**π‘ Problem Formulation:** You’re given a sorted list `sorted_list`

and a new element `new_element`

. You need to find the index at which to insert `new_element`

into `sorted_list`

so that the list remains sorted. For example, if `sorted_list = [1, 3, 4, 10]`

and `new_element = 2`

, the desired output is index `1`

.

## Method 1: Linear Search

A simple approach is to use a linear search. Starting from the beginning of the list, check each element until you find the position where the new element should be inserted. This method is straightforward but not the most efficient, with a time complexity of O(n).

Here’s an example:

def linear_search_insert(sorted_list, new_element): for i in range(len(sorted_list)): if sorted_list[i] >= new_element: return i return len(sorted_list) # Example usage idx = linear_search_insert([1, 3, 4, 10], 2) print(idx)

**Output:** 1

This code defines a function `linear_search_insert()`

that iterates over each element in the sorted list and returns the index where the new element is less than or equal to the current element. If it traverses the entire list, it returns the length of the list, indicating the new element should be placed at the end.

## Method 2: Bisect Module

Python’s built-in `bisect`

module provides an efficient way to handle insertion points in sorted lists. It uses a binary search algorithm to find the position with time complexity O(log n), making it more suitable for larger lists.

Here’s an example:

import bisect def bisect_insert(sorted_list, new_element): return bisect.bisect_left(sorted_list, new_element) # Example usage idx = bisect_insert([1, 3, 4, 10], 2) print(idx)

**Output:** 1

This code uses the `bisect_left()`

function from the `bisect`

module, which returns the index to insert the `new_element`

while maintaining the list order. It’s more efficient than linear search for larger lists due to its logarithmic time complexity.

## Method 3: List Comprehension with Enumerate

Combining list comprehension and the `enumerate()`

function can also yield the insertion index. This method is concise but, similar to the linear search, carries a linear time complexity for finding the element’s position.

Here’s an example:

sorted_list = [1, 3, 4, 10] new_element = 2 idx = next((i for i, x in enumerate(sorted_list) if x > new_element), len(sorted_list)) print(idx)

**Output:** 1

This one-liner uses a generator expression inside `next()`

with a default return value. It iterates over the list and stops at the element greater than `new_element`

, returning its index. It falls back to the length of the list if `new_element`

is the largest.

## Method 4: Using the numpy Library

For those using the scientific computing library numpy, `numpy.searchsorted()`

offers an efficient and concise way to find the insertion index. It assumes the list is already sorted and implements a binary search.

Here’s an example:

import numpy as np sorted_list = np.array([1, 3, 4, 10]) new_element = 2 idx = np.searchsorted(sorted_list, new_element) print(idx)

**Output:** 1

This snippet demonstrates how `numpy.searchsorted()`

finds the index where `new_element`

should be inserted in the sorted array. The function is robust and fast, utilizing numpy’s optimized C implementation.

## Bonus One-Liner Method 5: Using Built-in Function `filter()`

While less conventional, one can use the `filter()`

function with a lambda to find the insertion index. This method is interesting, though not as efficient as the binary search approaches.

Here’s an example:

sorted_list = [1, 3, 4, 10] new_element = 2 idx = len(list(filter(lambda x: x < new_element, sorted_list))) print(idx)

**Output:** 1

Using `filter()`

with a lambda, this example constructs a list of elements less than `new_element`

and finds the length of that list, which is the insertion index. It’s not the most efficient but is a clever use of built-in functions.

## Summary/Discussion

**Method 1: Linear Search.**Easy to understand. Best for small lists due to O(n) time complexity.**Method 2: Bisect Module.**Utilizes binary search. Most efficient for larger lists with O(log n) time complexity.**Method 3: List Comprehension with Enumerate.**Concise syntax. Has O(n) time complexity, similar to linear search.**Method 4: Using the numpy Library.**Best for people already using numpy in their projects. Very efficient for large, numeric datasets.**Bonus Method 5: Using**Interesting use of built-in functions. More of a novelty than a practical solution.`filter()`

.