{"id":1280,"date":"2020-06-10T17:51:16","date_gmt":"2020-06-10T17:51:16","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=1280"},"modified":"2023-11-27T04:44:41","modified_gmt":"2023-11-27T04:44:41","slug":"minimum-number-of-steps-to-make-two-strings-anagram","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/minimum-number-of-steps-to-make-two-strings-anagram\/","title":{"rendered":"Minimum Number of Steps to Make Two Strings Anagram"},"content":{"rendered":"

\"\"<\/p>\n

In the realm of string manipulation, anagrams hold a significant place due to their intriguing nature. An anagram is formed by rearranging the letters of one word or phrase to create another, using all the original letters exactly once. When it comes to making two strings anagrams of each other, the challenge lies in determining the minimum number of steps required to transform one string into another. These steps often involve altering, deleting, or adding characters within the strings to achieve the desired result. Understanding the concept of anagram transformation and the minimum steps involved provides valuable insights into string manipulation algorithms and problem-solving techniques. This article explores the intriguing puzzle of determining the minimum number of steps needed to convert one string into an anagram of another, unraveling various approaches and strategies to solve this computational challenge.<\/p>\n

How to Find the Minimum Number of Steps to Make Two Strings Anagram?How to Find the Minimum Number of Steps to Make Two Strings Anagram?<\/h2>\n

For given two string, Print minimum number of steps to make them anagram.<\/p>\n

<\/strong><\/u><\/a><\/p>\n

Test Case<\/h4>\n
Input:\n1\nexpertcoder zenithcoder\n\nOutput:\n4\n\nExplanation:\nIn 'expertcoder' and 'zenithcoder', 7 characters ('e','t','c','o','d','e','r') are common in both. \n\nCharacters which are not common are :\nIn 'expertcoder' : 'e','x','p','r'\nIn 'zenithcoder' : 'z','n','i','h'\n\nIf we convert, 'e' to 'z', 'x' to 'n', 'p' to 'i' and 'r' to 'h', both strings becomes Anagram.<\/code><\/pre>\n
Solving Approach :<\/h3>\n
Anagram:<\/strong> Two strings are anagram if both of them contain the same characters with a different arrangement. For example prepbytes<\/code> and bytesprep<\/code> are anagrams.<\/p>\n
    \n
  1. \n
    We can count a minimum number of changes if we can find if how many numbers of characters are different in one string from another string. We do it for both of the strings.<\/p>\n<\/li>\n
  2. \n
    We check this by scanning both strings, and increase value by 1<\/code> for the current character in 1st<\/sup> string and decrease the value by 1<\/code> for the current character in 2nd<\/sup> string. <\/p>\n<\/li>\n
  3. \n
    After scanning both strings, we find the absolute sum of the hash table where sum<\/code> represents the number of characters different in both strings.  <\/p>\n<\/li>\n
  4. \n
    After checking this out, we know a single change would affect two characters, if we change one character to another one which is extra in second string and missing in the first string.<\/p>\n<\/li>\n
  5. \n
    So we print half of that sum<\/code> as our answer.<\/p>\n<\/li>\n<\/ol>\n
    Example<\/h3>\n
      \n
    • Lets assume two strings to be ‘prepbuddy<\/code>‘ and ‘codepuddy<\/code>‘.<\/li>\n
    • As we defined in the definition of anagram, two strings must have the same number of characters in any order. So we first find the number of characters which are different in both strings.<\/li>\n
    • We can calculate such characters using a hash table.<\/li>\n
    • After scanning both strings our hash table becomes<\/li>\n<\/ul>\n
      \"\"<\/p>\n
        \n
      • \n
        1<\/code> in hash table means character is present in 1st<\/sup> string but not in 2nd<\/sup> string, and -1<\/code> in hash table means character is present in 2nd<\/sup> string but not in 1st<\/sup> string,.<\/p>\n<\/li>\n
      • \n
        Absolute sum of the hash table is 6<\/code> which means a total of 6<\/code> characters are different, half of it i.e. 3<\/code> is our final answer.<\/p>\n<\/li>\n<\/ul>\n
        Solutions<\/h3>\n
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