# Substring of length K having maximum frequency in the given string

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 = 5Output:ababaExplanation:

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 = 5Output:agoodExplanation:

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 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` |

**Output:**

ababa

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