Program to check whether we can form 24 by placing operators in python

Suppose we have a list of four numbers, each numbers are in range 1 to 9, in a fixed order. Now if we place the operators +, -, *, and / (/ denotes integer division) between the numbers, and group them with brackets, we have to check whether it is possible to get the value 24 or not.

So, if the input is like nums = [5, 3, 6, 8, 7], then the output will be True, as (5 * 3) - 6 + (8 + 7) = 24.

To solve this, we will follow these steps −

• Define a function recur() . This will take arr
• answer := a new list
• for i in range 0 to size of arr - 1, do
  • pre := recur(arr[from index 0 to i])
  • suf := recur(arr[from index i + 1 to end])
  • for each k in pre, do
    • for each j in suf, do
      • insert (k + j) at the end of answer
      • insert (k - j) at the end of answer
      • insert (k * j) at the end of answer
      • if j is not 0, then
        • insert (quotient of k / j) at the end of answer
• if size of answer is 0 and size of arr is 1, then
  • insert arr[0] at the end of answer
• return answer
• From the main method check whether 24 is in recur(nums) or not, if yes return True, otherwise false

Let us see the following implementation to get better understanding −

Example 

Live Demo

class Solution:
   def solve(self, nums):
      def recur(arr):
         answer = []
         for i in range(len(arr) - 1):
            pre, suf = recur(arr[: i + 1]), recur(arr[i + 1 :])
            for k in pre:
               for j in suf:
                  answer.append(k + j)
                  answer.append(k - j)
                  answer.append(k * j)
                  if j != 0:
                     answer.append(k // j)
         if len(answer) == 0 and len(arr) == 1:
            answer.append(arr[0])
         return answer
      return 24 in recur(nums)
ob = Solution()
nums = [5, 3, 6, 8, 7]
print(ob.solve(nums))

Input

[5, 3, 6, 8, 7]

Output

True

Arnab Chakraborty

Updated on: 02-Dec-2020

66 Views

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