{"id":25642,"date":"2022-12-11T12:56:50","date_gmt":"2022-12-11T07:26:50","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-duplicates-reduction\/"},"modified":"2022-12-11T12:56:50","modified_gmt":"2022-12-11T07:26:50","slug":"solved-duplicates-reduction","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-duplicates-reduction\/","title":{"rendered":"[Solved] Duplicates Reduction"},"content":{"rendered":"

[ad_1]
\n<\/p>\n

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<\/div>\n
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\n

Let’s get issue simple. We don’t need to delete duplicate character. We just remove the charachter at index 2. The string is “abcde”.<\/p>\n

StringBuilder sb = new StringBuilder(\"abcde\");\nchar one = sb.charAt(2); \/\/ comment1, we know it's 'c'\nsb = sb.deleteCharAt(2); \/\/ comment2, let's delete 'c'. What would \"abcde\" be?\n\/\/ It's so ugly to be \"ab de\", so designer makes it be \"abde\"\nchar two = sb.charAt(2); \/\/ comment3, now you will know the result ^v^\n<\/code><\/pre>\n
For comment2, how does it make “abcde” to “abde”? Let’s look into source of StringBuilder. StringBuilder is a subClass of AbstractStringBuilder. And deleteCharAt(int index) is implemented in AbstractStringBuilder.<\/p>\n
public AbstractStringBuilder deleteCharAt(int index) {\n    if ((index < 0) || (index >= count))\n        throw new StringIndexOutOfBoundsException(index);\n    System.arraycopy(value, index+1, value, index, count-index-1);\n    count--;\n    return this;\n}\n<\/code><\/pre>\n
Let’s omite index validation. What does System.arraycopy(value, index+1, value, index, count-index-1);<\/code> do? You can find it’s doc here<\/a>.<\/p>\n
public static void arraycopy(Object src,\n                             int srcPos,\n                             Object dest,\n                             int destPos,\n                             int length)\n<\/code><\/pre>\n
\n
Copies an array from the specified source array (which is “abcde”), beginning at the
\n  specified position (index+1<\/code>, which is 3), to the specified position (index<\/code>, which is 2) of the destination
\n  array (which is “abcde” too). A subsequence of array components are copied from the source
\n  array referenced by src<\/code> to the destination array referenced by dest<\/code> (src<\/code> and dest<\/code> are the same here, “abcde”).
\n  The number of components copied is equal to the length<\/code> argument (length<\/code> here is 5-2-1, it’s 2. So it will copy 2 chars). The
\n  components at positions srcPos<\/code> through srcPos+length-1<\/code> in the source
\n  array are copied into positions destPos<\/code> through destPos+length-1<\/code> (srcPos<\/code> here is 2+1, it’s 3. srcPos+length-1<\/code> is 3+2-1, it’s 4),
\n  respectively, of the destination array.<\/p>\n<\/blockquote>\n
So System.arraycopy(value, index+1, value, index, count-index-1);<\/code> means copy 2 chars from index 3, to override index 2, override from index 2 to 3. Copy ‘de’ to override ‘cd’. So after System.arraycopy(value, index+1, value, index, count-index-1);<\/code>, makes sb.value<\/code> to be “abdee”(Delete ‘l’ already.). then count–, means “abdee” remove last char, so it becomes “abde”. <\/p>\n
will the next \"l\" takes position of the previous one after deleting it?<\/code> Yes, it will.<\/p>\n
See? Coding is not so hard and not so mystic. And the direct way is code yourself and run it.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            <\/span> <\/p><\/div>\n<\/div>\n
[ad_2]<\/p>\n
solved Duplicates Reduction <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Let’s get issue simple. We don’t need to delete duplicate character. We just remove the charachter at index 2. The string is “abcde”. StringBuilder sb = new StringBuilder(“abcde”); char one = sb.charAt(2); \/\/ comment1, we know it’s ‘c’ sb = sb.deleteCharAt(2); \/\/ comment2, let’s delete ‘c’. What would “abcde” be? \/\/ It’s so ugly … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[323],"class_list":["post-25642","post","type-post","status-publish","format-standard","hentry","category-solved","tag-java"],"yoast_head":"\n[Solved] Duplicates Reduction - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-duplicates-reduction\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Duplicates Reduction - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Let’s get issue simple. We don’t need to delete duplicate character. We just remove the charachter at index 2. The string is “abcde”. StringBuilder sb = new StringBuilder("abcde"); char one = sb.charAt(2); \/\/ comment1, we know it's 'c' sb = sb.deleteCharAt(2); \/\/ comment2, let's delete 'c'. What would "abcde" be? \/\/ It's so ugly ... 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