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# Find The Minimum And Maximum Number Of Nodes Between Critical Points - Leetcode Solution

## Companies:

LeetCode: Find The Minimum And Maximum Number Of Nodes Between Critical Points Leetcode Solution

Difficulty: Medium

Topics: linked-list

Problem Statement:

Given a function f(x) defined on the real number line, we define a critical point as any point on the real number line where either:

- f'(x) = 0 (derivative of f(x) at x is equal to 0), or
- f(x) does not exist

Find the minimum and maximum number of nodes (i.e., local minima and maxima) between any two adjacent critical points.

Solution:

To solve this problem, we need to find all the critical points of the given function first. Then we need to find the minimum and maximum number of nodes (i.e., local minima and maxima) between any two adjacent critical points.

Step 1: Find all the critical points

To find all the critical points, we need to solve the equation f'(x) = 0 and also check if f(x) exists at any point. We can achieve this by using the following algorithm:

- Find all the points where f'(x) = 0.
- Find all the points where f(x) does not exist (i.e., where the denominator becomes zero in case of a fraction).
- Combine all the above points to get the set of all critical points.

Step 2: Find the minimum and maximum number of nodes between any two adjacent critical points

Once we have all the critical points, we can find the minimum and maximum number of nodes between any two adjacent critical points by using the following algorithm:

- Find the first critical point.
- Find the next critical point and count the number of nodes between them.
- Repeat the above step until we reach the last critical point.
- Keep track of the minimum and maximum number of nodes during the above process.

We need to take care of the following special cases:

- If there is only one critical point, then the minimum and maximum number of nodes both are 0.
- If there are no critical points, then the minimum and maximum number of nodes both are 1 (since there is no critical point to divide the function into separate regions).

Time Complexity:

Finding all the critical points requires calculating the derivative of the function, which takes O(n) time where n is the number of points on the real number line that we need to consider. Finding the minimum and maximum number of nodes requires going through all the critical points once, so it takes O(k) time where k is the number of critical points. Therefore, the overall time complexity of the algorithm is O(n + k).

Space Complexity:

The space required by the algorithm is O(k) to store the set of critical points.

Code:

Below is the Python code to solve this problem:

```
def find_critical_points(f):
critical_points = set()
for x in range(-1000, 1001):
if f(x-1) < f(x) and f(x) > f(x+1):
critical_points.add(x)
elif f(x) == None:
critical_points.add(x)
return critical_points
def find_min_and_max_nodes(f):
critical_points = find_critical_points(f)
if len(critical_points) == 1:
return 0, 0
elif len(critical_points) == 0:
return 1, 1
else:
nodes = []
prev_point = min(critical_points)
for point in sorted(critical_points)[1:]:
nodes.append(len(find_critical_points(lambda x: f(prev_point) + (f(point) - f(prev_point)) / (point - prev_point) * (x - prev_point))))
prev_point = point
return min(nodes), max(nodes)
```

The first function `find_critical_points()`

finds all the critical points of the given function f(x). It uses the brute-force approach of checking every point on the real number line to see if it satisfies the conditions of being a critical point.

The second function `find_min_and_max_nodes()`

finds the minimum and maximum number of nodes between any two adjacent critical points. It first checks for special cases when there is only one or zero critical points. Then it goes through all the critical points and finds the number of nodes between each pair of adjacent critical points. Finally, it returns the minimum and maximum number of nodes.

## Find The Minimum And Maximum Number Of Nodes Between Critical Points Solution Code

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