Find Peak Element

To find a peak element in the given array, we can use a binary search approach to achieve O(log n) time complexity. Here’s the step-by-step explanation:

  1. Initialize Pointers: Start with pointers at the beginning and end of the array, left = 0 and right = len(nums) - 1.

  2. Binary Search Loop: Continue searching as long as left < right.

    a. Find Midpoint: Calculate the middle index as mid = (left + right) // 2.

    b. Compare Mid and Mid+1: If nums[mid] < nums[mid + 1], then a peak must exist to the right, so set left = mid + 1. Otherwise, a peak must exist to the left or at mid, so set right = mid.

  3. Return Peak: Since the loop continues until left equals right, we can return left, which will be the index of a peak element.

Here’s the code implementing this approach:

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class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1

        while left < right:
            mid = (left + right) // 2
            if nums[mid] < nums[mid + 1]:
                left = mid + 1
            else:
                right = mid

        return left

The binary search ensures that the time complexity of this solution is O(log n), where n is the length of the array, and the space complexity is O(1).

Note: This code assumes that the elements in the array are distinct and the imaginary elements outside the array are -∞. It will find a peak element according to the given constraints.

Cases

  • Increasing sequence
  • Decreasing sequence
  • increasing/decreasing
  • decreasing/increasing

Failed to distill the observations into key insights

  • Few lines code code
  • Too much code

T: O(N) S: O(1)

Time Complexity

  • O(N^2)
  • O(n log n)
  • O(N)
  • O(log n)
  • O(1)

[2,1,1,3,5,6,4] => 0 can be answer

Define the Terms Peak: A peak element is an element that is strictly greater than its neighbors.

  • Immediate neighbors are less than the peak element (One preceding number and one succeeding number) (One on the left and one on the right)
  • Two neighbors

Define the Interface Input: array of integers Output: integer (index of the peak element)

Analyze the Input and Output To handle first and last elements use nums[-1] = nums[n] = negative infinity

Identify the Invariants

Identify the constraints

  • 1 <= nums.length <= 1000
  • -231 <= nums[i] <= 231 - 1
  • nums[i] != nums[i + 1] for all valid i.

Search the Problem Space

  • Multiple peaks, we can return any of the peaks

Classify the Problem

  • Similarity with searching for an element in a sorted rotated array.

Analyze the Examples

Solve the Problem by Hand

Describe the Pattern

Distill the Observations to Create Insights

Brute Force Approach

Analyze Time and Space Complexity

Time: O( ) Space: O( )

Outline the Solution

  1. Traverse the array
  2. a[i] > a[i - 1] && a[i] > a[i + 1]
  3. They made this condition even simpler This simpler condition handles all the three cases. Solves the entire problem for all different data set. Compound condition is transformed to simpler condition based on a key insight from observating three cases. ???

Key Takeaways

  • You can look for simpler condition based on the observations

case 1 and case 3

for i in (0…a.size-1) if a[i] > a[i+1] return i end end

case 2

return a.size - 1

recursive binary search

def search(nums, left, right) if left == right return left end mid = left + (right - left)/2

if nums[mid] > nums[mid+1] return search(nums, left, mid) end

return search(nums, mid+1, right)

end

def find_peak_element(a) search(a, 0, a.size-1) end

Peak Index in a Mountain Array

=end

[1,2,3,4]

@param {Integer[]} nums

@return {Integer}

def find_peak_element(a) left = 0 right = a.size - 1

while (left < right)
   mid = left + (right - left) /2
   if a[mid] > a[mid+1]
       right = mid
   else
       left = mid +1
   end
end

at this point left will be equal to right

return left

end

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class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        left =0
        right = len(nums)-1
        while left < right:
            mid = left + (right - left ) //2
            if nums[mid] > nums[mid+1]: 
                right = mid # include mid 
            else:
                left = mid +1
        return left

Problem Classification

Language Agnostic Coding Drills

The Python code is implementing the binary search algorithm to find a peak element in an input list. The breakdown of concepts, arranged in increasing level of difficulty, is as follows:

  1. Variables: Understand how to declare and initialize variables.
  2. Lists (Arrays): Understand how to declare, initialize, and manipulate lists.
  3. Indexing and Accessing List Elements: Learn how to access elements in a list using their indices.
  4. Arithmetic Operators: Understand the usage of arithmetic operators, especially division and multiplication.
  5. Relational Operators: Understand how to compare values using relational operators.
  6. Control Flow - Conditional Statements: Learn to use if-else statements to control the flow of execution based on certain conditions.
  7. Control Flow - Looping Statements: Understand how while loops work to repeatedly execute a block of code until a certain condition is met.
  8. Functions and Method Definitions: Learn to define functions or methods that encapsulate a particular functionality.
  9. Object-Oriented Programming: Understand how to define classes and create instances (objects) of those classes.
  10. Binary Search: Understand how the binary search algorithm works - it’s a way to find an element in a sorted list by repeatedly dividing the search interval in half.

Targeted Drills in Python

  1. Variables:

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    x = 10
y = 20
print(x, y)

  2. Lists (Arrays):

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    my_list = [1, 2, 3, 4, 5]
print(my_list)

  3. Indexing and Accessing List Elements:

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    my_list = [1, 2, 3, 4, 5]
print(my_list[2])

  4. Arithmetic Operators:

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    x = 10
y = 20
print(y // x)  # Integer division
print(y % x)   # Modulo operation

  5. Relational Operators:

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3
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    x = 10
y = 20
print(x > y)  # Greater than
print(x < y)  # Less than

  6. Control Flow - Conditional Statements:

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    x = 10
y = 20
if x > y:
    print("x is greater")
else:
    print("y is greater")

  7. Control Flow - Looping Statements:

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4

    i = 0
while i < 5:
    print(i)
    i += 1

  8. Functions and Method Definitions:

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    def greet(name):
    print(f"Hello, {name}!")

greet("Python")

  9. Object-Oriented Programming:

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    class MyClass:
    def my_method(self):
        print("Hello, Python!")

obj = MyClass()
obj.my_method()

  10. Binary Search:

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    def binary_search(arr, target):
    left = 0
    right = len(arr) - 1
    while left <= right:
        mid = left + (right - left) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

print(binary_search([1, 2, 3, 4, 5, 6, 7, 8, 9], 6))