# Kth most frequent Character in a given String

• Difficulty Level : Easy
• Last Updated : 25 Nov, 2021

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:
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:

Approach

Below is the implementation of the above approach.

## C++

 `// C++ program to find kth most frequent``// character in a string``#include ``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 > 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;``}`

## Python3

 `# Python 3 program to find kth most frequent``# character in a string` `# function to sort elements by frequency``def` `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 program``if` `__name__ ``=``=` `'__main__'``:``    ``s ``=` `"geeksforgeeks"``    ``k ``=` `3``    ``print``(sortByFreq(s, k))`
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)

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