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("Frequency of $ch = $frequency") } When you run the...
Java 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++){...
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....
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....
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 ...
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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...
The enumerate(some_string) function yields a new value i (A counter going up) and a character from the some_string in each iteration. It then sets the (just assigned) i key of the dictionary some_dict to that character. The unrolling of the loop can be simplified as: >>> i, some_...