Lower Bound in O(log n) Time Complexity using Python

Given a sorted array of integers nums, find the smallest index where we can place a given value such that the array remains sorted

Example 1:

Input: nums = [1, 2, 3, 5, 7], value = 4
Output: 3
Explanation: Placing the value 4 on the 4^th index we obtain nums = [1, 2, 3, 4, 5, 7]

Example 2:

Input: nums = [1, 2, 3], value = 2
Output: 1
Explanation: Placing the value 2 on the 1^st index we obtain nums = [1, 2, 2, 3]

Note:

Your algorithm should run in O(log n) time and use O(1) extra space.

Understanding the Problem

The core challenge of this problem is to find the correct position to insert a given value into a sorted array such that the array remains sorted. This is a common problem in computer science, often referred to as finding the "lower bound" or "insertion point".

Common applications include binary search algorithms, insertion operations in sorted data structures, and more.

Potential pitfalls include misunderstanding the requirement to maintain the sorted order and not optimizing the solution to run in O(log n) time.

Approach

To solve this problem, we can use a binary search algorithm. Binary search is ideal here because it allows us to find the insertion point in O(log n) time, which meets the problem's constraints.

Here's a step-by-step approach:

Initialize two pointers, left and right, to the start and end of the array, respectively.
While left is less than or equal to right, calculate the middle index mid.
If the value at mid is less than the target value, move the left pointer to mid + 1.
Otherwise, move the right pointer to mid - 1.
When the loop ends, left will be the smallest index where the target value can be inserted.

Algorithm

Let's break down the algorithm step-by-step:

Initialize left to 0 and right to the length of the array minus one.
Enter a while loop that continues as long as left is less than or equal to right.
Calculate the middle index mid as (left + right) // 2.
If nums[mid] is less than the target value, set left to mid + 1.
Otherwise, set right to mid - 1.
When the loop exits, left will be the correct insertion point.

Code Implementation

def lower_bound(nums, value):
    # Initialize left and right pointers
    left, right = 0, len(nums) - 1
    
    # Perform binary search
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] < value:
            left = mid + 1
        else:
            right = mid - 1
    
    # left is the smallest index where value can be inserted
    return left

# Example usage
nums1 = [1, 2, 3, 5, 7]
value1 = 4
print(lower_bound(nums1, value1))  # Output: 3

nums2 = [1, 2, 3]
value2 = 2
print(lower_bound(nums2, value2))  # Output: 1

Complexity Analysis

The time complexity of the binary search algorithm is O(log n) because we halve the search space with each iteration. The space complexity is O(1) because we only use a constant amount of extra space for the pointers and the middle index.

Edge Cases

Consider the following edge cases:

Inserting a value smaller than all elements in the array.
Inserting a value larger than all elements in the array.
Inserting a value that is already present in the array.

Our algorithm handles these cases effectively by adjusting the left and right pointers accordingly.

Testing

To test the solution comprehensively, consider the following test cases:

Empty array: nums = [], value = 1
Single element array: nums = [2], value = 1
Value smaller than all elements: nums = [2, 3, 4], value = 1
Value larger than all elements: nums = [2, 3, 4], value = 5
Value already present: nums = [2, 3, 4], value = 3

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Understand the problem requirements and constraints thoroughly.
Think about the properties of the data structure (e.g., sorted array) and how they can be leveraged.
Start with a naive solution and then optimize it.
Practice similar problems to improve your problem-solving skills.

Conclusion

In this blog post, we discussed how to find the smallest index to insert a value into a sorted array using a binary search algorithm. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing efficient algorithms and improving problem-solving skills.

Additional Resources

For further reading and practice, consider the following resources: