static int LongestKSubstr( string s, int k)
{
int i = 0;
int j = 0;
int ans = -1;
Dictionary< char , int > mp
= new Dictionary< char , int >();
while (j < s.Length) {
if (mp.ContainsKey(s[j])) {
mp[s[j]]++;
}
else {
mp[s[j]] = 1;
}
while (mp.Count > k) {
mp[s[i]]--;
if (mp[s[i]] == 0) {
mp.Remove(s[i]);
}
i++;
}
if (mp.Count == k) {
ans = Math.Max(ans, j - i + 1);
}
j++;
}
return ans;
}
public static void Main()
{
string s = "aabacbebebe" ;
int k = 3;
Console.WriteLine(LongestKSubstr(s, k));
}
} |
function longestKSubstr(s, k) {
let i = 0;
let j = 0;
let ans = -1;
let mp = new Map();
while (j < s.length) {
mp.set(s[j], (mp.get(s[j]) || 0) + 1);
while (mp.size > k) {
mp.set(s[i], mp.get(s[i]) - 1);
if (mp.get(s[i]) === 0) {
mp. delete (s[i]);
}
i++;
}
if (mp.size === k) {
ans = Math.max(ans, j - i + 1);
}
j++;
}
return ans;
} let s = "aabacbebebe" ;
let k = 3; console.log(longestKSubstr(s, k)); |
7
Time Complexity: O(|S|)
Space Complexity:O(|S|)