{"id":74540,"date":"2024-01-08T14:24:02","date_gmt":"2024-01-08T14:24:02","guid":{"rendered":"https:\/\/www.baeldung.com\/?p=171982"},"modified":"2024-01-08T14:24:02","modified_gmt":"2024-01-08T14:24:02","slug":"remove-characters-from-a-string-that-are-in-the-other-string","status":"publish","type":"post","link":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/2024\/01\/08\/remove-characters-from-a-string-that-are-in-the-other-string\/","title":{"rendered":"Remove Characters From a String That Are in the Other String"},"content":{"rendered":"

\"\"<\/p>\n

1. Overview<\/h2>\n
<\/div>\n

When we work with Java, we often encounter tasks that require precision and a collaborative effort between elements. Removing characters from a string based on their presence in another string is one such problem.<\/p>\n

In this tutorial, we’ll explore various techniques to achieve this task.<\/p>\n

2. Introduction to the Problem<\/h2>\n
<\/div>\n

As usual, an example can help us understand the problem quickly. Let’s say we have two strings:<\/p>\n

String STRING = "a b c d e f g h i j";\r\nString OTHER = "bdfhj";<\/code><\/pre>\n
Our goal is to eliminate characters from the STRING<\/em> string if they are present in the string OTHER<\/em><\/strong>.\u00a0<\/em>Thus, we expect to get this string as the result:<\/p>\n
"a  c  e  g  i "<\/code><\/pre>\n
We’ll learn various approaches to solving this problem in this tutorial. Also, we’ll unit test these solutions to verify whether they produce the expected result.<\/p>\n
3. Using Nested Loops<\/h2>\n
<\/div>\n
We know a string can be easily split into a char<\/em> array using the standard toCharArray()<\/em><\/a> method. So, a straightforward and classic approach is first converting the two strings to two char<\/em> arrays. Then, for each character in STRING<\/em>, we decide whether to remove it or not by checking if it’s present in OTHER<\/em>.<\/strong><\/p>\n
We can use nested for loops<\/a> to implement this logic:<\/p>\n
String nestedLoopApproach(String theString, String other) {\r\n    StringBuilder sb = new StringBuilder();\r\n    for (char c : theString.toCharArray()) {\r\n        boolean found = false;\r\n        for (char o : other.toCharArray()) {\r\n            if (c == o) {\r\n                found = true;\r\n                break;\r\n            }\r\n        }\r\n        if (!found) {\r\n            sb.append(c);\r\n        }\r\n    }\r\n    return sb.toString();\r\n}<\/code><\/pre>\n
It’s worth noting since Java strings are immutable objects, we use StringBuilder<\/em><\/a> instead of the ‘+’ operator<\/a> to concatenate strings to gain better performance<\/strong>.<\/p>\n
Next, let’s create a test:<\/p>\n
String result = nestedLoopApproach(STRING, OTHER);\r\nassertEquals("a  c  e  g  i ", result);<\/code><\/pre>\n
The test passes if we give it a run, so the method does the job.<\/p>\n
Since for each character in STRING,<\/em> we check through the string OTHER<\/em>, the time complexity<\/a> of this solution is O(n2<\/sup>)<\/strong>.<\/p>\n
4. Replacing the Inner Loop With the indexOf()<\/em> Method<\/h2>\n
<\/div>\n
In the nested loops solution, we created the boolean<\/em> flag found\u00a0<\/em>to store if the current character has been found in the OTHER\u00a0<\/em>String and then decided if we need to keep or discard this character by checking the found<\/em> flag.<\/p>\n
Java provides the String.indexOf()<\/em><\/a> method that allows us to locate a given character in a string. Further, if the string doesn’t contain the given character, the method returns -1<\/em><\/strong>.<\/p>\n
So, if we make use of the String.indexOf()\u00a0<\/em>method, the inner loop and the found<\/em> flag aren’t required:<\/p>\n
String loopAndIndexOfApproach(String theString, String other) {\r\n    StringBuilder sb = new StringBuilder();\r\n    for (char c : theString.toCharArray()) {\r\n        if (other.indexOf(c) == -1) {\r\n            sb.append(c);\r\n        }\r\n    }\r\n    return sb.toString();\r\n}<\/code><\/pre>\n
As we can see, this method’s code is easier to understand than the nested loops one, and it passes the test as well:<\/p>\n
String result = loopAndIndexOfApproach(STRING, OTHER);\r\nassertEquals("a  c  e  g  i ", result);<\/code><\/pre>\n
Although this implementation is compact and easy to read, as the String.indexOf()<\/em> method internally searches the target character through the string by a loop, its time complexity is still O(n2<\/sup>)<\/strong>.<\/p>\n
Next, let’s see if we can find a solution with lower time complexity.<\/p>\n
5. Using a HashSet<\/em><\/h2>\n
<\/div>\n
HashSet<\/em><\/a> is a commonly used collection data structure. It stores the elements in an internal HashMap<\/a>.<\/em><\/p>\n
Since the hash function’s time complexity is O(1), HashSet<\/em>‘s contains()<\/em> method is an O(1) operation.<\/strong><\/p>\n
Therefore, we can first store all characters in the OTHER<\/em> string in a HashSet <\/em>and then check each character from STRING<\/em> in the HashSet<\/em>:<\/p>\n
String hashSetApproach(String theString, String other) {\r\n    StringBuilder sb = new StringBuilder();\r\n    Set<Character> set = new HashSet<>(other.length());\r\n    for (char c : other.toCharArray()) {\r\n        set.add(c);\r\n    }\r\n    for (char i : theString.toCharArray()) {\r\n        if (set.contains(i)) {\r\n            continue;\r\n        }\r\n        sb.append(i);\r\n    }\r\n    return sb.toString();\r\n}<\/code><\/pre>\n
As the code above shows, the implementation is quite straightforward. Now, let’s delve into its performance.<\/p>\n
Initially, we iterate through one string to populate the Set<\/em> object, making it an O(n) operation. Subsequently, for each character in the other string, we utilize the set.contains()<\/em> method. This results in n<\/em> times O(1), becoming another O(n) complexity. Therefore, the entire solution comprises two O(n) operations.<\/strong><\/p>\n
However, since the factor of two is a constant, the overall time complexity of the solution remains O(n)<\/strong>. This stands out as a significant improvement compared to previous O(n2<\/sup>) solutions, demonstrating a considerably faster execution.<\/p>\n
Finally, if we test the hashSetApproach()<\/em> method, it gives the expected result:<\/p>\n
String result = hashSetApproach(STRING, OTHER);\r\nassertEquals("a  c  e  g  i ", result);<\/code><\/pre>\n
6. Conclusion<\/h2>\n
<\/div>\n
In this article, we explored three different approaches to removing characters from one string based on their presence in another.<\/p>\n
Furthermore, we conducted a performance analysis, explicitly focusing on time complexity. The results revealed that\u00a0both nested loops and loops utilizing indexOf()<\/em> exhibit equivalent time complexities, while solutions employing HashSet<\/em> to be the most efficient.<\/p>\n
As always, the complete source code for the examples is available over on GitHub<\/a>.<\/p>\n
\"\"<\/p>\n
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\"\"<\/p>\n
Explore three approaches to removing characters from one string if they’re present in another.<\/p>\n
<\/a>\u00a0<\/a>\u00a0<\/a>\u00a0<\/a>\u00a0<\/a>\u00a0<\/a>\u00a0<\/a>\u00a0<\/div>\n","protected":false},"author":263,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"site-container-style":"default","site-container-layout":"default","site-sidebar-layout":"default","disable-article-header":"default","disable-site-header":"default","disable-site-footer":"default","disable-content-area-spacing":"default","footnotes":""},"categories":[22],"tags":[61,122,127,129,124,128,125,132,131,133,126,130,123,66,94,88,97,56,64,65,60,112,40,75,95,104,33,120,105,101,98,115,30,29,41,86,70,69,68,72,71,26,118,108,87,46,55,48,52,54,51,50,83,62,58,57,19876,109,35,59,63,85,79,82,96,80,27,81,114,44,42,43,45,38,39,110,117,100,111,116,73,89,90,92,91,93,84,78,37,102,34,36,77,67,74,99,113,119,28,121,32,47,49,53,103,31,76],"class_list":["post-74540","post","type-post","status-publish","format-standard","hentry","category-mobile","tag-airpods","tag-anime","tag-anime-characters","tag-anime-cosplay","tag-anime-edits","tag-anime-merchandise","tag-anime-movies","tag-anime-news","tag-anime-recommendations","tag-anime-reviews","tag-anime-series","tag-anime-streaming","tag-animes","tag-app-store","tag-app-store-samsung","tag-appgallery","tag-appgallery-oneplus","tag-apple","tag-apple-music","tag-apple-tv","tag-apple-watch","tag-bbc-sport","tag-best-mobile-games","tag-bixby","tag-bixby-xiaomi","tag-champions-league","tag-cyberpunk","tag-cyberpunk-2077","tag-fantasy-football","tag-fifa","tag-football","tag-formula-1","tag-fortnite","tag-free-fire","tag-free-mobile-games","tag-freebuds-pro","tag-galaxy-a52","tag-galaxy-note-20","tag-galaxy-s21","tag-galaxy-watch-4","tag-galaxy-z-fold-3","tag-game","tag-games","tag-golf","tag-harmonyos","tag-how-to-backup-iphone","tag-how-to-factory-reset-iphone","tag-how-to-reset-iphone","tag-how-to-restore-iphone","tag-how-to-unlock-iphone","tag-how-to-unlock-iphone-5","tag-how-to-unlock-iphone-6","tag-huawei","tag-ios","tag-ipad","tag-iphone","tag-java-loops","tag-live-soccer","tag-lol","tag-macbook","tag-macos","tag-mate-40-pro","tag-mi-11-lite","tag-mi-home-security-camera-basic-1080p","tag-mi-home-security-camera-basic-1080p-huawei","tag-mi-smart-band-6","tag-minecraft","tag-miui","tag-mlb-scores","tag-mobile-game-design","tag-mobile-game-development","tag-mobile-game-marketing","tag-mobile-game-monetization","tag-mobile-games","tag-mobile-gaming","tag-nba-scores","tag-nba-standings","tag-nfl","tag-nfl-scores","tag-nhl-scores","tag-one-ui","tag-oneplus","tag-oneplus-9-pro","tag-oneplus-buds-pro","tag-oneplus-nord-ce-5g","tag-oxygenos","tag-p40-pro-plus","tag-poco-x3-pro","tag-pokemon","tag-premier-league","tag-pubg","tag-pubg-mobile","tag-redmi-note-10-pro","tag-samsung","tag-samsung-pay","tag-soccer","tag-sports","tag-steam","tag-steeam","tag-top-10-anime","tag-valorant","tag-when-do-the-iphone-7-come-out","tag-when-does-the-iphone-7-come-out","tag-when-is-the-iphone-7-coming-out","tag-world-cup","tag-xbox-series-x","tag-xiaomi"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/posts\/74540","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/users\/263"}],"replies":[{"embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/comments?post=74540"}],"version-history":[{"count":1,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/posts\/74540\/revisions"}],"predecessor-version":[{"id":74541,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/posts\/74540\/revisions\/74541"}],"wp:attachment":[{"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/media?parent=74540"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/categories?post=74540"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/tags?post=74540"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}