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Largest Number At Least Twice of Others

Easy

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Topics:

Arrays

You are given an integer array nums where the largest integer is unique.

Determine whether the largest element in the array is at least twice as much as every other number in the array. If it is, return the index of the largest element, or return -1 otherwise.

Example 1:

Input: nums = [3,6,1,0]
Output: 1
Explanation: 6 is the largest integer.
For every other number in the array x, 6 is at least twice as big as x.
The index of value 6 is 1, so we return 1.

Example 2:

Input: nums = [1,2,3,4]
Output: -1
Explanation: 4 is less than twice the value of 3, so we return -1.

Constraints:

2 <= nums.length <= 50
0 <= nums[i] <= 100
The largest element in nums is unique.

Largest Number At Least Twice of Others

Easy

Asked by:

4 views

Topics:

Arrays

You are given an integer array nums where the largest integer is unique.

Example 1:

Input: nums = [3,6,1,0]
Output: 1
Explanation: 6 is the largest integer.
For every other number in the array x, 6 is at least twice as big as x.
The index of value 6 is 1, so we return 1.

Example 2:

Input: nums = [1,2,3,4]
Output: -1
Explanation: 4 is less than twice the value of 3, so we return -1.

Constraints:

2 <= nums.length <= 50
0 <= nums[i] <= 100
The largest element in nums is unique.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input array contain negative numbers?
What should I return if no such number exists that is at least twice as large as every other number in the array?
Is it possible for the input array to be empty?
If there are multiple numbers that satisfy the condition, should I return the index of the first one encountered or is there a specific criteria for choosing among them?
Can the input array contain zero values?

Brute Force Solution

Approach

The brute force approach for this problem means checking every single number to see if it's the largest and at least twice as big as all the other numbers. We're just going to methodically compare each number against all the rest.

Here's how the algorithm would work step-by-step:

First, pick a number from the list.
Then, compare that number to every other number in the list.
For each comparison, check if the picked number is bigger than the other number, and also at least twice as big.
If the picked number fails to be at least twice as big as even one other number, it's not the answer. Move on to the next number in the original list.
If the picked number passes the test against every other number, we have found our answer and we can stop.

Code Implementation

def dominant_index(numbers):
    for potential_largest_index in range(len(numbers)):
        is_twice_as_large = True

        # Check if the current number is at least twice as large as all others
        for comparison_index in range(len(numbers)):

            #Skip comparison with itself
            if potential_largest_index == comparison_index:
                continue

            #Check if the picked number is at least twice as large
            if numbers[potential_largest_index] < 2 * numbers[comparison_index]:
                is_twice_as_large = False
                break

        # If it's at least twice as large, return the index
        if is_twice_as_large:
            return potential_largest_index

    #No such element exists
    return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input array of size n, selecting one number at a time. For each selected number, it compares it against all the remaining (n-1) numbers in the array. Therefore, in the worst-case scenario, every element could be compared against every other element in the array. This leads to roughly n * (n-1) comparisons. Simplifying this expression results in O(n²).

Space Complexity

O(1) – The provided brute force algorithm iterates through the input array but does not create any additional data structures that scale with the size of the input array, which we can denote as N. It only uses a few variables for comparisons and potentially an index variable to store the result. The number of variables used remains constant irrespective of the input size, therefore the algorithm has a constant auxiliary space complexity.

Optimal Solution

Approach

To find the largest number that's at least twice as big as all the others, we need to identify the largest number and then check if it meets the criteria. This avoids unnecessary comparisons and focuses directly on the most likely candidate.

Here's how the algorithm would work step-by-step:

First, go through the list and find the largest number.
At the same time, also keep track of the second largest number.
Once you have both the largest and the second largest numbers, check if the largest number is at least twice as big as the second largest number.
If it is, then the largest number is the answer.
If it's not, then there is no number in the list that meets the requirement.

Code Implementation

def largest_number_at_least_twice_of_others(numbers):
    largest_number = -1
    second_largest_number = -1
    largest_number_index = -1

    for index, number in enumerate(numbers):
        if number > largest_number:
            second_largest_number = largest_number
            largest_number = number
            largest_number_index = index
        elif number > second_largest_number:
            second_largest_number = number

    # Need to compare largest with the second largest number.
    if largest_number >= 2 * second_largest_number:
        
        return largest_number_index

    # If condition not met, no such element exists.
    else:
        return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array of size n once to find both the largest and second largest numbers. After this single pass, a constant-time comparison is performed to check if the largest number is at least twice the second largest. Therefore, the dominant operation is the initial linear scan of the input array. The time complexity is thus directly proportional to the input size n, resulting in O(n).

Space Complexity

O(1) – The provided solution stores the largest and second largest numbers. This requires storing at most two variables to hold these numbers, regardless of the input array's size, N. Therefore, the auxiliary space used is constant and independent of the input size. No other data structures are used that scale with the input. Thus, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return -1 immediately as there's no largest element to evaluate.
Array with only one element	Return 0 as the sole element satisfies the condition trivially being twice of nothing.
Array with two elements where the first element is twice the second	The problem asks for the index of the largest element so we check if nums[0] >= 2 * nums[1] and return 0.
Array contains negative numbers	The description did not mention that numbers can only be non-negative so the problem should handle negative values correctly and compare absolute values if the question states that the largest should be the largest absolute value.
Array contains zero	Zero should be handled correctly as any number greater than or equal to zero is twice zero.
Array contains large numbers close to the maximum integer value	Potential integer overflow issues when multiplying numbers; consider using a larger data type (e.g., long) or comparing ratios to avoid explicit multiplication if the largest can trigger an overflow.
No element in the array is at least twice any other element	Return -1 to indicate that no element satisfies the condition if after checking every element, no solution is found.
Input array size is at the maximum allowed limit (e.g., Integer.MAX_VALUE) which may cause resource exhaustion.	The solution should be designed to avoid excessive memory allocation or computationally expensive operations that could lead to timeouts.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input array contain negative numbers?
What should I return if no such number exists that is at least twice as large as every other number in the array?
Is it possible for the input array to be empty?
If there are multiple numbers that satisfy the condition, should I return the index of the first one encountered or is there a specific criteria for choosing among them?
Can the input array contain zero values?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, pick a number from the list.
Then, compare that number to every other number in the list.
For each comparison, check if the picked number is bigger than the other number, and also at least twice as big.
If the picked number fails to be at least twice as big as even one other number, it's not the answer. Move on to the next number in the original list.
If the picked number passes the test against every other number, we have found our answer and we can stop.

Code Implementation

def dominant_index(numbers):
    for potential_largest_index in range(len(numbers)):
        is_twice_as_large = True

        # Check if the current number is at least twice as large as all others
        for comparison_index in range(len(numbers)):

            #Skip comparison with itself
            if potential_largest_index == comparison_index:
                continue

            #Check if the picked number is at least twice as large
            if numbers[potential_largest_index] < 2 * numbers[comparison_index]:
                is_twice_as_large = False
                break

        # If it's at least twice as large, return the index
        if is_twice_as_large:
            return potential_largest_index

    #No such element exists
    return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, go through the list and find the largest number.
At the same time, also keep track of the second largest number.
Once you have both the largest and the second largest numbers, check if the largest number is at least twice as big as the second largest number.
If it is, then the largest number is the answer.
If it's not, then there is no number in the list that meets the requirement.

Code Implementation

def largest_number_at_least_twice_of_others(numbers):
    largest_number = -1
    second_largest_number = -1
    largest_number_index = -1

    for index, number in enumerate(numbers):
        if number > largest_number:
            second_largest_number = largest_number
            largest_number = number
            largest_number_index = index
        elif number > second_largest_number:
            second_largest_number = number

    # Need to compare largest with the second largest number.
    if largest_number >= 2 * second_largest_number:
        
        return largest_number_index

    # If condition not met, no such element exists.
    else:
        return -1

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return -1 immediately as there's no largest element to evaluate.
Array with only one element	Return 0 as the sole element satisfies the condition trivially being twice of nothing.
Array with two elements where the first element is twice the second	The problem asks for the index of the largest element so we check if nums[0] >= 2 * nums[1] and return 0.
Array contains negative numbers	The description did not mention that numbers can only be non-negative so the problem should handle negative values correctly and compare absolute values if the question states that the largest should be the largest absolute value.
Array contains zero	Zero should be handled correctly as any number greater than or equal to zero is twice zero.
Array contains large numbers close to the maximum integer value	Potential integer overflow issues when multiplying numbers; consider using a larger data type (e.g., long) or comparing ratios to avoid explicit multiplication if the largest can trigger an overflow.
No element in the array is at least twice any other element	Return -1 to indicate that no element satisfies the condition if after checking every element, no solution is found.
Input array size is at the maximum allowed limit (e.g., Integer.MAX_VALUE) which may cause resource exhaustion.	The solution should be designed to avoid excessive memory allocation or computationally expensive operations that could lead to timeouts.