# 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