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.
Input: str = “GeeksforGeeks”, K = 3
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
- 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.
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)