What is the time complexity of Largest Positive Integer That Exists With Its Negative?

The optimal solution for Largest Positive Integer That Exists With Its Negative runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the array a couple of times. The space complexity is O(n) due to the HashSet storing all elements.

What is the brute force solution for Largest Positive Integer That Exists With Its Negative?

The brute force approach for Largest Positive Integer That Exists With Its Negative is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every positive integer in the array and looks for its negative counterpart. This is straightforward but inefficient, as it involves comparing each element with every other element. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Largest Positive Integer That Exists With Its Negative?

Largest Positive Integer That Exists With Its Negative uses the following concepts: Array, Hash Table, Two Pointers, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Largest Positive Integer That Exists With Its Negative?

Always consider the data structures that can optimize your solution, like HashSets for existence checks. Clarify edge cases with the interviewer, such as arrays with only negative numbers. Explain your thought process clearly as you write code; it shows your understanding.

What are common mistakes when solving Largest Positive Integer That Exists With Its Negative?

Not using a HashSet for efficient lookups, leading to unnecessary nested loops. Overlooking the condition that the integer must be positive.

#2441

Largest Positive Integer That Exists With Its Negative

Easy

ArrayHash TableTwo PointersSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using a HashSet allows us to efficiently check for the existence of negative counterparts in constant time. This significantly reduces the number of comparisons needed.

⚙️

Algorithm

5 steps

1Step 1: Create a HashSet to store all elements from the array.
2Step 2: Initialize a variable 'max_k' to -1.
3Step 3: Loop through each element 'num' in the array.
4Step 4: If 'num' is positive and '-num' exists in the HashSet, update 'max_k' to the maximum of 'max_k' and 'num'.
5Step 5: After checking all elements, return 'max_k'. If 'max_k' is still -1, return -1.

solution.py9 lines

1# Full working Python code
2
3def largest_positive_integer(nums):
4    num_set = set(nums)
5    max_k = -1
6    for num in nums:
7        if num > 0 and -num in num_set:
8            max_k = max(max_k, num)
9    return max_k

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a couple of times. The space complexity is O(n) due to the HashSet storing all elements.

1Using a HashSet allows for O(1) average time complexity for lookups, making the solution efficient.
2The largest positive integer is always the focus, so we only need to track positive numbers.

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