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

• 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++ 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 frequency char 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 program int main() {     string str = "geeksforgeeks";     int k = 3;     cout << sortByFreq(str, k);     return 0; }

Output:

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)