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} ocuurs 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.
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; } |
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 subin 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 subthat 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 |
ababa
Time Complexity: O((N – K)*log(N – K))
Auxiliary Space: O(N – K)
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