{"id":25729,"date":"2021-12-04T09:19:48","date_gmt":"2021-12-04T03:49:48","guid":{"rendered":"https:\/\/python-programs.com\/?p=25729"},"modified":"2021-12-04T09:19:48","modified_gmt":"2021-12-04T03:49:48","slug":"python-program-to-find-number-of-possible-strings-with-no-consecutive-1s","status":"publish","type":"post","link":"https:\/\/python-programs.com\/python-program-to-find-number-of-possible-strings-with-no-consecutive-1s\/","title":{"rendered":"Python Program to Find Number of Possible Strings With No Consecutive 1s"},"content":{"rendered":"<p>Given an integer n, which represents the length of the string, or the number of bits in the final string.<\/p>\n<p>The task is to obtain a string with n bits and no consecutive 1s, which means that <strong>11(binary)<\/strong> is prohibited in the string.<\/p>\n<p>There are two possibilities for any string:<br \/>\nIf the first element is zero, the second element can be either zero or one.<br \/>\nIn the other instance, if the initial element is 1, the sole option for the next element is 0.<\/p>\n<p>Our task is to count the number of such possible strings that match the above-mentioned requirement.<\/p>\n<p><strong>Examples:<\/strong><\/p>\n<p><strong>Example1:<\/strong><\/p>\n<p><strong>Input:<\/strong><\/p>\n<pre>Given number = 2<\/pre>\n<p><strong>Output:<\/strong><\/p>\n<pre id=\"codeOutputPre\">The total number of possible strings for the given number { 2 } =  3<\/pre>\n<p><strong>Example2:<\/strong><\/p>\n<p><strong>Input:<\/strong><\/p>\n<pre>Given number = 3<\/pre>\n<p><strong>Output:<\/strong><\/p>\n<pre>The total number of possible strings for the given number { 3 } = 5<\/pre>\n<h2>Program to Find Number of Possible Strings With No Consecutive 1s in Python<\/h2>\n<p>This can be solved manually or via the recursion method. The recursion method is a superior, more efficient, and faster method of solving the problem.<\/p>\n<p>Before going into the problem just have a glance over the recursion.<\/p>\n<p>The two things are as follows:<\/p>\n<p>The first digit is 1, then the second bit is set to 0 and we check for n-2 places left in the final string.<br \/>\nThe first digit is 0, and we next look for n-1 places left in the string.<\/p>\n<p><strong>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 n\u00a0 \u00a0<\/strong> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<strong> number of possible strings<\/strong><\/p>\n<p>0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 0<\/p>\n<p>1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a02\u00a0 \u00a0(0, 1)<\/p>\n<p>2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 3\u00a0 \u00a0(00, 01, 10)<\/p>\n<p>We will consider the two cases in all circumstances when n is bigger than 2.<\/p>\n<ul>\n<li><a href=\"#Using_Recursion_(Static_Input)\">Using Recursion (Static Input)<\/a><\/li>\n<li><a href=\"#Using_Recursion_(User_Input)\">Using Recursion (User Input)<\/a><\/li>\n<\/ul>\n<h3 id=\"Using_Recursion_(Static_Input)\">Method #1: Using Recursion (Static Input)<\/h3>\n<p><strong>Approach:<\/strong><\/p>\n<ul>\n<li>Give the number as static input and store it in a variable.<\/li>\n<li>Create a recursive function say numberof_ways() which accepts the given number as an argument and returns the number of possible strings with no consecutive 1&#8217;s.<\/li>\n<li>Inside the function, check if the given number is equal to 0 using the if\u00a0conditional statement.<\/li>\n<li>If it is true, then return 0.<\/li>\n<li>Check if the given number is less than 3 using the if conditional statement.<\/li>\n<li>If it is true, then return the value given number+1.<\/li>\n<li>Return the value numberof_ways(gvn_n-1) + numberof_ways(gvn_n-2) (Recursive Logic).<\/li>\n<li>Pass the given number as an argument to the numberof_ways() function and store it in a variable.<\/li>\n<li>Print the above result.<\/li>\n<li>The Exit of the Program.<\/li>\n<\/ul>\n<p><strong>Below is the implementation:<\/strong><\/p>\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\"># Create a recursive function say numberof_ways() which accepts the given number\r\n# as an argument and returns the number of possible strings with no\r\n# consecutive 1's\r\n\r\n\r\ndef numberof_ways(gvn_n):\r\n     # Inside the function, check if the given number is equal to 0 using the if\r\n     # conditional statement.\r\n    if(gvn_n == 0):\r\n        # If it is true, then return 0.\r\n        return 0\r\n    # Check if the given number is less than 3 using the if\r\n    # conditional statement.\r\n    if(gvn_n &lt; 3):\r\n        # If it is true, then return the value given number+1\r\n        return gvn_n+1\r\n    # Return the value numberof_ways(gvn_n-1) + numberof_ways(gvn_n-2).\r\n    # (Recursive Logic)\r\n    return numberof_ways(gvn_n-1) + numberof_ways(gvn_n-2)\r\n\r\n\r\n# Give the number as static input and store it in a variable.\r\ngvn_n = 2\r\n# Pass the given number as an argument to the numberof_ways() function and\r\n# store it in a variable.\r\nrslt = numberof_ways(gvn_n)\r\n# Print the above result.\r\nprint(\r\n    \"The total number of possible strings for the given number {\", gvn_n, \"} = \", rslt)\r\n<\/pre>\n<p><strong>Output:<\/strong><\/p>\n<pre id=\"codeOutputPre\">The total number of possible strings for the given number { 2 } =  3<\/pre>\n<h3 id=\"Using_Recursion_(User_Input)\">Method #2: Using Recursion (User Input)<\/h3>\n<p><strong>Approach:<\/strong><\/p>\n<ul>\n<li>Give the number as user input using the int(input()) function and store it in a variable.<\/li>\n<li>Create a recursive function say numberof_ways() which accepts the given number as an argument and returns the number of possible strings with no consecutive 1&#8217;s.<\/li>\n<li>Inside the function, check if the given number is equal to 0 using the if\u00a0conditional statement.<\/li>\n<li>If it is true, then return 0.<\/li>\n<li>Check if the given number is less than 3 using the if conditional statement.<\/li>\n<li>If it is true, then return the value given number+1.<\/li>\n<li>Return the value numberof_ways(gvn_n-1) + numberof_ways(gvn_n-2) (Recursive Logic).<\/li>\n<li>Pass the given number as an argument to the numberof_ways() function and store it in a variable.<\/li>\n<li>Print the above result.<\/li>\n<li>The Exit of the Program.<\/li>\n<\/ul>\n<p><strong>Below is the implementation:<\/strong><\/p>\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\"># Create a recursive function say numberof_ways() which accepts the given number\r\n# as an argument and returns the number of possible strings with no\r\n# consecutive 1's\r\n\r\n\r\ndef numberof_ways(gvn_n):\r\n     # Inside the function, check if the given number is equal to 0 using the if\r\n     # conditional statement.\r\n    if(gvn_n == 0):\r\n        # If it is true, then return 0.\r\n        return 0\r\n    # Check if the given number is less than 3 using the if\r\n    # conditional statement.\r\n    if(gvn_n &lt; 3):\r\n        # If it is true, then return the value given number+1\r\n        return gvn_n+1\r\n    # Return the value numberof_ways(gvn_n-1) + numberof_ways(gvn_n-2).\r\n    # (Recursive Logic)\r\n    return numberof_ways(gvn_n-1) + numberof_ways(gvn_n-2)\r\n\r\n\r\n# Give the number as user input using the int(input()) function and store it in a variable.\r\ngvn_n = int(input(\"Enter some random number = \"))\r\n# Pass the given number as an argument to the numberof_ways() function and\r\n# store it in a variable.\r\nrslt = numberof_ways(gvn_n)\r\n# Print the above result.\r\nprint(\r\n    \"The total number of possible strings for the given number {\", gvn_n, \"} = \", rslt)\r\n<\/pre>\n<p><strong>Output:<\/strong><\/p>\n<pre>Enter some random number = 3\r\nThe total number of possible strings for the given number { 3 } = 5<\/pre>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Given an integer n, which represents the length of the string, or the number of bits in the final string. The task is to obtain a string with n bits and no consecutive 1s, which means that 11(binary) is prohibited in the string. There are two possibilities for any string: If the first element is &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/python-programs.com\/python-program-to-find-number-of-possible-strings-with-no-consecutive-1s\/\"> <span class=\"screen-reader-text\">Python Program to Find Number of Possible Strings With No Consecutive 1s<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","jetpack_publicize_message":"","jetpack_is_tweetstorm":false,"jetpack_publicize_feature_enabled":true},"categories":[5],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v18.9 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Python Program to Find Number of Possible Strings With No Consecutive 1s - Python Programs<\/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:\/\/python-programs.com\/python-program-to-find-number-of-possible-strings-with-no-consecutive-1s\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Python Program to Find Number of Possible Strings With No Consecutive 1s - Python Programs\" \/>\n<meta property=\"og:description\" content=\"Given an integer n, which represents the length of the string, or the number of bits in the final string. 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