How do you solve Largest Number At Least Twice of Others on LeetCode?

Largest Number At Least Twice of Others (LeetCode #747) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The largest element is guaranteed to be unique, simplifying the problem.

What is the time complexity of Largest Number At Least Twice of Others?

The optimal solution for Largest Number At Least Twice of Others runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only loop through the array once. The space complexity is O(1) since we use a constant amount of extra space.

What is the brute force solution for Largest Number At Least Twice of Others?

The brute force approach for Largest Number At Least Twice of Others is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves scanning through the array to find the largest element and then checking if it is at least twice as large as every other element. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Largest Number At Least Twice of Others?

Largest Number At Least Twice of Others uses the following concepts: Array, Sorting. The recommended patterns to study are: Array, Sorting.

What are the interview tips for Largest Number At Least Twice of Others?

Always clarify the constraints and assumptions before starting. Think about edge cases, such as arrays with only two elements. Practice explaining your thought process as you code.

What are common mistakes when solving Largest Number At Least Twice of Others?

Not checking all elements correctly against the maximum value. Assuming the largest number is always at the end of the array.

#747

Largest Number At Least Twice of Others

Easy

ArraySortingArraySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution improves efficiency by combining the two scans into one. We find the maximum and check the condition in a single pass, reducing the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables to track the maximum value, its index, and a flag for the condition check.
2Step 2: Loop through the array once to find the maximum value and check if it is at least twice as large as every other number.
3Step 3: If the condition fails for any number, return -1. If it passes for all, return the index of the maximum value.

solution.py14 lines

1# Full working Python code
2nums = [3, 6, 1, 0]
3max_value = nums[0]
4max_index = 0
5for i in range(1, len(nums)):
6    if nums[i] > max_value:
7        max_value = nums[i]
8        max_index = i
9for num in nums:
10    if num != max_value and max_value < 2 * num:
11        print(-1)
12        break
13else:
14    print(max_index)

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array once. The space complexity is O(1) since we use a constant amount of extra space.

1The largest element is guaranteed to be unique, simplifying the problem.
2We can combine finding the maximum and checking the condition in one pass for efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.