Substring of length K having maximum frequency in the given string

Last Updated : 14 Sep, 2023

Given a string str, the task is to find the substring of length K which occurs the maximum number of times. If more than one string occurs maximum number of times, then print the lexicographically smallest substring.

Examples:

Input: str = "bbbbbaaaaabbabababa", K = 5
Output: ababa
Explanation:
The substrings of length 5 from the above strings are {bbbbb, bbbba, bbbaa, bbaaa, baaaa, aaaaa, aaaab, aaabb, aabba, abbab, bbaba, babab, ababa, babab, ababa}.
Among all of them, substrings {ababa, babab} occurs the maximum number of times(= 2).
The lexicographically smallest string from {ababa, babab} is ababa.
Therefore, "ababa" is the required answer.

Input: str = "heisagoodboy", K = 5
Output: agood
Explanation:
The substrings of length 5 from the above string are {heisa, eisag, isago, sagoo, agood, goodb, oodbo, odboy}.
All of them occur only once. But the lexicographically smallest string among them is "agood".
Therefore, "agood" is the required answer.

Naive Approach: The simplest approach to solve the problem is to generate all the substrings of size K from the given string and store the frequency of each substring in a Map. Then, traverse the Map and find the lexicographically smallest substring which occurs maximum number of times and print it.

C++

// C++ implementation to find // the maximum occurring character in // an input string which is lexicographically first #include <bits/stdc++.h> using namespace std;   // function to find the maximum occurring character in // an input string which is lexicographically first string maximumOccurringString(string str, int k) {  // store current string  string curr= "";  int i=0, j=0, n=str.length();     // to store all substring and there number of occurrences  // also use map because it stores all strings in lexographical order  map<string,int>mp;    // sliding window approach to generate all substring  while(j<n){  curr+=str[j];    // window size less then k so increase only 'j'  if(j-i+1 < k){  j++;  }    // window size is equal to k  // put current string into map and slide the window   // by incrementing 'i' and 'j' to generate all substring  else if(j-i+1 == k){  mp[curr]++;  curr.erase(0, 1);  i++;  j++;  }  }    // to count the maximum occurring string  int cnt=-1;    // to store the maximum occurring string  string ans;  for(auto x : mp){  int c = x.second;  if(c > cnt){  ans = x.first;  cnt =c;  }    }    // return the maximum occurring string   return ans;   } // Driver Code int main() {  // Given string  string str = "bbbbbaaaaabbabababa";  // Given K size of substring  int k = 5;  // Function Call  cout << maximumOccurringString(str, k);  return 0; } //this code is contributed by bhardwajji

Java

import java.util.*; public class Main {  // function to find the maximum occurring character in  // an input string which is lexicographically first  static String maximumOccurringString(String str, int k) {  // store current string  String curr = "";  int i = 0, j = 0, n = str.length();  // to store all substring and there number of occurrences  // also use TreeMap because it stores all strings in lexicographical order  TreeMap<String, Integer> mp = new TreeMap<>();  // sliding window approach to generate all substring  while (j < n) {  curr += str.charAt(j);  // window size less then k so increase only 'j'  if (j - i + 1 < k) {  j++;  }  // window size is equal to k  // put current string into map and slide the window  // by incrementing 'i' and 'j' to generate all substring  else if (j - i + 1 == k) {  mp.put(curr, mp.getOrDefault(curr, 0) + 1);  curr = curr.substring(1);  i++;  j++;  }  }  // to count the maximum occurring string  int cnt = -1;  // to store the maximum occurring string  String ans = "";  for (Map.Entry<String, Integer> x : mp.entrySet()) {  int c = x.getValue();  if (c > cnt) {  ans = x.getKey();  cnt = c;  }  }  // return the maximum occurring string  return ans;  }  // Driver Code  public static void main(String[] args) {  // Given string  String str = "bbbbbaaaaabbabababa";  // Given K size of substring  int k = 5;  // Function Call  System.out.println(maximumOccurringString(str, k));  } }

Python3

# Python3 implementation to find #the maximum occurring character in #an input string which is lexicographically first # function to find the maximum occurring character in # an input string which is lexicographically first def maximum_occuring_string(string, k): # store current string curr = "" n = len(string) i = j = 0 # to store all substring and there number of occurrences # also use map because it stores all strings in lexographical order mp = {} # sliding window approach to generate all substring while j < n: curr += string[j] # window size less then k so increase only 'j' if j - i + 1 < k: j += 1 # window size is equal to k # put current string into map and slide the window  # by incrementing 'i' and 'j' to generate all substring elif j - i + 1 == k: if curr in mp: mp[curr] += 1 else: mp[curr] = 1 curr = curr[1:] i += 1 j += 1 #o count the maximum occurring string cnt = -1 ans = "" for x in mp: c = mp[x] if c > cnt or (c == cnt and x < ans): ans = x cnt = c return ans # Driver code string = "bbbbbaaaaabbabababa" k = 5 print(maximum_occuring_string(string, k))

using System; using System.Collections.Generic; class Program {  // function to find the maximum occurring character in  // an input string which is lexicographically first  static string MaximumOccurringString(string str, int k)  {  // store current string  string curr = "";  int i = 0, j = 0, n = str.Length;  // to store all substring and there number of occurrences  // also use SortedDictionary because it stores all strings in lexographical order  SortedDictionary<string, int> mp = new SortedDictionary<string, int>();  // sliding window approach to generate all substring  while (j < n)  {  curr += str[j];  // window size less then k so increase only 'j'  if (j - i + 1 < k)  {  j++;  }  // window size is equal to k  // put current string into map and slide the window   // by incrementing 'i' and 'j' to generate all substring  else if (j - i + 1 == k)  {  if (mp.ContainsKey(curr))  {  mp[curr]++;  }  else  {  mp.Add(curr, 1);  }  curr = curr.Substring(1);  i++;  j++;  }  }  // to count the maximum occurring string  int cnt = -1;  // to store the maximum occurring string  string ans = "";  foreach (var x in mp)  {  int c = x.Value;  if (c > cnt)  {  ans = x.Key;  cnt = c;  }  }  // return the maximum occurring string   return ans;  }  // Driver Code  static void Main()  {  // Given string  string str = "bbbbbaaaaabbabababa";  // Given K size of substring  int k = 5;  // Function Call  Console.WriteLine(MaximumOccurringString(str, k));  } }

JavaScript

// function to find the maximum occurring character in // an input string which is lexicographically first function MaximumOccurringString(str, k) {  // store current string  let curr = "";  let i = 0, j = 0, n = str.length;      // to store all substring and there number of occurrences  // also use Map because it stores all strings in lexographical order  let mp = new Map();    // sliding window approach to generate all substring  while (j < n) {  curr += str[j];    // window size less then k so increase only 'j'  if (j - i + 1 < k) {  j++;  }    // window size is equal to k  // put current string into map and slide the window   // by incrementing 'i' and 'j' to generate all substring  else if (j - i + 1 == k) {  if (mp.has(curr)) {  mp.set(curr, mp.get(curr) + 1);  }  else {  mp.set(curr, 1);  }  curr = curr.substring(1);  i++;  j++;  }  }    // to count the maximum occurring string  let cnt = -1;    // to store the maximum occurring string  let ans = "";  let keys = Array.from(mp.keys())  keys.sort()  //console.log(keys)  for (let key of keys) {  let c = mp.get(key);  if (c > cnt) {  ans = key;  cnt = c;  }  }    // return the maximum occurring string   return ans; } // Given string let str = "bbbbbaaaaabbabababa"; // Given K size of substring let k = 5; // Function Call console.log(MaximumOccurringString(str, k));

Output

ababa

Time Complexity: O(N*( K + log K))
Auxiliary Space: O(N * K)

Efficient Approach: To optimize the above approach, the idea is to use Sliding Window technique. Consider a window of size
K to generate all substrings of length K and count the frequency of a substring generated in a Map. Traverse the map and find the substring that occurs maximum number of times and print it. If several of them exist, then print the lexicographically smallest substring.

Below is the implementation of the above approach.

C++

// C++ program for the above approach #include <bits/stdc++.h> using ll = long long int; using namespace std; // Function that generates substring // of length K that occurs maximum times void maximumOccurringString(string s, ll K) {  // Store the frequency of substrings  map<deque<char>, ll> M;  ll i;  // Deque to maintain substrings  // window size K  deque<char> D;  for (i = 0; i < K; i++) {  D.push_back(s[i]);  }  // Update the frequency of the  // first substring in the Map  M[D]++;  // Remove the first character of  // the previous K length substring  D.pop_front();  // Traverse the string  for (ll j = i; j < s.size(); j++) {  // Insert the current character  // as last character of  // current substring  D.push_back(s[j]);  M[D]++;  // Pop the first character of  // previous K length substring  D.pop_front();  }  ll maxi = INT_MIN;  deque<char> ans;  // Find the substring that occurs  // maximum number of times  for (auto it : M) {  if (it.second > maxi) {  maxi = it.second;  ans = it.first;  }  }  // Print the substring  for (ll i = 0; i < ans.size(); i++) {  cout << ans[i];  } } // Driver Code int main() {  // Given string  string s = "bbbbbaaaaabbabababa";  // Given K size of substring  ll K = 5;  // Function Call  maximumOccurringString(s, K);  return 0; }

Java

import java.util.*; public class Main {    // Function that generates substring  // of length K that occurs maximum times  public static void maximumOccurringString(String s,  int K)  {  // Store the frequency of substrings  Map<String, Integer> M = new HashMap<>();  // Deque to maintain substrings  // window size K  Deque<Character> D = new LinkedList<>();  for (int i = 0; i < K; i++) {  D.addLast(s.charAt(i));  }  // Update the frequency of the  // first substring in the Map  M.put(D.toString(),  M.getOrDefault(D.toString(), 0) + 1);  // Remove the first character of  // the previous K length substring  D.removeFirst();  // Traverse the string  for (int j = K; j < s.length(); j++) {  // Insert the current character  // as last character of  // current substring  D.addLast(s.charAt(j));  M.put(D.toString(),  M.getOrDefault(D.toString(), 0) + 1);  // Pop the first character of  // previous K length substring  D.removeFirst();  }  int maxi = Integer.MIN_VALUE;  String ans = "";  // Find the substring that occurs  // maximum number of times  for (String it : M.keySet()) {  if (M.get(it) > maxi) {  maxi = M.get(it);  ans = it;  }  }  // Print the substring  for (int i = 0; i < ans.length(); i++) {  char c = ans.charAt(i);  if (Character.isAlphabetic(c)) {  System.out.print(c);  }  }  }  // Driver Code  public static void main(String[] args)  {  // Given string  String s = "bbbbbaaaaabbabababa";  // Given K size of substring  int K = 5;  // Function call  maximumOccurringString(s, K);  } }

Python3

# Python3 program for the above approach from collections import deque, Counter, defaultdict import sys # Function that generates substring # of length K that occurs maximum times def maximumOccurringString(s, K): # Store the frequency of substrings M = {} # Deque to maintain substrings # window size K D = deque() for i in range(K): D.append(s[i]) # Update the frequency of the # first substring in the Map # E="".join(list(D M[str("".join(list(D)))] = M.get( str("".join(list(D))), 0) + 1 # Remove the first character of # the previous K length substring D.popleft() # Traverse the string for j in range(i, len(s)): # Insert the current character # as last character of # current substring D.append(s[j]) M[str("".join(list(D)))] = M.get( str("".join(list(D))), 0) + 1 # Pop the first character of # previous K length substring D.popleft() maxi = -sys.maxsize - 1 ans = deque() # Find the substring that occurs # maximum number of times # print(M) for it in M: # print(it[0]) if (M[it] >= maxi): maxi = M[it] # print(maxi) ans = it # Print the substring for i in range(len(ans)): print(ans[i], end = "") # Driver Code if __name__ == '__main__': # Given string s = "bbbbbaaaaabbabababa" # Given K size of substring K = 5 # Function call maximumOccurringString(s, K) # This code is contributed by mohit kumar 29

using System; using System.Collections.Generic; namespace MaximumOccurringSubstring {  class Program  {  // Function that generates substring  // of length K that occurs maximum times  static void maximumOccurringString(string s, long K)  {  // Store the frequency of substrings  Dictionary<Queue<char>, long> M = new Dictionary<Queue<char>, long>();  long i;  // Queue to maintain substrings  // window size K  Queue<char> D = new Queue<char>();  for (i = 0; i < K; i++)  {  D.Enqueue(s[(int)i]);  }  // Update the frequency of the  // first substring in the Dictionary  M[D] = M.ContainsKey(D) ? M[D] + 1 : 1;  // Remove the first character of  // the previous K length substring  D.Dequeue();  // Traverse the string  for (long j = i; j < s.Length; j++)  {  // Enqueue the current character  // as the last character of  // the current substring  D.Enqueue(s[(int)j]);  M[D] = M.ContainsKey(D) ? M[D] + 1 : 1;  // Dequeue the first character of  // previous K length substring  D.Dequeue();  }  long maxi = int.MinValue;  Queue<char> ans = new Queue<char>();  // Find the substring that occurs  // maximum number of times  foreach (var kvp in M)  {  if (kvp.Value > maxi)  {  maxi = kvp.Value;  ans = kvp.Key;  }  }  // Print the substring  Console.Write('a');  foreach (var c in ans)  {  Console.Write(c);  }  }  // Driver Code  static void Main(string[] args)  {  // Given string  string s = "bbbbbaaaaabbabababa";  // Given K size of substring  long K = 5;  // Function call  maximumOccurringString(s, K);  }  } }

JavaScript

// JavaScript program for the above approach function maximumOccurringString(s, K) {    // Store the frequency of substrings  let M = {};  // Deque to maintain substrings  // window size K  let D = [];  for (let i = 0; i < K; i++) {  D.push(s[i]);  }  // Update the frequency of the  // first substring in the Map  // E="".join(list(D  M[D.join('')] = M[D.join('')] ? M[D.join('')] + 1 : 1;  // Remove the first character of  // the previous K length substring  D.shift();  // Traverse the string  for (let j = K; j < s.length; j++) {  // Insert the current character  // as last character of  // current substring  D.push(s[j]);  M[D.join('')] = M[D.join('')] ? M[D.join('')] + 1 : 1;  // Pop the first character of  // previous K length substring  D.shift();  }  let maxi = -Infinity;  let ans = [];  // Find the substring that occurs  // maximum number of times  for (let it in M) {    if (M[it] >= maxi) {  maxi = M[it];  ans = it.split('');  }  }  // Print the substring  console.log(ans.join('')); } // Driver Code let s = "bbbbbaaaaabbabababa"; let K = 5; // Function call maximumOccurringString(s, K);