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In this tutorial, we’ll explore various methods of merging the contents of two arrays<\/a> and subsequently eliminating the duplicates. It is similar to a union operation:<\/p>\n For this, we’ll consider two arrays of integers.\u00a0Basically, if the following are the two arrays:<\/p>\n Then the result should be:<\/p>\n Normally, this requirement can be implemented by using Set<\/em> or Stream<\/em> API, explained in the later sections. But those libraries are resource-intensive. Hence, we’ll focus more on a traditional approach by using basic array operations while iterating through it.<\/strong><\/p>\n First, let’s define a function that would merge the arrays. Then, we’ll create a method to remove the duplicates from it.<\/strong> Alternatively, we can remove the duplicates from the arrays and then merge them. But this won’t make much of a difference with respect to performance.<\/p>\n Hence, let’s begin with the method that merges two arrays:<\/p>\n Going forward, we’re going to use the above method for merging the arrays in all the sections.<\/p>\n Now, let’s implement the method to remove the duplicates from the merged arrays:<\/p>\n In the above method, removeDuplicate()<\/em>, the withoutDuplicates<\/em> array is populated with the elements of the merged array only if the element is not pre-existing in it.<\/p>\n By using the above methods, let’s define mergeAndRemoveDuplicates()<\/em>:<\/p>\n Let’s see how that works:<\/p>\n It proves we achieved the expected result. Interestingly, this approach preserves the order of the array elements as well.<\/p>\n Since removing duplicates involves comparing each element in the merged array with all other elements, the time complexity of this approach is close to O(n x n).<\/strong><\/p>\n Let’s consider a scenario where the elements are already sorted in the array<\/strong>. Also, in the second array, the first element is equal to or greater than the last element of the first array.<\/p>\n In such a case, we can use the following method to get rid of duplicate elements:<\/p>\n This approach is more efficient because it only compares adjacent elements<\/strong>. If they are not equal, they are added to the uniqueArray<\/em> array. In contrast, earlier, the isElementPresent()<\/em> method compared the array element with all the other elements in the array.<\/p>\n Let’s see removeDuplicateOnSortedArray()<\/em> in action:<\/p>\n In a worst-case scenario, the time complexity of this approach is O(n).<\/strong><\/p>\n The Set<\/i><\/a> doesn’t allow duplicate elements. Hence, this feature helps remove duplicates<\/strong>.<\/p>\n Let’s have a look at the method mergeAndRemoveDuplicatesUsingSet()<\/em>:<\/p>\n While adding the array elements into the Set<\/em> uniqueInts<\/em>, the elements are ignored if they are already present init. At the end, getArrayFromSet()<\/em> converts the Set<\/em> into an array.<\/p>\n Let’s see how the above method behaves:<\/p>\n It’s evident. that the arrays no longer preserve the order of the elements.<\/p>\n The execution time with this approach is dependent on the number of array elements. Hence, its time complexity is O(n).<\/strong><\/p>\n The powerful Stream<\/em><\/a> API provides a very concise and declarative way of merging and removing duplicates from the arrays.<\/strong> For further details, let’s check out the method mergeAndRemoveDuplicatesUsingStream()<\/em>:<\/p>\n Firstly, the arrays are converted individually into Stream<\/em>. Then, the boxed method wraps the int<\/em> elements in the arrays as Integer<\/em>. Further, the Stream<\/em> pipeline merges the arrays and then eliminates the duplicates from them.<\/p>\n Let’s see how it works:<\/p>\n It’s evident that the Stream approach preserves the order of the elements in the array.<\/p>\n The dominant factor in deciding the time complexity is the distinct()\u00a0<\/em>method in the stream, which is O(n). Overall, we can say that the time complexity of this approach is O(n).<\/strong><\/p>\n In this tutorial, we examined various ways to merge two arrays and then remove the duplicates from them. The most concise approach uses Stream<\/em> API. The set\u00a0<\/em>doesn’t preserve the order of the elements inserted in them.\u00a0However, Set<\/em> is extremely effective in removing duplicates. The other two approaches preserve the order of the elements in the array.<\/p>\n It is worth noting that Stream<\/em> and Set<\/em> are quite resource-intensive, and hence, it is advisable to avoid them wherever possible.<\/strong><\/p>\n As usual, the code examples are available over on GitHub<\/a>.<\/p>\n <\/a><\/p>\n <\/p><\/div>\n\n \u7acb\u523b\u8a3b\u518a\u53ca\u5831\u540d\u96fb\u8166\u88dc\u7fd2\u8ab2\u7a0b\u5427! <\/a> www.ITSec.vip<\/a> www.Sraa.com.hk<\/a> www.ITSec.hk<\/a> www.Penetrationtest.hk<\/a> www.ITSeceu.uk<\/a> Explore various methods of merging the contents of two arrays and eliminating the duplicates.<\/p>\n <\/a><\/p>\n \u00a0<\/p><\/div>\n","protected":false},"author":259,"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":[61,122,127,129,124,128,125,132,131,133,126,130,123,66,94,88,97,56,64,65,60,112,40,75,95,104,33,120,105,101,98,115,30,29,41,86,70,69,68,72,71,26,118,108,87,46,55,48,52,54,51,50,83,62,58,57,2480,2481,2482,109,35,59,63,85,79,82,96,80,27,81,114,44,42,43,45,38,39,110,117,100,111,116,73,89,90,92,91,93,84,78,37,102,34,36,77,67,74,99,2483,113,119,28,121,32,47,49,53,103,31,76],"class_list":["post-5427","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-array","tag-java-set","tag-java-streams","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-sorting","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\/5427","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\/259"}],"replies":[{"embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/comments?post=5427"}],"version-history":[{"count":1,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/posts\/5427\/revisions"}],"predecessor-version":[{"id":5428,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/posts\/5427\/revisions\/5428"}],"wp:attachment":[{"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/media?parent=5427"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/categories?post=5427"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gamefootballmobileanimeiphone.com\/index.php\/wp-json\/wp\/v2\/tags?post=5427"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
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2. Approach with Basic Array Operations<\/h2>\n
<\/div>\n2.1. Approach for Unsorted Arrays<\/h3>\n
<\/div>\nstatic int[] mergeArrays(int[] arr1, int[] arr2) {\r\n int[] mergedArrays = new int[arr1.length + arr2.length];\r\n System.arraycopy(arr1, 0, mergedArrays, 0, arr1.length);\r\n System.arraycopy(arr2, 0, mergedArrays, arr1.length, arr2.length);\r\n return mergedArrays;\r\n}<\/code><\/pre>\nstatic int[] removeDuplicate(int[] arr) {\r\n int[] withoutDuplicates = new int[arr.length];\r\n int i = 0;\r\n for (int element : arr) {\r\n if (!isElementPresent(withoutDuplicates, element)) {\r\n withoutDuplicates[i] = element;\r\n i++;\r\n }\r\n }\r\n int[] truncatedArray = new int[i];\r\n System.arraycopy(withoutDuplicates, 0, truncatedArray, 0, i);\r\n return truncatedArray;\r\n}\r\nstatic boolean isElementPresent(int[] arr, int element) {\r\n for (int el : arr) {\r\n if (el == element) {\r\n return true;\r\n }\r\n }\r\n return false;\r\n}<\/code><\/pre>\npublic static int[] mergeAndRemoveDuplicates(int[] arr1, int[] arr2) {\r\n return removeDuplicate(mergeArrays(arr1, arr2));\r\n}<\/code><\/pre>\n@Test\r\npublic void givenNoLibraryAndUnSortedArrays_whenArr1andArr2_thenMergeAndRemoveDuplicates() {\r\n int[] arr1 = {3, 2, 1, 4, 5, 6, 8, 7, 9};\r\n int[] arr2 = {8, 9, 10, 11, 12, 13, 15, 14, 15, 14, 16, 17};\r\n int[] expectedArr = {3, 2, 1, 4, 5, 6, 8, 7, 9, 10, 11, 12, 13, 15, 14, 16, 17};\r\n int[] mergedArr = MergeArraysAndRemoveDuplicate.mergeAndRemoveDuplicates(arr1, arr2);\r\n assertArrayEquals(expectedArr, mergedArr);\r\n}<\/code><\/pre>\n2.2. Approach for Sorted Arrays<\/h3>\n
<\/div>\npublic static int[] removeDuplicateOnSortedArray(int[] arr) {\r\n int[] uniqueArray = new int[arr.length];\r\n uniqueArray[0] = arr[0];\r\n int uniqueCount = 1;\r\n for (int i = 1; i < arr.length; i++) {\r\n if (arr[i] != arr[i - 1]) {\r\n uniqueArray[uniqueCount] = arr[i];\r\n uniqueCount++;\r\n }\r\n }\r\n int[] truncatedArray = new int[uniqueCount];\r\n System.arraycopy(uniqueArray, 0, truncatedArray, 0, uniqueCount);\r\n return truncatedArray;\r\n}<\/code><\/pre>\n@Test\r\npublic void givenNoLibraryAndSortedArrays_whenArr1andArr2_thenMergeAndRemoveDuplicates() {\r\n int[] arr1 = {1, 2, 3, 4, 5, 5, 6, 7, 7, 8};\r\n int[] arr2 = {8, 9, 10, 11, 12, 13, 14, 15, 15, 16, 17};\r\n int[] expectedArr = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17};\r\n int[] mergedArr = MergeArraysAndRemoveDuplicate.mergeAndRemoveDuplicatesOnSortedArray(arr1, arr2);\r\n assertArrayEquals(expectedArr, mergedArr);\r\n}<\/code><\/pre>\n3. Merge Arrays Using Set<\/em><\/h2>\n
<\/div>\npublic static int[] mergeAndRemoveDuplicatesUsingSet(int[] arr1, int[] arr2) {\r\n int[] mergedArr = mergeArrays(arr1, arr2);\r\n Set<Integer> uniqueInts = new HashSet<>();\r\n for (int el : mergedArr) {\r\n uniqueInts.add(el);\r\n }\r\n return getArrayFromSet(uniqueInts);\r\n}<\/code><\/pre>\n@Test\r\npublic void givenSet_whenArr1andArr2_thenMergeAndRemoveDuplicates() {\r\n int[] arr1 = {3, 2, 1, 4, 5, 6, 8, 7, 9};\r\n int[] arr2 = {8, 9, 10, 11, 12, 13, 15, 14, 15, 14, 16, 17};\r\n int[] expectedArr = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17};\r\n int[] mergedArr = MergeArraysAndRemoveDuplicate.mergeAndRemoveDuplicatesUsingSet(arr1, arr2);\r\n assertArrayEquals(expectedArr, mergedArr);\r\n}<\/code><\/pre>\n4. Merge Arrays Using Stream<\/em><\/h2>\n
<\/div>\npublic static int[] mergeAndRemoveDuplicatesUsingStream(int[] arr1, int[] arr2) {\r\n Stream<Integer> s1 = Arrays.stream(arr1).boxed();\r\n Stream<Integer> s2 = Arrays.stream(arr2).boxed();\r\n return Stream.concat(s1, s2)\r\n .distinct()\r\n .mapToInt(Integer::intValue)\r\n .toArray();\r\n}<\/code><\/pre>\n@Test\r\npublic void givenStream_whenArr1andArr2_thenMergeAndRemoveDuplicates() {\r\n int[] arr1 = {3, 2, 1, 4, 5, 6, 8, 7, 9};\r\n int[] arr2 = {8, 9, 10, 11, 12, 13, 15, 14, 15, 14, 16, 17};\r\n int[] expectedArr = {3, 2, 1, 4, 5, 6, 8, 7, 9, 10, 11, 12, 13, 15, 14, 16, 17};\r\n int[] mergedArr = MergeArraysAndRemoveDuplicate.mergeAndRemoveDuplicatesUsingStream(arr1, arr2);\r\n assertArrayEquals(expectedArr, mergedArr);\r\n}<\/code><\/pre>\n5. Conclusion<\/h2>\n
<\/div>\n<\/p>\n
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