Learn to write asimple Java program that finds the duplicate characters in a String. This can be a possibleJava interview questionwhile the interviewer may evaluate our coding skills. We can use the given code t
To determine that a word is duplicate, we are mainitaining aHashSet. If theSet.add()method returnfalse, the it means that word is already present in the set and thus it is duplicate. List<String>wordsList=Arrays.stream(sentence.split(" ")).collect(Collectors.toList());Set<String>tempS...
Java interview may surprise you sometimes. There are so many similar questions you may get in an Interview like Create your own contains() method in java, find duplicate char from String, etc. In this tutorial we will create simple way to find duplicate character from String. package com.cru...
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In Java 8 Stream, filter withSet.Add()is the fastest algorithm to find duplicate elements, because it loops only one time. Set<T> items =newHashSet<>();returnlist.stream() .filter(n -> !items.add(n)) .collect(Collectors.toSet());Copy ...
Write a Java program to find the first non-repeating character in a string. Visual Presentation: Sample Solution: Java Code: // Importing necessary Java utilities.importjava.util.*;// Define a class named Main.publicclassMain{// Main method to execute the program.publicstaticvoidmain(String[]...
Now,we aim to find the keys with duplicate values, grouping different developers that use the same operating system in aSet: Map<String, Set<String>> EXPECTED = Map.of( "Linux", Set.of("Kai", "David"), "Windows", Set.of("Saajan", "Kevin"), ...
Write a Java program to find duplicate values in an array of string values.Pictorial Presentation:Sample Solution:Java Code:// Define a class named Exercise13. public class Exercise13 { // The main method where the program execution starts. public static void main(String[] args) { // ...
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public int findDuplicate(int[] nums) { int slow = 0; int fast = 0; // 找到快慢指针相遇的地方 do{ slow = nums[slow]; fast = nums[nums[fast]]; } while(slow != fast); int find = 0; // 用一个新指针从头开始,直到和慢指针相遇 ...