Largest Number By Two Times in Python

Suppose we have a list of numbers; we have to check whether the largest number is bigger than the second-largest number by more than two times. As an example, if the list is like [3, 9, 6], then it will return false, as 9 is not bigger than 12 (2 times 6). When the list is [6, 3, 15], it will return true, as 15 is bigger than 12 (2 times 6).

To solve this, we will follow these steps −

• if size of nums < 2, then
  • return False
• p_max := minimum of nums[0] and nums[1]
• c_max := maximum of nums[0] and nums[1]
• for i in range 2 to size of nums, do
  • if nums[i] > p_max, then
    • if nums[i] > c_max, then
      • p_max := c_max
      • c_max := nums[i]
    • otherwise,
      • p_max := nums[i]
• return c_max > p_max * 2

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, nums):
      if len(nums) < 2:
         return False
         p_max = min(nums[0], nums[1])
         c_max = max(nums[0], nums[1])
         for i in range(2, len(nums)):
            if nums[i] > p_max:
               if nums[i] > c_max:
                  p_max = c_max
                  c_max = nums[i]
               else:
                  p_max = nums[i]
         return c_max > p_max * 2
ob = Solution()
nums = [3,6,15]
print(ob.solve(nums))

Input

[3,6,15]

Output

None

Arnab Chakraborty

Updated on: 23-Sep-2020

162 Views

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