{"id":367,"date":"2020-03-11T14:29:00","date_gmt":"2020-03-11T14:29:00","guid":{"rendered":"https:\/\/danwritescode.com\/?p=367"},"modified":"2020-06-24T14:39:47","modified_gmt":"2020-06-24T14:39:47","slug":"missing-integer-codility-100-correct-javascript-solution","status":"publish","type":"post","link":"https:\/\/danwritescode.com\/missing-integer-codility-100-correct-javascript-solution\/","title":{"rendered":"Missing Integer &#8211; Codility 100% Correct Javascript Solution"},"content":{"rendered":"\n<p>The Missing Integer problem in the 4th Codility lesson is marked as medium but I think the solution is pretty straight forward.<\/p>\n\n\n\n<p>The complexity of my solution is <strong>O(N)<\/strong> or <strong>O(N * log(N))<\/strong>.<\/p>\n\n\n\n<p>First, we check which is the max value of the array. If it is below 0 we immediately return 1.<\/p>\n\n\n\n<p>We then fill an array of length max with the number of occurrences of each positive element from the A array.<\/p>\n\n\n\n<p>Lastly, we check for the first appearance of 0 in the array of occurrence and return the value we were looking for. <\/p>\n\n\n\n<pre class=\"wp-block-code\"><code lang=\"javascript\" class=\"language-javascript\">.function solution(A) {\n  let max = Math.max.apply(null,A);\n  if (max &lt; 0) return 1;\n  let B = new Array(max).fill(0);\n  for (let i = 0; i &lt; A.length; i++) {\n    if (A[i] > 0) {\n      B[A[i] - 1]++;\n    }\n  }\n  let index = B.indexOf(0);\n  if ((index === -1)) {\n    return max + 1;\n  } else {\n    return index + 1;\n  }\n}<\/code><\/pre>\n\n\n\n<p><a href=\"https:\/\/app.codility.com\/demo\/results\/trainingBU5NKK-29J\/\" target=\"_blank\" aria-label=\"Click here (opens in a new tab)\" rel=\"noreferrer noopener\" class=\"rank-math-link\">Click here<\/a> to see the report of this algo.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Missing Integer problem in the 4th Codility lesson is marked as medium but I think the solution is pretty straight forward. The complexity of my solution is O(N) or O(N * log(N)). First, we check which is the max value of the array. If it is below 0 we immediately return 1. We then fill an array of length max with the number of occurrences of each positive element from the A array. Lastly, we check for the first appearance of 0 in the array of occurrence and return the value we were looking for. Click here to see <\/p>\n","protected":false},"author":1,"featured_media":368,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13,5,6],"tags":[],"_links":{"self":[{"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/posts\/367"}],"collection":[{"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/comments?post=367"}],"version-history":[{"count":0,"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/posts\/367\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/media\/368"}],"wp:attachment":[{"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/media?parent=367"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/categories?post=367"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/danwritescode.com\/wp-json\/wp\/v2\/tags?post=367"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}