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Find the element before which all the elements are smaller than it, and after which all are greater in Python
Suppose we have an array, we have to find an element before which all elements are less than it, and after which all are greater than it. Finally, return the index of the element, if there is no such element, then return -1.
So, if the input is like A - [6, 2, 5, 4, 7, 9, 11, 8, 10], then the output will be 4.
To solve this, we will follow these steps −
n := size of arr
maximum_left := an array of size n
maximum_left[0] := -infinity
for i in range 1 to n, do
maximum_left[i] := maximum of maximum_left[i-1], arr[i-1]
minimum_right := infinity
for i in range n-1 to -1, decrease by 1, do
if maximum_left[i] < arr[i] and minimum_right > arr[i], then
return i
minimum_right := minimum of minimum_right, arr[i]
return -1
Example
Let us see the following implementation to get better understanding −
def get_element(arr):
n = len(arr)
maximum_left = [None] * n
maximum_left[0] = float('-inf')
for i in range(1, n):
maximum_left[i] = max(maximum_left[i-1], arr[i-1])
minimum_right = float('inf')
for i in range(n-1, -1, -1):
if maximum_left[i] < arr[i] and minimum_right > arr[i]:
return i
minimum_right = min(minimum_right, arr[i])
return -1
arr = [6, 2, 5, 4, 7, 9, 11, 8, 10]
print(get_element(arr))Input
[6, 2, 5, 4, 7, 9, 11, 8, 10]
Output
4
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