How do you solve Find Closest Person on LeetCode?

Find Closest Person (LeetCode #3516) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Distance comparison is the key to solving the problem.

What is the time complexity of Find Closest Person?

The optimal solution for Find Closest Person runs in O(1) time and uses O(1) space. Both time and space complexities are constant since we only perform a fixed number of calculations.

What is the brute force solution for Find Closest Person?

The brute force approach for Find Closest Person is Brute Force, with O(1) time complexity and O(1) space complexity. Calculate the distance of both Person 1 and Person 2 from Person 3 and compare. The one with the shorter distance reaches first. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Find Closest Person?

Find Closest Person uses the following concepts: Math. The recommended patterns to study are: Math, Absolute Difference.

What are the interview tips for Find Closest Person?

Clarify the problem requirements before coding. Think through edge cases, like equal distances. Explain your thought process as you code.

What are common mistakes when solving Find Closest Person?

Confusing the distance calculations. Not considering the case where both distances are equal.

#3516

Find Closest Person

Easy

MathMathAbsolute Difference

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(1)Space O(1)

Directly compute distances and compare them in constant time, ensuring efficiency.

⚙️

Algorithm

3 steps

1Step 1: Calculate distance from Person 1 to Person 3: |x - z|.
2Step 2: Calculate distance from Person 2 to Person 3: |y - z|.
3Step 3: Compare the distances and return the result based on who is closer.

solution.py9 lines

1def find_closest_person(x, y, z):
2    dist1 = abs(x - z)
3    dist2 = abs(y - z)
4    if dist1 < dist2:
5        return 1
6    elif dist2 < dist1:
7        return 2
8    else:
9        return 0

ℹ

Complexity note: Both time and space complexities are constant since we only perform a fixed number of calculations.

1Distance comparison is the key to solving the problem.
2Both persons move at the same speed, simplifying the logic.

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