Example: Find Frequency of Character fun main(args: Array<String>) { val str = "This website is awesome." val ch = 'e' var frequency = 0 for (i in 0..str.length - 1) { if (ch == str[i]) { ++frequency } } println
3 + * Description : Find Frequency of a charachter in a string 4 + * Author : Stephin Mathew 5 + * Version : 1.0 6 + * Date :25/09/2023 7 + */ 8 + 1 9 package test; 2 10 import java.util.Scanner; 3 11 public class FrequencyOfCharacter { 0...
classSolution {publicString frequencySort(String s) {for(charc:s.toCharArray()){ map.put(c, map.getOrDefault(c,0) + 1); } p.addAll(map.entrySet());while(!p.isEmpty()){ Map.Entry<Character, Integer> e =p.poll();for(inti = 0; i < e.getValue().intValue(); i++){ res....
public String frequencySort(String s) { Map<Character, Integer> map = new HashMap<>(); for (int i = 0; i < s.length(); i++) { map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0)+1); } PriorityQueue<Map.Entry<Character, Integer>> queue = new PriorityQueue<>((a, b)...
AC Java: 1 public class Solution { 2 public String frequencySort(String s) { 3 if(s == null || s.length() == 0){ 4 return s; 5 } 6 7 HashMap<Character, Integer> freqMap = new HashMap<Character, Integer>(); 8 for(char c : s.toCharArray()){ 9 freqMap.put(c, freqMap...
Firstly, enter the size of the array that you are concerned with. The array size, in this case, is 10. With that, you need to enter the elements of the array as well. The elements entered in this array are as follows: 1 2 3 4 4 3 2 1 1 4 ...
Java实现 时间O(n) - no need to sort anything 空间O(n) 1classSolution {2publicString frequencySort(String s) {3HashMap<Character, Integer> map =newHashMap<>();4for(charc : s.toCharArray()) {5map.put(c, map.getOrDefault(c, 0) + 1);6}78List<Character>[] bucket =newList[s....
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void repeatedCharacter() { var arr = "DeepakSharma"; Map<String, int> map = Map(); for (int i = 0; i < arr.length; i++) { if (map.containsKey(arr[i])) { map[arr[i]] = map[arr[i]]! + 1; } else { map[arr[i]] = 1; } } print("O/P -> $map"); } ...
For this problem, each frequency should be bounded in 0-len. (len is the length of the string). To achieve the O(n), use bucket sort. To put the same character together, use a arraylist for each bucket unit. import java.util.ArrayList;publicclassSolution{publicStringfrequencySort(Strings...