**π‘ Problem Formulation:** Given an array of numbers, the goal is to determine if there exists a number that is at least twice as large as every other number in the array. If such a number exists, the function should return its index, otherwise return -1. For instance, given an input of `[3, 6, 1]`

, the output should be `1`

as the number ‘6’ is two times larger than the rest of the numbers.

## Method 1: Using a Loop to Compare Elements

In this method, we iterate through the list of numbers to find the largest number, then iterate again to ensure it is at least twice as large as all other numbers in the list. This is a brute force approach, and is straightforward to implement.

Here’s an example:

def find_largest_twice(nums): max_num = max(nums) max_index = nums.index(max_num) for num in nums: if max_num < 2 * num and max_num != num: return -1 return max_index print(find_largest_twice([3, 6, 1]))

Output:

1

This code snippet finds the maximum number in the array, then checks whether this number is at least twice as large as all the other numbers. If any number violates this condition, `-1`

is returned; otherwise, the index of the maximum number is returned.

## Method 2: Sorting and Comparing the Two Largest Numbers

We can sort the array, which allows us to directly compare the two largest numbers once, rather than comparing the largest number with all other numbers in the array. This can be more efficient for large arrays.

Here’s an example:

def find_largest_twice(nums): nums_sorted = sorted(nums, reverse=True) if nums_sorted[0] >= 2 * nums_sorted[1]: return nums.index(nums_sorted[0]) return -1 print(find_largest_twice([3, 6, 1]))

Output:

1

This code snippet sorts the array in reverse order and then compares the largest number with the second largest number. If the condition is met, it returns the index of the largest number from the original array; otherwise, it returns `-1`

.

## Method 3: Optimized Linear Pass

This method involves making a single pass through the list to find both the largest number and the second-largest number, then performing the comparison. This method combines efficiency and simplicity.

Here’s an example:

def find_largest_twice(nums): m = second = idx = 0 for i, num in enumerate(nums): if num > m: second = m m, idx = num, i elif num > second: second = num return idx if m >= 2 * second else -1 print(find_largest_twice([3, 6, 1]))

Output:

1

The code performs a single pass through the list, maintaining the largest and second-largest numbers. The index of the largest number is recorded, and the comparison to check if it’s at least twice as large as the second-largest number is made at the end.

## Method 4: Using List Comprehension and Max Function

This method makes use of Python’s list comprehension feature and the `max`

function to find the largest number that is at least twice as large as the others in a concise manner.

Here’s an example:

def find_largest_twice(nums): max_num = max(nums) if all(max_num >= 2*x for x in nums if x != max_num): return nums.index(max_num) return -1 print(find_largest_twice([3, 6, 1]))

Output:

1

This snippet uses list comprehension to check the condition for all elements except for the maximum one. Then, if the condition is met, the index of the maximum number is returned. If not, `-1`

is returned.

## Bonus One-Liner Method 5: Expressive Python Conditional

A one-liner solution that leverages Python’s max function along with a conditional expression for a concise yet readable implementation.

Here’s an example:

find_largest_twice = lambda nums: nums.index(max(nums)) if max(nums) >= 2*sorted(nums)[-2] else -1 print(find_largest_twice([3, 6, 1]))

Output:

1

This one-liner uses a lambda function to first sort the list, then checks if the largest number is at least twice as large as the second largest using a conditional expression, returning the corresponding index or `-1`

.

## Summary/Discussion

**Method 1:**Brute Force Comparison. Simple to understand and implement. Inefficient for larger arrays.**Method 2:**Sorting and Comparing. More efficient for large arrays. Extra space required for sorting and additional complexity.**Method 3:**Optimized Linear Pass. Best trade-off between complexity and performance. May be slightly harder to understand.**Method 4:**Using List Comprehension. Readable and concise. Potentially less efficient due to the creation of intermediate lists.**Method 5:**Expressive Python One-Liner. Extremely concise. May sacrifice readability for experienced Python programmers.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.