{"id":11297,"date":"2021-03-01T02:24:53","date_gmt":"2021-03-01T02:24:53","guid":{"rendered":"https:\/\/www.softwareeverydayblog.com\/?p=11297"},"modified":"2021-03-28T10:19:50","modified_gmt":"2021-03-28T10:19:50","slug":"programming-problem-paint-fence","status":"publish","type":"post","link":"https:\/\/www.softwareeverydayblog.com\/?p=11297","title":{"rendered":"[Programming Problem] Paint Fence"},"content":{"rendered":"<p>You are painting a fence of n posts with k different colors. You must paint the posts following these rules:<\/p>\n<ul>\n<li>Every post must be painted exactly one color.<\/li>\n<li>At most one pair of adjacent fence posts can have the same color.<\/li>\n<\/ul>\n<p>Given the two integers n and k, return the number of ways you can paint the fence.<\/p>\n<p>[<a href=\"https:\/\/leetcode.com\/problems\/paint-fence\/\" rel=\"noopener\" target=\"_blank\">Link<\/a>]<\/p>\n<p>The recursive solution times out! So let&#8217;s think of a dynamic programming solution!<\/p>\n<p><strong>Single Fence<\/strong><br \/>\nopt(0) = k options<\/p>\n<p><strong>Double Fence<\/strong><br \/>\nopt(1) = k^2 options<\/p>\n<p><strong>Third Fence<\/strong><br \/>\nWhen first two fences are same color: k * (k-1) = opt(0) * (k-1)<br \/>\nWhen first two fences are diff color: k * (k-1) * k = k^2 * (k-1) = opt(1) * (k-1)<br \/>\nopt(2) = opt(0) * (k-1) + opt(1) * (k-1)<\/p>\n<p><strong>Nth Fence??<\/strong><br \/>\nopt(i) = opt(i-2) * (k-1) + opt(i-1) * (k-1)<\/p>\n<pre lang=\"javascript\">\r\n\/**\r\n * @param {number} n\r\n * @param {number} k\r\n * @return {number}\r\n *\/\r\n\r\nvar numWays = function(n, k) {\r\n    if (n===0) return 0;\r\n    if (n===1) return k;\r\n\r\n    const dp = new Array(n);\r\n    dp[0] = k;\r\n    dp[1] = k*k;\r\n    \r\n    for (let i = 2; i < n ; i++ ) {\r\n        dp[i] = (dp[i-2] * (k-1)) + (dp[i-1] * (k-1));\r\n    }\r\n    \r\n    return dp[n-1];\r\n};\r\n\r\n<\/pre>\n<p>The recursive solution (this will timeout).<\/p>\n<pre lang=\"javascript\">\r\n\/**\r\n * @param {number} n\r\n * @param {number} k\r\n * @return {number}\r\n *\/\r\n\r\nvar numWays = function(n, k) {\r\n    if (n <= 2) return Math.pow(k, n);\r\n    return countForOneColor([], 0, n, k);\r\n};\r\n\r\nvar countForOneColor = function(curr, level, n, k) {\r\n    if (level === n) return 1;\r\n    let ret = 0;\r\n    for (let i = 0 ; i < k ; i++) {\r\n        if (level >= 2 && curr[level-1] === curr[level-2] && i === curr[level-1]) continue;\r\n        curr[level] = i;\r\n        ret += countForOneColor(curr, level+1, n, k);\r\n        curr.pop();\r\n    }\r\n    return ret;\r\n}\r\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>You are painting a fence of n posts with k different colors. You must paint the posts following these rules: Every post must be painted exactly one color. At most one pair of adjacent fence posts can have the same color. Given the two integers n and k, return the number of ways you can [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-11297","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=\/wp\/v2\/posts\/11297","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=11297"}],"version-history":[{"count":5,"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=\/wp\/v2\/posts\/11297\/revisions"}],"predecessor-version":[{"id":11301,"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=\/wp\/v2\/posts\/11297\/revisions\/11301"}],"wp:attachment":[{"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.softwareeverydayblog.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}