Kth most frequent Character in a given String
Given a string str and an integer K, the task is to find the K-th most frequent character in the string. If there are multiple characters that can account as K-th most frequent character then, print any one of them.
Examples:
Input: str = “GeeksforGeeks”, K = 3
Output: f
Explanation:
K = 3, here ‘e’ appears 4 times
& ‘g’, ‘k’, ‘s’ appears 2 times
& ‘o’, ‘f’, ‘r’ appears 1 time.
Any output from ‘o’ (or) ‘f’ (or) ‘r’ will be correct.
Input: str = “trichotillomania”, K = 2
Output: l
Approach
- The idea is to Use the Characters as key in Hashmap and store their occurrences in the string.
- Sort the Hashmap and find the K-th character.
Below is the implementation of the above approach.
C++
// C++ program to find kth most frequent// character in a string#include <bits/stdc++.h>using namespace std;// Used for sorting by frequency.bool sortByVal(const pair<char, int>& a, const pair<char, int>& b){ return a.second > b.second;}// function to sort elements by frequencychar sortByFreq(string str, int k){ // Store frequencies of characters unordered_map<char, int> m; for (int i = 0; i < str.length(); ++i) m[str[i]]++; // Copy map to vector vector<pair<char, int> > v; copy(m.begin(), m.end(), back_inserter(v)); // Sort the element of array by frequency sort(v.begin(), v.end(), sortByVal); // Find k-th most frequent item. Please note // that we need to consider only distinct int count = 0; for (int i = 0; i < v.size(); i++) { // Increment count only if frequency is // not same as previous if (i == 0 || v[i].second != v[i - 1].second) count++; if (count == k) return v[i].first; } return -1;}// Driver programint main(){ string str = "geeksforgeeks"; int k = 3; cout << sortByFreq(str, k); return 0;} |
Python3
# Python 3 program to find kth most frequent# character in a string# function to sort elements by frequencydef sortByFreq(s, k): # Store frequencies of characters m=dict() for c in s: m=m.get(c,0)+1 # Copy map to vector v=list(m.items()) # Sort the element of array by frequency v.sort(key=lambda x:x[1],reverse= True) # Find k-th most frequent item. Please note # that we need to consider only distinct count = 0 for i in range(len(v)): # Increment count only if frequency is # not same as previous if (i == 0 or v[i][1] != v[i - 1][1]): count+=1 if (count == k): return v[i][0] return -1# Driver programif __name__ == '__main__': s = "geeksforgeeks" k = 3 print(sortByFreq(s, k)) |
r
Time Complexity: O(NlogN) Please note that this is an upper bound on time complexity. If we consider alphabet size as constant (for example lower case English alphabet size is 26), we can say time complexity as O(N). The vector size would never be more that alphabet size.
Auxiliary Space: O(N)
