{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-04-06T14:01:22.000Z","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":"2026-04-06T00:00:00.000Z","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":782,"title":"Some Assembly Required","description":"The input to this function is a matrix of real numbers.  Your job is to assemble the rows of the matrix into one large row that contains all of the individual rows of the matrix, and to make this output row as short as possible.  You accomplish this by joining the rows together at the points at which they overlap.  To help with the task, you can flip rows if you need to do so.\r\n\r\nFor example:\r\n\r\n  input  = [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3]\r\n\r\n  output = [1 1 2 3 8 4 5 6 7]\r\n\r\nExplanation:\r\nThe [1 1 2 3] is the first row of the input.\r\n\r\n[3 8 4 5] is the third row (flipped) so the 3 at the end of the third row overlaps with the 3 at the end of the first row.\r\n\r\n[4 5 6 7] is the second row, with the [4 5] overlapping with the flipped [5 4] from the third row.\r\n\r\nOther than the mirrored version of the solution ([7 6 5 4 8 3 2 1 1] in the above example), each solution will be unique.  Flipped versions of the entered value of y_correct will be tested for as well, in case your code comes up with that version of the correct answer.\r\n\r\nGood luck, and happy hunting.","description_html":"\u003cp\u003eThe input to this function is a matrix of real numbers.  Your job is to assemble the rows of the matrix into one large row that contains all of the individual rows of the matrix, and to make this output row as short as possible.  You accomplish this by joining the rows together at the points at which they overlap.  To help with the task, you can flip rows if you need to do so.\u003c/p\u003e\u003cp\u003eFor example:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003einput  = [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3]\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003eoutput = [1 1 2 3 8 4 5 6 7]\r\n\u003c/pre\u003e\u003cp\u003eExplanation:\r\nThe [1 1 2 3] is the first row of the input.\u003c/p\u003e\u003cp\u003e[3 8 4 5] is the third row (flipped) so the 3 at the end of the third row overlaps with the 3 at the end of the first row.\u003c/p\u003e\u003cp\u003e[4 5 6 7] is the second row, with the [4 5] overlapping with the flipped [5 4] from the third row.\u003c/p\u003e\u003cp\u003eOther than the mirrored version of the solution ([7 6 5 4 8 3 2 1 1] in the above example), each solution will be unique.  Flipped versions of the entered value of y_correct will be tested for as well, in case your code comes up with that version of the correct answer.\u003c/p\u003e\u003cp\u003eGood luck, and happy hunting.\u003c/p\u003e","function_template":"function y = assemble_this(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nx= [1 2 ; 2 3 ; 3 4 ; 4 5];\r\ny_correct=[1 2 3 4 5];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx= [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3];\r\ny_correct=[1 1 2 3 8 4 5 6 7];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx=[2 3 ; 4 2 ; 3 1 ; 1 5 ; 5 9];\r\ny_correct=[9 5 1 3 2 4];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx=[10:-1:6 ; 1:5 ; 5:0.25:6];\r\ny_correct=[1:4 5:0.25:6 7:10];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx=[8 16 24 ; 2 4 8 ; 6 4 2];\r\ny_correct=[6 4 2 4 8 16 24];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\ny=ceil(rand*7)+5;\r\nry=[rand(1,y) y];\r\nfry=fliplr([y ry(1:end-1)]);\r\natf=assemble_this([fry ; ry]);\r\ny_correct=[y fry];\r\nassert(isequal(atf,y_correct)||isequal(atf,fliplr(y_correct)))\r\n%%\r\nt=rand(1,2);\r\nx=[8 16 24 ; 2 4 8 ; 6 4 2 ; 24 t];\r\nat=assemble_this(x);\r\ny_correct=[fliplr(t) 24 16 8 4 2 4 6];\r\nassert(isequal(at,y_correct)||isequal(at,fliplr(y_correct)))\r\n%%\r\nk=5+ceil(8*rand);\r\nx=randperm(k);\r\ny=randperm(k)+k;\r\nat=assemble_this([x x ; x y]);\r\ny_correct=[x x y];\r\nassert(isequal(at,y_correct)||isequal(at,fliplr(y_correct)))\r\n","published":true,"deleted":false,"likes_count":9,"comments_count":6,"created_by":1615,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":130,"test_suite_updated_at":"2018-01-08T18:23:27.000Z","rescore_all_solutions":true,"group_id":19,"created_at":"2012-06-21T18:36:35.000Z","updated_at":"2026-04-03T20:25:05.000Z","published_at":"2012-06-21T18:43:48.000Z","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\u003eThe input to this function is a matrix of real numbers. Your job is to assemble the rows of the matrix into one large row that contains all of the individual rows of the matrix, and to make this output row as short as possible. You accomplish this by joining the rows together at the points at which they overlap. To help with the task, you can flip rows if you need to do so.\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\u003eFor 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[input  = [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3]\\n\\noutput = [1 1 2 3 8 4 5 6 7]]]\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\u003eExplanation: The [1 1 2 3] is the first row of the input.\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\u003e[3 8 4 5] is the third row (flipped) so the 3 at the end of the third row overlaps with the 3 at the end of the first row.\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\u003e[4 5 6 7] is the second row, with the [4 5] overlapping with the flipped [5 4] from the third row.\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\u003eOther than the mirrored version of the solution ([7 6 5 4 8 3 2 1 1] in the above example), each solution will be unique. Flipped versions of the entered value of y_correct will be tested for as well, in case your code comes up with that version of the correct answer.\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\u003eGood luck, and happy hunting.\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":782,"title":"Some Assembly Required","description":"The input to this function is a matrix of real numbers.  Your job is to assemble the rows of the matrix into one large row that contains all of the individual rows of the matrix, and to make this output row as short as possible.  You accomplish this by joining the rows together at the points at which they overlap.  To help with the task, you can flip rows if you need to do so.\r\n\r\nFor example:\r\n\r\n  input  = [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3]\r\n\r\n  output = [1 1 2 3 8 4 5 6 7]\r\n\r\nExplanation:\r\nThe [1 1 2 3] is the first row of the input.\r\n\r\n[3 8 4 5] is the third row (flipped) so the 3 at the end of the third row overlaps with the 3 at the end of the first row.\r\n\r\n[4 5 6 7] is the second row, with the [4 5] overlapping with the flipped [5 4] from the third row.\r\n\r\nOther than the mirrored version of the solution ([7 6 5 4 8 3 2 1 1] in the above example), each solution will be unique.  Flipped versions of the entered value of y_correct will be tested for as well, in case your code comes up with that version of the correct answer.\r\n\r\nGood luck, and happy hunting.","description_html":"\u003cp\u003eThe input to this function is a matrix of real numbers.  Your job is to assemble the rows of the matrix into one large row that contains all of the individual rows of the matrix, and to make this output row as short as possible.  You accomplish this by joining the rows together at the points at which they overlap.  To help with the task, you can flip rows if you need to do so.\u003c/p\u003e\u003cp\u003eFor example:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003einput  = [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3]\r\n\u003c/pre\u003e\u003cpre class=\"language-matlab\"\u003eoutput = [1 1 2 3 8 4 5 6 7]\r\n\u003c/pre\u003e\u003cp\u003eExplanation:\r\nThe [1 1 2 3] is the first row of the input.\u003c/p\u003e\u003cp\u003e[3 8 4 5] is the third row (flipped) so the 3 at the end of the third row overlaps with the 3 at the end of the first row.\u003c/p\u003e\u003cp\u003e[4 5 6 7] is the second row, with the [4 5] overlapping with the flipped [5 4] from the third row.\u003c/p\u003e\u003cp\u003eOther than the mirrored version of the solution ([7 6 5 4 8 3 2 1 1] in the above example), each solution will be unique.  Flipped versions of the entered value of y_correct will be tested for as well, in case your code comes up with that version of the correct answer.\u003c/p\u003e\u003cp\u003eGood luck, and happy hunting.\u003c/p\u003e","function_template":"function y = assemble_this(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nx= [1 2 ; 2 3 ; 3 4 ; 4 5];\r\ny_correct=[1 2 3 4 5];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx= [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3];\r\ny_correct=[1 1 2 3 8 4 5 6 7];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx=[2 3 ; 4 2 ; 3 1 ; 1 5 ; 5 9];\r\ny_correct=[9 5 1 3 2 4];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx=[10:-1:6 ; 1:5 ; 5:0.25:6];\r\ny_correct=[1:4 5:0.25:6 7:10];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\nx=[8 16 24 ; 2 4 8 ; 6 4 2];\r\ny_correct=[6 4 2 4 8 16 24];\r\nassert(isequal(assemble_this(x),y_correct)||isequal(assemble_this(x),fliplr(y_correct)))\r\n%%\r\ny=ceil(rand*7)+5;\r\nry=[rand(1,y) y];\r\nfry=fliplr([y ry(1:end-1)]);\r\natf=assemble_this([fry ; ry]);\r\ny_correct=[y fry];\r\nassert(isequal(atf,y_correct)||isequal(atf,fliplr(y_correct)))\r\n%%\r\nt=rand(1,2);\r\nx=[8 16 24 ; 2 4 8 ; 6 4 2 ; 24 t];\r\nat=assemble_this(x);\r\ny_correct=[fliplr(t) 24 16 8 4 2 4 6];\r\nassert(isequal(at,y_correct)||isequal(at,fliplr(y_correct)))\r\n%%\r\nk=5+ceil(8*rand);\r\nx=randperm(k);\r\ny=randperm(k)+k;\r\nat=assemble_this([x x ; x y]);\r\ny_correct=[x x y];\r\nassert(isequal(at,y_correct)||isequal(at,fliplr(y_correct)))\r\n","published":true,"deleted":false,"likes_count":9,"comments_count":6,"created_by":1615,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":130,"test_suite_updated_at":"2018-01-08T18:23:27.000Z","rescore_all_solutions":true,"group_id":19,"created_at":"2012-06-21T18:36:35.000Z","updated_at":"2026-04-03T20:25:05.000Z","published_at":"2012-06-21T18:43:48.000Z","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\u003eThe input to this function is a matrix of real numbers. Your job is to assemble the rows of the matrix into one large row that contains all of the individual rows of the matrix, and to make this output row as short as possible. You accomplish this by joining the rows together at the points at which they overlap. To help with the task, you can flip rows if you need to do so.\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\u003eFor 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[input  = [1 1 2 3 ; 4 5 6 7 ; 5 4 8 3]\\n\\noutput = [1 1 2 3 8 4 5 6 7]]]\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\u003eExplanation: The [1 1 2 3] is the first row of the input.\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\u003e[3 8 4 5] is the third row (flipped) so the 3 at the end of the third row overlaps with the 3 at the end of the first row.\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\u003e[4 5 6 7] is the second row, with the [4 5] overlapping with the flipped [5 4] from the third row.\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\u003eOther than the mirrored version of the solution ([7 6 5 4 8 3 2 1 1] in the above example), each solution will be unique. 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