{"id":93,"date":"2023-07-21T21:10:29","date_gmt":"2023-07-21T20:10:29","guid":{"rendered":"https:\/\/techedges.in\/?page_id=93"},"modified":"2023-07-21T21:10:29","modified_gmt":"2023-07-21T20:10:29","slug":"explanation-of-word-break-problem-in-easy-fashion","status":"publish","type":"page","link":"https:\/\/techedges.in\/index.php\/explanation-of-word-break-problem-in-easy-fashion\/","title":{"rendered":"Explanation of Word Break Problem in Easy Fashion:"},"content":{"rendered":"\n

The Word Break Problem is a classic algorithmic challenge that involves determining if a given non-empty string can be segmented into a space-separated sequence of words from a dictionary. In simpler terms, we want to find out if we can form the input string by using words from a given set of valid words.<\/p>\n\n\n\n

Step-by-Step Approach:<\/strong><\/p>\n\n\n\n

    \n
  1. We are given a non-empty string and a list of valid words in the dictionary.<\/li>\n\n\n\n
  2. We need to check if the input string can be broken down into a sequence of words such that each word is a valid word from the dictionary.<\/li>\n\n\n\n
  3. To solve this problem, we use a technique called Dynamic Programming.<\/li>\n\n\n\n
  4. We create a DP array, where each element dp[i]<\/code> represents whether the substring from index 0<\/code> to i<\/code> in the input string can be segmented into valid words or not.<\/li>\n\n\n\n
  5. The DP array is initialized with False<\/code> values.<\/li>\n\n\n\n
  6. We iterate through the input string, considering all possible substrings from index 0<\/code> to i<\/code>.<\/li>\n\n\n\n
  7. At each index i<\/code>, we check if the substring from 0<\/code> to i<\/code> is a valid word or if it can be formed by combining valid words from previous substrings.<\/li>\n\n\n\n
  8. If either of these conditions is true, we update dp[i]<\/code> to True<\/code>, indicating that the substring from 0<\/code> to i<\/code> can be segmented into valid words.<\/li>\n\n\n\n
  9. After completing the iteration, the value of dp[n-1]<\/code>, where n<\/code> is the length of the input string, will represent whether the entire string can be segmented or not.<\/li>\n<\/ol>\n\n\n\n

    Example:<\/strong>
    Suppose we have the following input:<\/p>\n\n\n\n

      \n
    • String: “leetcode”<\/li>\n\n\n\n
    • Dictionary: [“leet”, “code”]<\/li>\n<\/ul>\n\n\n\n

      Explanation:<\/strong>
      We can break down the input string “leetcode” into the valid words “leet” and “code.” Thus, the output will be True<\/code>.<\/p>\n\n\n\n

      Python Solution:<\/strong><\/p>\n\n\n\n

      def word_break(s, word_dict):\n    n = len(s)\n    dp = [False] * n\n\n    for i in range(n):\n        if s[:i + 1] in word_dict:\n            dp[i] = True\n        else:\n            for j in range(i):\n                if dp[j] and s[j + 1:i + 1] in word_dict:\n                    dp[i] = True\n                    break\n\n    return dp[-1]\n\nif __name__ == \"__main__\":\n    input_string = \"leetcode\"\n    dictionary = [\"leet\", \"code\"]\n    print(word_break(input_string, dictionary))<\/code><\/pre>\n\n\n\n
      Explanation of Python Solution:<\/strong>
      The Python solution implements the Word Break Problem using Dynamic Programming. The word_break<\/code> function takes the input string s<\/code> and a list of valid words word_dict<\/code> as inputs and returns True<\/code> if the string can be segmented into valid words, and False<\/code> otherwise.<\/p>\n\n\n\n
      The provided example demonstrates the input string “leetcode” and a dictionary containing the valid words “leet” and “code.” Since we can break down the input string into “leet” and “code,” the function will return True<\/code>.<\/p>\n\n\n\n
      By using the Dynamic Programming technique, we can efficiently solve the Word Break Problem and determine if a given string can be segmented into valid words from the dictionary.<\/p>\n\n\n\n
      Explanation of Word Break Problem in Easy Fashion:<\/strong><\/p>\n\n\n\n
      The Word Break Problem is a classic algorithmic challenge that involves determining if a given non-empty string can be segmented into a space-separated sequence of words from a dictionary. In simpler terms, we want to find out if we can form the input string by using words from a given set of valid words.<\/p>\n\n\n\n
      Step-by-Step Approach:<\/strong><\/p>\n\n\n\n
        \n
      1. We are given a non-empty string and a list of valid words in the dictionary.<\/li>\n\n\n\n
      2. We need to check if the input string can be broken down into a sequence of words such that each word is a valid word from the dictionary.<\/li>\n\n\n\n
      3. To solve this problem, we use a technique called Dynamic Programming.<\/li>\n\n\n\n
      4. We create a DP array, where each element dp[i]<\/code> represents whether the substring from index 0<\/code> to i<\/code> in the input string can be segmented into valid words or not.<\/li>\n\n\n\n
      5. The DP array is initialized with False<\/code> values.<\/li>\n\n\n\n
      6. We iterate through the input string, considering all possible substrings from index 0<\/code> to i<\/code>.<\/li>\n\n\n\n
      7. At each index i<\/code>, we check if the substring from 0<\/code> to i<\/code> is a valid word or if it can be formed by combining valid words from previous substrings.<\/li>\n\n\n\n
      8. If either of these conditions is true, we update dp[i]<\/code> to True<\/code>, indicating that the substring from 0<\/code> to i<\/code> can be segmented into valid words.<\/li>\n\n\n\n
      9. After completing the iteration, the value of dp[n-1]<\/code>, where n<\/code> is the length of the input string, will represent whether the entire string can be segmented or not.<\/li>\n<\/ol>\n\n\n\n
        Example:<\/strong>
        Suppose we have the following input:<\/p>\n\n\n\n
          \n
        • String: “leetcode”<\/li>\n\n\n\n
        • Dictionary: [“leet”, “code”]<\/li>\n<\/ul>\n\n\n\n
          Explanation:<\/strong>
          We can break down the input string “leetcode” into the valid words “leet” and “code.” Thus, the output will be True<\/code>.<\/p>\n\n\n\n
          Python Solution:<\/strong><\/p>\n\n\n\n
          def word_break(s, word_dict):\n    n = len(s)\n    dp = [False] * n\n\n    for i in range(n):\n        if s[:i + 1] in word_dict:\n            dp[i] = True\n        else:\n            for j in range(i):\n                if dp[j] and s[j + 1:i + 1] in word_dict:\n                    dp[i] = True\n                    break\n\n    return dp[-1]\n\nif __name__ == \"__main__\":\n    input_string = \"leetcode\"\n    dictionary = [\"leet\", \"code\"]\n    print(word_break(input_string, dictionary))<\/code><\/pre>\n\n\n\n
          Explanation of Python Solution:<\/strong>
          The Python solution implements the Word Break Problem using Dynamic Programming. The word_break<\/code> function takes the input string s<\/code> and a list of valid words word_dict<\/code> as inputs and returns True<\/code> if the string can be segmented into valid words, and False<\/code> otherwise.<\/p>\n\n\n\n
          The provided example demonstrates the input string “leetcode” and a dictionary containing the valid words “leet” and “code.” Since we can break down the input string into “leet” and “code,” the function will return True<\/code>.<\/p>\n\n\n\n
          By using the Dynamic Programming technique, we can efficiently solve the Word Break Problem and determine if a given string can be segmented into valid words from the dictionary.<\/p>\n","protected":false},"excerpt":{"rendered":"
          The Word Break Problem is a classic algorithmic challenge that involves determining if a given non-empty string can be segmented into a space-separated sequence of words from a dictionary. In simpler terms, we want to find out if we can form the input string by using words from a given set of valid words. Step-by-Step […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"saved_in_kubio":false,"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"aioseo_notices":[],"kubio_ai_page_context":{"short_desc":"","purpose":"general"},"_links":{"self":[{"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/pages\/93"}],"collection":[{"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/comments?post=93"}],"version-history":[{"count":1,"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/pages\/93\/revisions"}],"predecessor-version":[{"id":94,"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/pages\/93\/revisions\/94"}],"wp:attachment":[{"href":"https:\/\/techedges.in\/index.php\/wp-json\/wp\/v2\/media?parent=93"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}