Given a string S of length N and an integer K, the task is to find the frequency of the lexicographically Kth smallest character present in the given string.
Examples:
Input: S = “neveropen”, K = 3
Output: 4
Explanation: The lexicographically 3rd smallest character in S is ‘e’. Frequency of ‘e’ in S = 4.Input: S = “abcdabcd”, K = 4
Output: 2
Explanation: The lexicographically 4th smallest lexicographical character in S is ‘b’. Frequency of ‘b’ in S = 2.
Approach: The idea is to sort the string in ascending order of their ASCII values. Now, find the (K – 1)th character in the sorted string and find its frequency. Follow the steps below to solve the problem:
- Sort the given string S in ascending order.
- Store the (K – 1)th character in a variable ch, i.e. S[K – 1].
- Find the frequency of the character ch in S and print that value.
Below is the implementation of the approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the frequency of// the lexicographically Kth smallest// charactervoid KthCharacter( string S, int N, int K){ // Convert the string to // array of characters char strarray[N + 1]; strcpy(strarray, S.c_str()); // Sort the array in ascending order sort(strarray, strarray + N); // Store the Kth character char ch = strarray[K - 1]; // Store the frequency of // the K-th character int count = 0; // Count the frequency of // the K-th character for (auto c : strarray) { if (c == ch) count++; } // Print the count cout << count;}// Driver Codeint main(){ string S = "neveropen"; int N = S.length(); int K = 3; KthCharacter(S, N, K); return 0;}// This code is contributed by sanjoy_62. |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find the frequency of // the lexicographically Kth smallest // character public static void KthCharacter( String S, int N, int K) { // Convert the string to // array of characters char strarray[] = S.toCharArray(); // Sort the array in ascending order Arrays.sort(strarray); // Store the Kth character char ch = strarray[K - 1]; // Store the frequency of // the K-th character int count = 0; // Count the frequency of // the K-th character for (char c : strarray) { if (c == ch) count++; } // Print the count System.out.println(count); } // Driver Code public static void main(String[] args) { String S = "neveropen"; int N = S.length(); int K = 3; KthCharacter(S, N, K); }} |
Python3
# Python program for the above approach# Function to find the frequency of# the lexicographically Kth smallest# characterdef KthCharacter(S, N, K): # Convert the string to # array of characters strarray = [char for char in S]; # Sort the array in ascending order strarray.sort(); # Store the Kth character ch = strarray[K - 1]; # Store the frequency of # the K-th character count = 0; # Count the frequency of # the K-th character for c in strarray: if (c == ch): count += 1; # Print the count print(count);# Driver Codeif __name__ == '__main__': S = "neveropen"; N = len(S); K = 3; KthCharacter(S, N, K);# This code is contributed by 29AjayKumar |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to find the frequency of // the lexicographically Kth smallest // character public static void KthCharacter( string S, int N, int K) { // Convert the string to // array of characters char[] strarray = S.ToCharArray(); // Sort the array in ascending order Array.Sort(strarray); // Store the Kth character char ch = strarray[K - 1]; // Store the frequency of // the K-th character int count = 0; // Count the frequency of // the K-th character foreach (char c in strarray) { if (c == ch) count++; } // Print the count Console.Write(count); } // Driver Code public static void Main() { string S = "neveropen"; int N = S.Length; int K = 3; KthCharacter(S, N, K); }}// This code is contributed by splevel62. |
Javascript
<script>// Javascript program for the above approach// Function to find the frequency of// the lexicographically Kth smallest// characterfunction KthCharacter(S, N, K){ // Convert the string to // array of characters var strarray = S.split(''); strarray.sort(); // Store the Kth character var ch = strarray[K - 1]; // Store the frequency of // the K-th character var count = 0; // Count the frequency of // the K-th character strarray.forEach(c => { if (c == ch) count++; }); // Print the count document.write(count);}// Driver Codevar S = "neveropen";var N = S.length;var K = 3;KthCharacter(S, N, K);// This code is contributed by famously</script> |
Output:
4
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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