Program to check whether given list is in valid state or not in Python

PythonServer Side ProgrammingProgramming

Suppose we have a list of numbers called nums, we have to check whether every number can be grouped using one of the following rules: 1. Contiguous pairs (a, a) 2. Contiguous triplets (a, a, a) 3. Contiguous triplets (a, a + 1, a + 2)

So, if the input is like nums = [7, 7, 3, 4, 5], then the output will be True, as We can group [7, 7] together and [3, 4, 5] together.

To solve this, we will follow these steps −

  • n := size of nums

  • dp := a list of size n+1, first value is True, others are False

  • for i in range 2 to n, do

    • if i >= 2 and dp[i − 2] is not 0, then

      • if nums[i − 1] is same as nums[i − 2], then

        • dp[i] := True

    • if i >= 3 and dp[i − 3] is not 0, then

      • if (nums[i − 1], nums[i − 2], nums[i − 3]) are same or (nums[i − 1], nums[i − 2] + 1, nums[i − 3] + 2 are same), then

        • dp[i] := True

  • return dp[n]

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, nums):
      n = len(nums)
      dp = [True] + [False] * n
      for i in range(2, n + 1):
         if i >= 2 and dp[i − 2]:
            if nums[i − 1] == nums[i − 2]:
               dp[i] = True
         if i >= 3 and dp[i − 3]:
            if (nums[i − 1] == nums[i − 2] == nums[i − 3]) or (nums[i − 1] == nums[i − 2] + 1 == nums[i − 3] + 2):
               dp[i] = True
         return dp[n]
ob = Solution()
nums = [8, 8, 4, 5, 6]
print(ob.solve(nums))

Input

[8, 8, 4, 5, 6]

Output

True
raja
Published on 21-Oct-2020 16:00:34
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