{"id":624,"date":"2020-06-10T07:43:41","date_gmt":"2020-06-10T07:43:41","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=624"},"modified":"2022-03-30T11:35:54","modified_gmt":"2022-03-30T11:35:54","slug":"mike-and-exam","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/mike-and-exam\/","title":{"rendered":"Mike and Exam"},"content":{"rendered":"

\"\"<\/p>\n

CONCEPTS USED:<\/h3>\n
\n

Recursion<\/p>\n<\/blockquote>\n

DIFFICULTY LEVEL:<\/h3>\n
\n

Medium<\/p>\n<\/blockquote>\n

PROBLEM STATEMENT(<\/code>SIMPLIFIED)<\/code>:<\/h3>\n
\n

Given an array containing N<\/code> elements and an integer K<\/code>, find the number of ways to calculate the value of K<\/code> using array elements.<\/p>\n<\/blockquote>\n

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

NOTE<\/strong>:<\/p>\n

\n
    \n
  1. \n

    The array contains both (-ve)<\/code> and (+ve)<\/code> integers. <\/p>\n<\/li>\n

  2. \n

    Use only Addition<\/code> and Subtraction<\/code> to accomplish the task.<\/p>\n<\/li>\n<\/ol>\n<\/blockquote>\n

    For Example :<\/h4>\n
    Input : \n\nN = 4, K = 2\nA = [1 3 2 6] \n\nOutput : 5\n\nExplanation : All combinations that give 2 as output are - \n\n1st : +(2) = 2\n\n2nd : -(1) - (3) + (6) = 2\n\n3rd : -(1) + (3) = 2\n\n4th : +(1) - (2) + (3) = 2\n\n5th : +(1) - (2) - (3) + (6) = 2<\/code><\/pre>\n
    SOLVING APPROACH:<\/h3>\n
      \n
    1. \n
      The idea is to start processing array elements from (i=0)<\/code> to (i=N-1)<\/code>, with K<\/code> as the required value.  <\/p>\n<\/li>\n
    2. \n
      Check if at any point the value of i<\/code> becomes (>= N)<\/code> and the required value of K<\/code> is not equal to 0<\/code>, simply return 0<\/code>. <\/p>\n<\/li>\n
    3. \n
      Else recursively check for the three possible cases :- <\/p>\n
        \n
      1. Consider current element and add it to K<\/code> (Addition Operation<\/code>)  <\/li>\n
      2. Consider current element and subtract it from K<\/code> (Subtaction Operation<\/code>) <\/li>\n
      3. Don’t consider the current element.<\/li>\n
      4. At any point if the value of K<\/code> becomes 0<\/code>, return 1. <\/li>\n<\/ol>\n<\/li>\n
      5. \n
        Finally return the total count<\/code> of the three cases.<\/p>\n<\/li>\n<\/ol>\n
        SOLUTIONS:<\/h3>\n\t\t\t\t\t\t