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Program to find local peak element indices from a list of numbers in Python
Suppose we have a list of numbers called nums whose length is at least 2. We have to find the index of every peak in the list. The list is sorted in ascending order. An index i is a peak when −
nums[i] > nums[i + 1] when i = 0
nums[i] > nums[i - 1] when i = n - 1
nums[i - 1] < nums[i] > nums[i + 1] else
So, if the input is like nums = [5, 6, 7, 6, 9], then the output will be [2, 4], as the element at index 2 is 7 which is larger than two neighbors, and item at index 4 is 9, this is larger than its left item.
To solve this, we will follow these steps −
ans := a new list
n := size of nums
if n is same as 1, then
return ans
for each index i and number num in nums, do
if i > 0 and i < n - 1, then
if nums[i - 1] < num > nums[i + 1], then
insert i at the end of ans
if i is same as 0, then
if num > nums[i + 1], then
insert i at the end of ans
if i is same as n - 1, then
if num > nums[i - 1], then
insert i at the end of ans
return ans
Example
Let us see the following implementation to get better understanding
def solve(nums): ans = [] n = len(nums) if n == 1: return ans for i, num in enumerate(nums): if i > 0 and i < n - 1: if nums[i - 1] < num > nums[i + 1]: ans.append(i) if i == 0: if num > nums[i + 1]: ans.append(i) if i == n - 1: if num > nums[i - 1]: ans.append(i) return ans nums = [5, 6, 7, 6, 9] print(solve(nums))
Input
[5, 6, 7, 6, 9]
Output
[2, 4]
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