Java String Exercises: Remove duplicate characters from a given string presents in another given string
Java String: Exercise-41 with Solution
Write a Java program to remove duplicate characters from a given string presents in another given string.
Pictorial Presentation:
Sample Solution-1:
Java Code:
import java.util.*;
public class Main {
public static void main(String[] args) {
String str1 = "the quick brown fox";
String str2 = "queen";
System.out.println("The given string is: " + str1);
System.out.println("The given mask string is: " + str2);
char arr[] = new char[str1.length()];
char[] mask = new char[256];
for (int i = 0; i < str2.length(); i++)
mask[str2.charAt(i)]++;
System.out.println("\nThe new string is: ");
for (int i = 0; i < str1.length(); i++) {
if (mask[str1.charAt(i)] == 0)
System.out.print(str1.charAt(i));
}
}
}
Sample Output:
The given string is: the quick brown fox The given mask string is: queen The new string is: th ick brow fox
Flowchart:
Sample Solution-2:
Main.java Code:
//MIT License: https://bit.ly/35gZLa3
import java.util.Arrays;
import java.util.Map;
import java.util.concurrent.TimeUnit;
public class Main {
private static final String TEXT = "Be strong, be fearless, be beautiful. "
+ "And believe that anything is possible when you have the right "
+ "people there to support you. ";
// Ӝ -> Unicode: \u04DC, Code Point: 1244
// 💕 -> Unicode: \uD83D\uDC95, Code Point: 128149
// 🎼 -> \uD83C\uDFBC, Code Point: 127932
// 😍 ->\uD83D\uDE0D, Code Point: 128525
private static final String TEXT_CP = TEXT + "😍 I love 💕 you Ӝ so much 💕 😍 🎼🎼🎼!";
public static void main(String[] args) {
System.out.println("Input text: \n" + TEXT + "\n");
System.out.println("\n\nASCII or 16 bits Unicode characters (less than 65,535 (0xFFFF)) examples:\n");
System.out.println("HashMap based solution:");
long startTimeV1 = System.nanoTime();
Map<Character, Integer> duplicatesV1 = Strings.countDuplicateCharactersV1(TEXT);
displayExecutionTime(System.nanoTime()-startTimeV1);
System.out.println(Arrays.toString(duplicatesV1.entrySet().toArray()));
// or: duplicatesV1.forEach( (k, v) -> System.out.print(k + "="+ v + ", "));
System.out.println();
System.out.println("Java 8, functional-style solution:");
long startTimeV2 = System.nanoTime();
Map<Character, Integer> duplicatesV2 = Strings.countDuplicateCharactersV2(TEXT);
displayExecutionTime(System.nanoTime()-startTimeV2);
System.out.println(Arrays.toString(duplicatesV2.entrySet().toArray()));
// or: duplicatesV2.forEach( (k, v) -> System.out.print(k + "="+ v + ", "));
System.out.println("\n--------------------------------------\n");
System.out.println("Input text: \n" + TEXT_CP + "\n");
System.out.println("\n\nIncluding Unicode surrogate pairs examples:\n");
System.out.println("HashMap based solution:");
long startTimeV3 = System.nanoTime();
Map<String, Integer> duplicatesV3 = Strings.countDuplicateCharactersVCP1(TEXT_CP);
displayExecutionTime(System.nanoTime()-startTimeV3);
System.out.println(Arrays.toString(duplicatesV3.entrySet().toArray()));
// or: duplicatesV3.forEach( (k, v) -> System.out.print(k + "="+ v + ", "));
System.out.println();
System.out.println("Java 8, functional-style solution:");
long startTimeV4 = System.nanoTime();
Map<String, Integer> duplicatesV4 = Strings.countDuplicateCharactersVCP2(TEXT_CP);
displayExecutionTime(System.nanoTime()-startTimeV4);
System.out.println(Arrays.toString(duplicatesV4.entrySet().toArray()));
// or: duplicatesV4.forEach( (k, v) -> System.out.print(k + "="+ v + ", "));
}
private static void displayExecutionTime(long time) {
System.out.println("Execution time: " + time + " ns" + " (" +
TimeUnit.MILLISECONDS.convert(time, TimeUnit.NANOSECONDS) + " ms)");
}
}
Strings.java Code:
//MIT License: https://bit.ly/35gZLa3
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.stream.Collectors;
public final class Strings {
private Strings() {
throw new AssertionError("Cannot be instantiated");
}
public static Map<Character, Integer> countDuplicateCharactersV1(String str) {
if (str == null || str.isEmpty()) {
// or throw IllegalArgumentException
return Collections.emptyMap();
}
Map<Character, Integer> result = new HashMap<>();
// or use for(char ch: str.toCharArray()) { ... }
for (int i = 0; i < str.length(); i++) {
char ch = str.charAt(i);
result.compute(ch, (k, v) -> (v == null) ? 1 : ++v);
}
return result;
}
public static Map<String, Integer> countDuplicateCharactersVCP1(String str) {
if (str == null || str.isEmpty()) {
// or throw IllegalArgumentException
return Collections.emptyMap();
}
Map<String, Integer> result = new HashMap<>();
for (int i = 0; i < str.length(); i++) {
/*
String ch = String.valueOf(Character.toChars(str.codePointAt(i)));
if (i < str.length() - 1 && str.codePointCount(i, i + 2) == 1) {
i++;
}
*/
// or, like this (this code produce the same result as the commented code above
int cp = str.codePointAt(i);
String ch = String.valueOf(Character.toChars(cp));
if(Character.charCount(cp) == 2) { // 2 means a suroggate pair
i++;
}
result.compute(ch, (k, v) -> (v == null) ? 1 : ++v);
}
return result;
}
public static Map<Character, Integer> countDuplicateCharactersV2(String str) {
if (str == null || str.isEmpty()) {
// or throw IllegalArgumentException
return Collections.emptyMap();
}
Map<Character, Integer> result = str.chars()
.mapToObj(c -> (char) c)
.collect(Collectors.groupingBy(c -> c, Collectors.counting()));
return result;
}
public static Map<String, Integer> countDuplicateCharactersVCP2(String str) {
if (str == null || str.isEmpty()) {
// or throw IllegalArgumentException
return Collections.emptyMap();
}
Map<String, Integer> result = str.codePoints()
.mapToObj(c -> String.valueOf(Character.toChars(c)))
.collect(Collectors.groupingBy(c -> c, Collectors.counting()));
return result;
}
}
Sample Output:
Input text: Be strong, be fearless, be beautiful. And believe that anything is possible when you have the right people there to support you. ASCII or 16 bits Unicode characters (less than 65,535 (0xFFFF)) examples: HashMap based solution: Execution time: 89574632 ns (89 ms) [ =22, A=1, a=5, b=5, B=1, d=1, e=17, f=2, g=3, h=7, i=6, l=5, ,=2, .=2, n=5, o=7, p=5, r=5, s=7, t=10, u=5, v=2, w=1, y=3] Java 8, functional-style solution: Execution time: 232078057 ns (232 ms) [ =22, A=1, a=5, b=5, B=1, d=1, e=17, f=2, g=3, h=7, i=6, l=5, ,=2, .=2, n=5, o=7, p=5, r=5, s=7, t=10, u=5, v=2, w=1, y=3] -------------------------------------- Input text: Be strong, be fearless, be beautiful. And believe that anything is possible when you have the right people there to support you. 😍 I love 💕 you Ӝ so much 💕 😍 🎼🎼🎼! Including Unicode surrogate pairs examples: HashMap based solution: Execution time: 1478360 ns (1 ms) [A=1, B=1, I=1, 🎼=3, Ӝ=1, =32, !=1, a=5, b=5, c=1, 💕=2, d=1, e=18, f=2, g=3, h=8, i=6, 😍=2, l=6, ,=2, m=1, .=2, n=5, o=10, p=5, r=5, s=8, t=10, u=7, v=3, w=1, y=4] Java 8, functional-style solution: Execution time: 2797489 ns (2 ms) [A=1, B=1, I=1, 🎼=3, Ӝ=1, =32, !=1, a=5, b=5, c=1, 💕=2, d=1, e=18, f=2, g=3, h=8, i=6, 😍=2, l=6, ,=2, m=1, .=2, n=5, o=10, p=5, r=5, s=8, t=10, u=7, v=3, w=1, y=4]
Flowchart:
Java Code Editor:
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Java: Tips of the Day
How to sort an ArrayList?
Collections.sort(testList); Collections.reverse(testList);
That will do what you want. Remember to import Collections though!
Ref: https://bit.ly/32urdSe
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