{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2025-12-14T01:33:56.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2025-12-14T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":1779,"title":"Oh Zero Zero Zero!!!","description":"Hello all,\r\nSo you have to find the largest section of zeros in a vector and then find the length of those zeros and there starting position...\r\nFor example:\r\n \r\n x = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\r\n %then the output is:\r\n LP = [9 10] %[Length Position]\r\n \r\n %Or another example:\r\n \r\n x = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\r\n %then the output is:\r\n LP = [6 9]\r\n \r\n %Or another example:\r\n \r\n x = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\r\n %then the output is:\r\n LP = [7 3];\r\n\r\nHave Fun!\r\n","description_html":"\u003cp\u003eHello all,\r\nSo you have to find the largest section of zeros in a vector and then find the length of those zeros and there starting position...\r\nFor example:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003ex = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\r\n%then the output is:\r\nLP = [9 10] %[Length Position]\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003e%Or another example:\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003ex = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\r\n%then the output is:\r\nLP = [6 9]\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003e%Or another example:\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003ex = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\r\n%then the output is:\r\nLP = [7 3];\r\n\u003c/pre\u003e\u003cp\u003eHave Fun!\u003c/p\u003e","function_template":"function y = LengthAndPosnZeros(x)\r\n y = x;\r\nend","test_suite":"%%\r\nx = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\r\nLP = [9 10] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\r\nLP = [6 9]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\r\nLP = [7 3];\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 2 0 0];\r\nLP = [2 3] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 2 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0];\r\nLP = [9 3] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 0 0 0 0 0 0 0 0 0 1];\r\nLP = [9 2] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [111 541 0 45 3 0 0 0 15 26 0 4 84 3 84 0 9];\r\nLP = [3 6] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 0 1];\r\nLP = [1 2] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n\r\n","published":true,"deleted":false,"likes_count":15,"comments_count":1,"created_by":15013,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":540,"test_suite_updated_at":"2013-08-08T19:35:25.000Z","rescore_all_solutions":false,"group_id":13,"created_at":"2013-08-08T18:54:26.000Z","updated_at":"2026-03-24T00:57:27.000Z","published_at":"2013-08-08T19:35:25.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHello all, So you have to find the largest section of zeros in a vector and then find the length of those zeros and there starting position... For example:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[x = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\\n%then the output is:\\nLP = [9 10] %[Length Position]\\n\\n%Or another example:\\n\\nx = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\\n%then the output is:\\nLP = [6 9]\\n\\n%Or another example:\\n\\nx = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\\n%then the output is:\\nLP = [7 3];]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHave Fun!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":1740,"title":"Find number patterns in rows of a matrix...","description":"So I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a *single row*, so all numbers in a *single rows* are unique. So you need to find up to 2, 3 or 4 numbers that are found in a *single row* that repeat in *other rows* and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:\r\n \r\n %p is the given matrix\r\n p = [1 2 3 4 5;\r\n 2 3 4 5 1;\r\n 3 4 6 7 8;\r\n 1 2 4 3 5;\r\n 4 6 3 2 1];\r\n %v is the amount of elements per row that are being checked to see if they repeat in another row\r\n v = 4;\r\n %Output:\r\n out =\r\n \r\n 1 2 3 4 4\r\n 1 2 3 5 3\r\n 1 2 4 5 3\r\n 1 3 4 5 3\r\n 2 3 4 5 3\r\n %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.\r\n \r\n So more examples...\r\n \r\n p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 4;\r\n \r\n out =\r\n 1 2 3 4 2\r\n %so any combination of 1,2,3 and 4 appear 2 times in matrix p\r\n \r\n and another:\r\n \r\n p = [1 2 3 4 9;\r\n 2 3 1 4 8;\r\n 3 4 2 7 1]\r\n v = 3;\r\n \r\n out =\r\n 1 2 3 3\r\n 1 2 4 3\r\n 1 3 4 3\r\n 2 3 4 3\r\n %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found\r\n \r\n last example:\r\n \r\n p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 2;\r\n \r\n out =\r\n 1 2 2\r\n 1 3 2\r\n 1 4 2\r\n 2 3 2\r\n 2 4 2\r\n 3 4 2\r\n %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.\r\n\r\nNOTE: only display any row number combos that occur at least 2 times or more!\r\n\r\nI hope I gave enough instructions and explanation, if not please let me know and I will add to it! Thanks and good luck!","description_html":"\u003cp\u003eSo I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a \u003cb\u003esingle row\u003c/b\u003e, so all numbers in a \u003cb\u003esingle rows\u003c/b\u003e are unique. So you need to find up to 2, 3 or 4 numbers that are found in a \u003cb\u003esingle row\u003c/b\u003e that repeat in \u003cb\u003eother rows\u003c/b\u003e and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:\u003c/p\u003e\u003cpre\u003e %p is the given matrix\r\n p = [1 2 3 4 5;\r\n 2 3 4 5 1;\r\n 3 4 6 7 8;\r\n 1 2 4 3 5;\r\n 4 6 3 2 1];\r\n %v is the amount of elements per row that are being checked to see if they repeat in another row\r\n v = 4;\r\n %Output:\r\n out =\u003c/pre\u003e\u003cpre\u003e 1 2 3 4 4\r\n 1 2 3 5 3\r\n 1 2 4 5 3\r\n 1 3 4 5 3\r\n 2 3 4 5 3\r\n %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.\u003c/pre\u003e\u003cpre\u003e So more examples...\u003c/pre\u003e\u003cpre\u003e p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 4;\u003c/pre\u003e\u003cpre\u003e out =\r\n 1 2 3 4 2\r\n %so any combination of 1,2,3 and 4 appear 2 times in matrix p\u003c/pre\u003e\u003cpre\u003e and another:\u003c/pre\u003e\u003cpre\u003e p = [1 2 3 4 9;\r\n 2 3 1 4 8;\r\n 3 4 2 7 1]\r\n v = 3;\u003c/pre\u003e\u003cpre\u003e out =\r\n 1 2 3 3\r\n 1 2 4 3\r\n 1 3 4 3\r\n 2 3 4 3\r\n %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found\u003c/pre\u003e\u003cpre\u003e last example:\u003c/pre\u003e\u003cpre\u003e p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 2;\u003c/pre\u003e\u003cpre\u003e out =\r\n 1 2 2\r\n 1 3 2\r\n 1 4 2\r\n 2 3 2\r\n 2 4 2\r\n 3 4 2\r\n %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.\u003c/pre\u003e\u003cp\u003eNOTE: only display any row number combos that occur at least 2 times or more!\u003c/p\u003e\u003cp\u003eI hope I gave enough instructions and explanation, if not please let me know and I will add to it! Thanks and good luck!\u003c/p\u003e","function_template":"function out = findRowPattern(p,v)\r\n out = [p v];\r\nend","test_suite":"%%\r\np = [1 2 3 4 5;\r\n 2 3 4 5 1;\r\n 3 4 6 7 8;\r\n 1 2 4 3 5;\r\n 4 6 3 2 1];\r\nv = 4;\r\nout = [1 2 3 4 4;\r\n 1 2 3 5 3;\r\n 1 2 4 5 3;\r\n 1 3 4 5 3;\r\n 2 3 4 5 3]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\nv = 4;\r\nout = [1 2 3 4 2]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [1 2 3 4 9;\r\n 2 3 1 4 8;\r\n 3 4 2 7 1]\r\nv = 3;\r\nout = [1 2 3 3;\r\n 1 2 4 3;\r\n 1 3 4 3;\r\n 2 3 4 3]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\nv = 2;\r\nout = [1 2 2;\r\n 1 3 2;\r\n 1 4 2;\r\n 2 3 2;\r\n 2 4 2;\r\n 3 4 2]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [15 23 68 49 88;\r\n 69 58 78 21 35;\r\n 10 23 21 35 88;\r\n 99 58 63 24 10;\r\n 64 28 14 33 58;\r\n 85 69 21 45 55;\r\n 99 24 76 49 33;\r\n 89 69 33 98 21;\r\n 99 10 21 55 58;\r\n 35 68 69 44 21;\r\n 21 69 35 46 33];\r\nv = 3;\r\nout = [21 35 69 3;\r\n 10 58 99 2;\r\n 21 33 69 2]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [15 23 68 49 88;\r\n 69 58 78 21 35;\r\n 10 23 21 35 88;\r\n 99 58 63 24 10;\r\n 64 28 14 33 58;\r\n 85 69 21 45 55;\r\n 99 24 76 49 33;\r\n 89 69 33 98 21;\r\n 99 10 21 55 58;\r\n 35 68 69 44 21;\r\n 21 69 35 46 33];\r\nv = 2;\r\n\r\nout = [21 69 5;\r\n 21 35 4;\r\n 35 69 3;\r\n 10 21 2;\r\n 10 58 2;\r\n 10 99 2;\r\n 21 33 2;\r\n 21 55 2;\r\n 21 58 2;\r\n 23 88 2;\r\n 24 99 2;\r\n 33 69 2;\r\n 58 99 2]\r\nassert(isequal(findRowPattern(p,v),out))","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":15013,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":11,"test_suite_updated_at":"2013-07-23T17:11:41.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2013-07-23T16:51:20.000Z","updated_at":"2025-06-22T11:25:55.000Z","published_at":"2013-07-23T17:11:41.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003esingle row\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e, so all numbers in a\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003esingle rows\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e are unique. So you need to find up to 2, 3 or 4 numbers that are found in a\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003esingle row\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e that repeat in\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eother rows\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ %p is the given matrix\\n p = [1 2 3 4 5;\\n 2 3 4 5 1;\\n 3 4 6 7 8;\\n 1 2 4 3 5;\\n 4 6 3 2 1];\\n %v is the amount of elements per row that are being checked to see if they repeat in another row\\n v = 4;\\n %Output:\\n out =\\n\\n 1 2 3 4 4\\n 1 2 3 5 3\\n 1 2 4 5 3\\n 1 3 4 5 3\\n 2 3 4 5 3\\n %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.\\n\\n So more examples...\\n\\n p = [1 2 3 4 9;\\n 2 3 1 4 8]\\n v = 4;\\n\\n out =\\n 1 2 3 4 2\\n %so any combination of 1,2,3 and 4 appear 2 times in matrix p\\n\\n and another:\\n\\n p = [1 2 3 4 9;\\n 2 3 1 4 8;\\n 3 4 2 7 1]\\n v = 3;\\n\\n out =\\n 1 2 3 3\\n 1 2 4 3\\n 1 3 4 3\\n 2 3 4 3\\n %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found\\n\\n last example:\\n\\n p = [1 2 3 4 9;\\n 2 3 1 4 8]\\n v = 2;\\n\\n out =\\n 1 2 2\\n 1 3 2\\n 1 4 2\\n 2 3 2\\n 2 4 2\\n 3 4 2\\n %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eNOTE: only display any row number combos that occur at least 2 times or more!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eI hope I gave enough instructions and explanation, if not please let me know and I will add to it! Thanks and good luck!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":1779,"title":"Oh Zero Zero Zero!!!","description":"Hello all,\r\nSo you have to find the largest section of zeros in a vector and then find the length of those zeros and there starting position...\r\nFor example:\r\n \r\n x = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\r\n %then the output is:\r\n LP = [9 10] %[Length Position]\r\n \r\n %Or another example:\r\n \r\n x = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\r\n %then the output is:\r\n LP = [6 9]\r\n \r\n %Or another example:\r\n \r\n x = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\r\n %then the output is:\r\n LP = [7 3];\r\n\r\nHave Fun!\r\n","description_html":"\u003cp\u003eHello all,\r\nSo you have to find the largest section of zeros in a vector and then find the length of those zeros and there starting position...\r\nFor example:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003ex = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\r\n%then the output is:\r\nLP = [9 10] %[Length Position]\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003e%Or another example:\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003ex = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\r\n%then the output is:\r\nLP = [6 9]\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003e%Or another example:\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003ex = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\r\n%then the output is:\r\nLP = [7 3];\r\n\u003c/pre\u003e\u003cp\u003eHave Fun!\u003c/p\u003e","function_template":"function y = LengthAndPosnZeros(x)\r\n y = x;\r\nend","test_suite":"%%\r\nx = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\r\nLP = [9 10] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\r\nLP = [6 9]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\r\nLP = [7 3];\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 2 0 0];\r\nLP = [2 3] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 2 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0];\r\nLP = [9 3] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 0 0 0 0 0 0 0 0 0 1];\r\nLP = [9 2] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [111 541 0 45 3 0 0 0 15 26 0 4 84 3 84 0 9];\r\nLP = [3 6] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n%%\r\nx = [1 0 1];\r\nLP = [1 2] %[Length Position]\r\nassert(isequal(LengthAndPosnZeros(x),LP))\r\n\r\n","published":true,"deleted":false,"likes_count":15,"comments_count":1,"created_by":15013,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":540,"test_suite_updated_at":"2013-08-08T19:35:25.000Z","rescore_all_solutions":false,"group_id":13,"created_at":"2013-08-08T18:54:26.000Z","updated_at":"2026-03-24T00:57:27.000Z","published_at":"2013-08-08T19:35:25.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHello all, So you have to find the largest section of zeros in a vector and then find the length of those zeros and there starting position... For example:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[x = [1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 9 8 7 6 5 4 3 2 1];\\n%then the output is:\\nLP = [9 10] %[Length Position]\\n\\n%Or another example:\\n\\nx = [1 0 3 49 3 2 232 3 0 0 0 0 0 0 8 290 0 0 0 12 323 34];\\n%then the output is:\\nLP = [6 9]\\n\\n%Or another example:\\n\\nx = [1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0];\\n%then the output is:\\nLP = [7 3];]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eHave Fun!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"},{"id":1740,"title":"Find number patterns in rows of a matrix...","description":"So I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a *single row*, so all numbers in a *single rows* are unique. So you need to find up to 2, 3 or 4 numbers that are found in a *single row* that repeat in *other rows* and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:\r\n \r\n %p is the given matrix\r\n p = [1 2 3 4 5;\r\n 2 3 4 5 1;\r\n 3 4 6 7 8;\r\n 1 2 4 3 5;\r\n 4 6 3 2 1];\r\n %v is the amount of elements per row that are being checked to see if they repeat in another row\r\n v = 4;\r\n %Output:\r\n out =\r\n \r\n 1 2 3 4 4\r\n 1 2 3 5 3\r\n 1 2 4 5 3\r\n 1 3 4 5 3\r\n 2 3 4 5 3\r\n %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.\r\n \r\n So more examples...\r\n \r\n p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 4;\r\n \r\n out =\r\n 1 2 3 4 2\r\n %so any combination of 1,2,3 and 4 appear 2 times in matrix p\r\n \r\n and another:\r\n \r\n p = [1 2 3 4 9;\r\n 2 3 1 4 8;\r\n 3 4 2 7 1]\r\n v = 3;\r\n \r\n out =\r\n 1 2 3 3\r\n 1 2 4 3\r\n 1 3 4 3\r\n 2 3 4 3\r\n %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found\r\n \r\n last example:\r\n \r\n p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 2;\r\n \r\n out =\r\n 1 2 2\r\n 1 3 2\r\n 1 4 2\r\n 2 3 2\r\n 2 4 2\r\n 3 4 2\r\n %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.\r\n\r\nNOTE: only display any row number combos that occur at least 2 times or more!\r\n\r\nI hope I gave enough instructions and explanation, if not please let me know and I will add to it! Thanks and good luck!","description_html":"\u003cp\u003eSo I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a \u003cb\u003esingle row\u003c/b\u003e, so all numbers in a \u003cb\u003esingle rows\u003c/b\u003e are unique. So you need to find up to 2, 3 or 4 numbers that are found in a \u003cb\u003esingle row\u003c/b\u003e that repeat in \u003cb\u003eother rows\u003c/b\u003e and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:\u003c/p\u003e\u003cpre\u003e %p is the given matrix\r\n p = [1 2 3 4 5;\r\n 2 3 4 5 1;\r\n 3 4 6 7 8;\r\n 1 2 4 3 5;\r\n 4 6 3 2 1];\r\n %v is the amount of elements per row that are being checked to see if they repeat in another row\r\n v = 4;\r\n %Output:\r\n out =\u003c/pre\u003e\u003cpre\u003e 1 2 3 4 4\r\n 1 2 3 5 3\r\n 1 2 4 5 3\r\n 1 3 4 5 3\r\n 2 3 4 5 3\r\n %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.\u003c/pre\u003e\u003cpre\u003e So more examples...\u003c/pre\u003e\u003cpre\u003e p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 4;\u003c/pre\u003e\u003cpre\u003e out =\r\n 1 2 3 4 2\r\n %so any combination of 1,2,3 and 4 appear 2 times in matrix p\u003c/pre\u003e\u003cpre\u003e and another:\u003c/pre\u003e\u003cpre\u003e p = [1 2 3 4 9;\r\n 2 3 1 4 8;\r\n 3 4 2 7 1]\r\n v = 3;\u003c/pre\u003e\u003cpre\u003e out =\r\n 1 2 3 3\r\n 1 2 4 3\r\n 1 3 4 3\r\n 2 3 4 3\r\n %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found\u003c/pre\u003e\u003cpre\u003e last example:\u003c/pre\u003e\u003cpre\u003e p = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\n v = 2;\u003c/pre\u003e\u003cpre\u003e out =\r\n 1 2 2\r\n 1 3 2\r\n 1 4 2\r\n 2 3 2\r\n 2 4 2\r\n 3 4 2\r\n %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.\u003c/pre\u003e\u003cp\u003eNOTE: only display any row number combos that occur at least 2 times or more!\u003c/p\u003e\u003cp\u003eI hope I gave enough instructions and explanation, if not please let me know and I will add to it! Thanks and good luck!\u003c/p\u003e","function_template":"function out = findRowPattern(p,v)\r\n out = [p v];\r\nend","test_suite":"%%\r\np = [1 2 3 4 5;\r\n 2 3 4 5 1;\r\n 3 4 6 7 8;\r\n 1 2 4 3 5;\r\n 4 6 3 2 1];\r\nv = 4;\r\nout = [1 2 3 4 4;\r\n 1 2 3 5 3;\r\n 1 2 4 5 3;\r\n 1 3 4 5 3;\r\n 2 3 4 5 3]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\nv = 4;\r\nout = [1 2 3 4 2]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [1 2 3 4 9;\r\n 2 3 1 4 8;\r\n 3 4 2 7 1]\r\nv = 3;\r\nout = [1 2 3 3;\r\n 1 2 4 3;\r\n 1 3 4 3;\r\n 2 3 4 3]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [1 2 3 4 9;\r\n 2 3 1 4 8]\r\nv = 2;\r\nout = [1 2 2;\r\n 1 3 2;\r\n 1 4 2;\r\n 2 3 2;\r\n 2 4 2;\r\n 3 4 2]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [15 23 68 49 88;\r\n 69 58 78 21 35;\r\n 10 23 21 35 88;\r\n 99 58 63 24 10;\r\n 64 28 14 33 58;\r\n 85 69 21 45 55;\r\n 99 24 76 49 33;\r\n 89 69 33 98 21;\r\n 99 10 21 55 58;\r\n 35 68 69 44 21;\r\n 21 69 35 46 33];\r\nv = 3;\r\nout = [21 35 69 3;\r\n 10 58 99 2;\r\n 21 33 69 2]\r\nassert(isequal(findRowPattern(p,v),out))\r\n%%\r\np = [15 23 68 49 88;\r\n 69 58 78 21 35;\r\n 10 23 21 35 88;\r\n 99 58 63 24 10;\r\n 64 28 14 33 58;\r\n 85 69 21 45 55;\r\n 99 24 76 49 33;\r\n 89 69 33 98 21;\r\n 99 10 21 55 58;\r\n 35 68 69 44 21;\r\n 21 69 35 46 33];\r\nv = 2;\r\n\r\nout = [21 69 5;\r\n 21 35 4;\r\n 35 69 3;\r\n 10 21 2;\r\n 10 58 2;\r\n 10 99 2;\r\n 21 33 2;\r\n 21 55 2;\r\n 21 58 2;\r\n 23 88 2;\r\n 24 99 2;\r\n 33 69 2;\r\n 58 99 2]\r\nassert(isequal(findRowPattern(p,v),out))","published":true,"deleted":false,"likes_count":0,"comments_count":0,"created_by":15013,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":11,"test_suite_updated_at":"2013-07-23T17:11:41.000Z","rescore_all_solutions":false,"group_id":1,"created_at":"2013-07-23T16:51:20.000Z","updated_at":"2025-06-22T11:25:55.000Z","published_at":"2013-07-23T17:11:41.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eSo I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003esingle row\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e, so all numbers in a\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003esingle rows\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e are unique. So you need to find up to 2, 3 or 4 numbers that are found in a\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003esingle row\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e that repeat in\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:rPr\u003e\u003cw:b/\u003e\u003c/w:rPr\u003e\u003cw:t\u003eother rows\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[ %p is the given matrix\\n p = [1 2 3 4 5;\\n 2 3 4 5 1;\\n 3 4 6 7 8;\\n 1 2 4 3 5;\\n 4 6 3 2 1];\\n %v is the amount of elements per row that are being checked to see if they repeat in another row\\n v = 4;\\n %Output:\\n out =\\n\\n 1 2 3 4 4\\n 1 2 3 5 3\\n 1 2 4 5 3\\n 1 3 4 5 3\\n 2 3 4 5 3\\n %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.\\n\\n So more examples...\\n\\n p = [1 2 3 4 9;\\n 2 3 1 4 8]\\n v = 4;\\n\\n out =\\n 1 2 3 4 2\\n %so any combination of 1,2,3 and 4 appear 2 times in matrix p\\n\\n and another:\\n\\n p = [1 2 3 4 9;\\n 2 3 1 4 8;\\n 3 4 2 7 1]\\n v = 3;\\n\\n out =\\n 1 2 3 3\\n 1 2 4 3\\n 1 3 4 3\\n 2 3 4 3\\n %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found\\n\\n last example:\\n\\n p = [1 2 3 4 9;\\n 2 3 1 4 8]\\n v = 2;\\n\\n out =\\n 1 2 2\\n 1 3 2\\n 1 4 2\\n 2 3 2\\n 2 4 2\\n 3 4 2\\n %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eNOTE: only display any row number combos that occur at least 2 times or more!\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eI hope I gave enough instructions and explanation, if not please let me know and I will add to it! 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