{"id":4936,"date":"2022-08-25T05:56:44","date_gmt":"2022-08-25T00:26:44","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-count-the-os-from-a-given-string-sayxooo\/"},"modified":"2022-08-25T05:56:44","modified_gmt":"2022-08-25T00:26:44","slug":"solved-count-the-os-from-a-given-string-sayxooo","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-count-the-os-from-a-given-string-sayxooo\/","title":{"rendered":"[Solved] Count the O’s from a Given String, say=”*XOO*O”"},"content":{"rendered":"

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\n<\/p>\n

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The problem with your code is this check:<\/p>\n

if (ch == 'O') {\n    if (!((inp.charAt(i + 1)) == 'X' || (inp.charAt(i - 1)) == 'X')) {\n        count++\n    }\n}\n<\/code><\/pre>\n
Here you check for only a case where there is O not surrounded by X. But this check doesn’t work for cases like *OOXOO*. You’ll get 2 but the result is 0. So the solution is to do it in loop or using regexp. Examples are shown below.<\/p>\n
Also you did that:<\/p>\n
inp = \".\" + inp + \".\";\n<\/code><\/pre>\n
which I think is not needed. I’m guessing you did that because you were getting ArrayIndexOutOfBoundsException<\/code> while looping for i = 0<\/code>. The solution you proposed works, but you could also add check in the loop for (i>0)<\/code> or start from 1 and check behind + front (see examples below).<\/p>\n
If you don’t require a loop with array then you can use a simple regex:<\/p>\n
testString = testString.replaceAll(\"[O]+X\",\"X\").replaceAll(\"X[O]+\", \"X\");\n<\/code><\/pre>\n
The whole test-program:<\/p>\n
public class Main {\n\n    public static void main(final String args[]) throws IOException\n    {\n        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));\n        System.out.print(\"Enter Test String: \");\n        String testString = br.readLine();\n        testString = \".\" + testString + \".\";\n\n        testString = testString.replaceAll(\"[O]+X\",\"X\").replaceAll(\"X[O]+\", \"X\");\n\n        int count = testString.length() - testString.replace(\"O\", \"\").length();\n        System.out.println(\"Result string:  \" + testString);\n        System.out.println(\"Total 'O' left:  \" + count);\n    }\n}\n<\/code><\/pre>\n
Test1:<\/strong><\/p>\n
Enter Test String: *OOX*XOX*OO*XO*O*X*OX*XO*O*\nResult string:  .*X*XX*OO*X*O*X*X*X*O*.\nTotal 'O' left:  4\n<\/code><\/pre>\n
Test2:<\/strong><\/p>\n
Enter Test String: *OOXOO*OO\nResult string:  .*X*OO.\nTotal 'O' left:  2\n<\/code><\/pre>\n
Test3:<\/strong><\/p>\n
Enter Test String: *O*OXOO*OO\nResult string:  .*O*X*OO.\nTotal 'O' left:  3\n<\/code><\/pre>\n
EDIT<\/strong>
\nBelow there are 3 types of solutions to your problem:<\/p>\n
public class Main {\n\n    private BufferedReader userInput;\n    private String testString;\n\n    public static void main(final String args[])\n    {\n        BufferedReader userInput = new BufferedReader(new InputStreamReader(System.in));  \/\/get the reader.\n        Main test = new Main(userInput); \/\/create an object of a test class that will use this reader\n        test.getTestString(); \/\/ask user for the input\n        test.removeOsWithRegex();  \/\/first method - regex\n        test.removeOsByReplacingWithX(); \/\/second method - in loop change all O's to X's\n        test.removeOsByInLoopDeletion(); \/\/third method - delete all O's step by step in a loop.\n    }\n\n    public Main(final BufferedReader userInput)\n    {\n        this.userInput = userInput;  \/\/constructor\n    }\n\n    \/\/getting the test string.\n    private void getTestString()\n    {\n        System.out.print(\"Enter Test String: \");\n        try\n        {\n            testString = userInput.readLine();\n            testString = \".\" + testString + \".\";\n        }\n        catch (IOException e)\n        {\n            throw new IllegalStateException(\"Something wrong happened: \" + e.getMessage() + \". Shutting down.\");\n        }\n    }\n\n    \/\/removing with regex\n    private void removeOsWithRegex()\n    {\n        System.out.println(\"Removing 'O' with regexp from: \" + testString);\n\n        \/\/change all parts of testString where there are Os followed by X to X.\n        \/\/do the same with parts of testString where there is X followed by one or more O.\n        String resultString = testString.replaceAll(\"[O]+X\",\"X\").replaceAll(\"X[O]+\", \"X\");\n        \/\/the number of Os is the length of the string minus length of the string without Os.\n        int count = resultString.length() - resultString.replace(\"O\", \"\").length();\n        System.out.println(\"Result string:  \" + resultString);\n        System.out.println(\"Total 'O' left:  \" + count);\n    }\n\n    \/\/2nd method\n    public void removeOsByReplacingWithX()\n    {\n        System.out.println(\"Removing 'O' by replacing them with X: \" + testString);\n\n        byte[] byteArray = testString.getBytes();\n        boolean changesMade;\n        \/\/create loop\n        do\n        {\n            changesMade = false;\n            for (int i = 1; i < byteArray.length - 1; i++)\n            {\n                \/\/if X was found change adjacent Os to X\n                if (byteArray[i] == 'X')\n                {\n                    if (byteArray[i-1] == 'O')\n                    {\n                        byteArray[i-1] = 'X';\n                        changesMade = true;\n                    }\n                    if (byteArray[i+1] == 'O')\n                    {\n                        byteArray[i+1] = 'X';\n                        changesMade = true;\n                    }\n                }\n            }\n        } while(changesMade);  \/\/do this as long as there is something to change.\n\n        String resultString = new String(byteArray);\n\n        int count = resultString.length() - resultString.replace(\"O\", \"\").length();\n        System.out.println(\"Result string:  \" + resultString);\n        System.out.println(\"Total 'O' left:  \" + count);\n    }\n\n\n    \/\/3rd method\n    public void removeOsByInLoopDeletion()\n    {\n        System.out.println(\"Removing 'O' by deleting them in loop: \" + testString);\n\n        byte[] byteArray = testString.getBytes();\n        boolean changesMade;\n        \/\/create a loop\n        do\n        {\n            changesMade = false;\n            for (int i = 1; i < byteArray.length - 1; i++)\n            {\n                \/\/if there is X\n                if (byteArray[i] == 'X')\n                {\n                    \/\/if next is O then shorten the array by cutting the O\n                    if (byteArray[i+1] == 'O')\n                    {\n                        byteArray = shortenByteArray(byteArray, i+1); \/\/i+1 is the position of O\n                        changesMade = true;\n                        break;\n                    }\n                }\n                \/\/on the other hand if O was wound\n                if (byteArray[i] == 'O')\n                {\n                    \/\/if X is next, shorten it by cutting O from current position\n                    if (byteArray[i+1] == 'X')\n                    {\n                        byteArray = shortenByteArray(byteArray, i); \/\/i is current position.\n                        changesMade = true;\n                        break;\n                    }\n                }\n            }\n        } while(changesMade);\n\n        String resultString = new String(byteArray);\n\n        int count = resultString.length() - resultString.replace(\"O\", \"\").length();\n        System.out.println(\"Result string:  \" + resultString);\n        System.out.println(\"Total 'O' left:  \" + count);\n    }\n\n    \/\/cutting one element from byte array\n    private byte[] shortenByteArray(final byte[] byteArray, final Integer startIndex)\n    {\n        byte[] newByteArray = new byte[byteArray.length-1];\n        for (int i = 0; i < newByteArray.length; i++)\n        {\n            \/\/rewrite all before the item to cut\n            if (i < startIndex)\n            {\n                newByteArray[i] = byteArray[i];\n            }\n            else\n            {\n                \/\/ommit the element to cut and move all others by 1.\n                newByteArray[i] = byteArray[i+1];\n            }\n        }\n        return newByteArray;\n    }\n}\n<\/code><\/pre>\n
The results are the same for all three cases for all 3 tests I’ve supplied earlier.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            3<\/span> <\/p><\/div>\n<\/div>\n
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solved Count the O’s from a Given String, say=”*XOO*O” <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] The problem with your code is this check: if (ch == ‘O’) { if (!((inp.charAt(i + 1)) == ‘X’ || (inp.charAt(i – 1)) == ‘X’)) { count++ } } Here you check for only a case where there is O not surrounded by X. But this check doesn’t work for cases like *OOXOO*. You’ll … 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-4936","post","type-post","status-publish","format-standard","hentry","category-solved","tag-java"],"yoast_head":"\n[Solved] Count the O's from a Given String, say="*XOO*O" - 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-count-the-os-from-a-given-string-sayxooo\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Count the O's from a Given String, say="*XOO*O" - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] The problem with your code is this check: if (ch == 'O') { if (!((inp.charAt(i + 1)) == 'X' || (inp.charAt(i - 1)) == 'X')) { count++ } } Here you check for only a case where there is O not surrounded by X. 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