Find Indices of Local Peaks in Python

Suppose we have a list of numbers called nums. We have to find the index of every peak element in the nums, sorted in ascending order. An index i of a peak element when all of these three conditions are satisfied: 1. The next number on its right that's different than nums[i] doesn't present or must be smaller than nums[i] 2. The previous number on its left that's different than nums[i] doesn't present or must be smaller than nums[i] 3. There is at least one number on its left side or right side that is different from nums[i].

So, if the input is like nums = [5, 8, 8, 8, 6, 11, 11], then the output will be [1, 2, 3, 5, 6], because the plateau of 8s are considered as peaks [1,2,3]. And 11s are also, [5, 6].

To solve this, we will follow these steps −

• n := size of nums
• ans := a new list
• i := 0
• while i < n, do
  • i0 := i
  • while i < n and nums[i] is same as nums[i0], do
    • i := i + 1
  • if (i0 is 0 or nums[i0] > nums[i0 - 1]) and (i is n or nums[i0] > nums[i]), then
    • if i0 is not 0 or i is not n, then
      • insert a (list from i0 to i-1) at the end of ans
• return ans

Example

Let us see the following implementation to get better understanding −

def solve(nums):
   n = len(nums)
   ans = []
   i = 0
   while i < n:
      i0 = i
      while i < n and nums[i] == nums[i0]:
         i += 1
      if (i0 == 0 or nums[i0] > nums[i0 - 1]) and (i == n or nums[i0] > nums[i]):
         if i0 != 0 or i != n:
            ans.extend(range(i0, i))
   return ans

nums = [5, 8, 8, 8, 6, 11, 11]
print(solve(nums))

Input

[5, 8, 8, 8, 6, 11, 11]

Output

[1, 2, 3, 5, 6]