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++ 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;
} |
// Java program to find kth most frequent // character in a string import java.io.*;
import java.util.*;
class imp3 {
// Used for sorting by frequency.
static class pair implements Comparable<pair> {
char ch;
int freq;
pair( char ch, int freq)
{
this .ch = ch;
this .freq = freq;
}
public int compareTo(pair a)
{
return a.freq - this .freq;
}
}
// function to sort elements by frequency
static char sortByFreq(String str, int k)
{
// Store frequencies of characters
HashMap<Character, Integer> m = new HashMap<>();
for ( int i = 0 ; i < str.length(); ++i) {
char ch = str.charAt(i);
int freq = m.getOrDefault(str.charAt(i), 0 ) + 1 ;
m.put(ch, freq);
}
// Copy map to array
pair[] v = new pair[m.size()];
int idx = 0 ;
for (Character ch : m.keySet()) {
v[idx++] = new pair(ch, m.get(ch));
}
// Sort the element of array by frequency
Arrays.sort(v);
// Find k-th most frequent item. Please note
// that we need to consider only distinct
int count = 0 ;
for ( int i = 0 ; i < v.length; i++) {
// Increment count only if frequency is
// not same as previous
if (i == 0 || v[i].freq != v[i - 1 ].freq)
count++;
if (count == k)
return v[i].ch;
}
return '.' ;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
int k = 3 ;
System.out.println(sortByFreq(str, k));
}
} // This code is contributed by Karandeep1234 |
# 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))
|
// C# code for the above approach using System;
using System.Collections.Generic;
using System.Linq;
class imp3 {
// Used for sorting by frequency.
class pair : IComparable<pair> {
public char ch;
public int freq;
public pair( char ch, int freq)
{
this .ch = ch;
this .freq = freq;
}
public int CompareTo(pair a)
{
return a.freq - this .freq;
}
}
// function to sort elements by frequency
static char sortByFreq( string str, int k)
{
// Store frequencies of characters
Dictionary< char , int > m = new Dictionary< char , int >();
for ( int i = 0; i < str.Length; ++i) {
char ch = str[i];
int freq = m.GetValueOrDefault(str[i], 0) + 1;
m[ch] = freq;
}
// Copy map to array
pair[] v = new pair[m.Count];
int idx = 0;
foreach ( char ch in m.Keys) {
v[idx++] = new pair(ch, m[ch]);
}
// Sort the element of array by frequency
Array.Sort(v);
// Find k-th most frequent item. Please note
// that we need to consider only distinct
int count = 0;
for ( int i = 0; i < v.Length; i++) {
// Increment count only if frequency is
// not same as previous
if (i == 0 || v[i].freq != v[i - 1].freq)
count++;
if (count == k)
return v[i].ch;
}
return '.' ;
}
static void Main( string [] args)
{
string str = "geeksforgeeks" ;
int k = 3;
Console.WriteLine(sortByFreq(str, k));
}
} // This code is contributed by Prince Kumar |
<script> // JavaScript program to find kth most frequent // character in a string // Used for sorting by frequency. function sortByVal(a,b)
{ return b[1] - a[1];
} // function to sort elements by frequency function sortByFreq(str,k)
{ // Store frequencies of characters
let m = new Map();
for (let i = 0; i < str.length; ++i){
if (m.has(str[i])){
m.set(str[i],m.get(str[i])+1);
}
else m.set(str[i],1);
}
// Copy map to vector
let v = [];
for (let [x,y] of m){
v.push([x,y]);
}
// Sort the element of array by frequency
v.sort(sortByVal);
// Find k-th most frequent item. Please note
// that we need to consider only distinct
let count = 0;
for (let i = 0; i < v.length; i++) {
// Increment count only if frequency is
// not same as previous
if (i == 0 || v[i][1] != v[i - 1][1])
count++;
if (count == k)
return v[i][0];
}
return -1;
} // Driver program let str = "geeksforgeeks" ;
let k = 3; document.write(sortByFreq(str, k)); // This code is contributed by shinjanpatra. </script> |
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)