LeetCode 613: Shortest Distance in a Line Solution in Python – A Step-by-Step Guide

What Is LeetCode 613: Shortest Distance in a Line?

In LeetCode 613: Shortest Distance in a Line, you’re given a table point with a single column x, each row representing a point’s coordinate on a 1D line. Your task is to find the shortest distance between any two distinct points, where the distance is simply the absolute difference |x1 - x2|. For example, points at 1 and 4 have a distance of 3. This problem tests your ability to compute distances efficiently, typically posed as an SQL query with a self-join, but we’ll solve it in Python for a coding twist.

Problem Statement

• Input: A list of records (or table rows) with [x].
• Output: An integer—the shortest distance between any two distinct points.
• Rules:
• Distance: |x1 - x2|.
• Points are distinct (no self-distance).
• At least two points exist.

Constraints

• Table has at least 2 rows.
• -10⁴ ≤ x ≤ 10⁴.
• Number of rows ≤ 10⁵.

Examples

• Input:
``` x 1 4 ```
• Output: 3 (Distance between 1 and 4)
• Input:
``` x -1 0 2 ```
• Output: 1 (Distance between -1 and 0 or 0 and 2)
• Input:
``` x 5 5 ```
• Output: 0 (If duplicates allowed, but typically distinct in context)

These examples set the stage—let’s solve it!

Understanding the Problem: Measuring Gaps on a Line

To solve LeetCode 613: Shortest Distance in a Line in Python, we need to find the minimum absolute difference between any two distinct x-coordinates in a list. A naive approach might compare every pair, but we can optimize since points are on a 1D line. We’ll use:

• Best Solution (Sorting with Adjacent Differences): O(n log n) time, O(1) space—fast and efficient.
• Alternative Solution (Brute-Force Pairwise): O(n²) time, O(1) space—simple but slow.

Let’s start with the sorting solution, breaking it down for beginners.

Best Solution: Sorting with Adjacent Differences

Why This Is the Best Solution

The sorting with adjacent differences approach is the top choice for LeetCode 613 because it’s efficient—O(n log n) time with O(1) space—and leverages a key insight: on a 1D line, the shortest distance must occur between two adjacent points after sorting. It avoids checking all pairs, making it scalable up to 10⁵ points. It’s like lining up points and measuring gaps next door!

How It Works

Think of this as organizing points on a number line:

• Step 1: Extract and Sort:
• Pull x-values into a list and sort them.
• Step 2: Check Adjacent Pairs:
• Iterate through sorted list, compute differences between neighbors.
• Step 3: Track Minimum:
• Keep the smallest difference found.
• Step 4: Return Result:
• Output the minimum distance.

It’s like a ruler sliding along sorted points—check the gaps!

Step-by-Step Example

Example:

Copy codex
1
4

• Sort: [1, 4].
• Differences:
• 4 - 1 = 3.
• Min Distance: 3.
• Output: 3.

Example:

Copy codex
-1
0
2

• Sort: [-1, 0, 2].
• Differences:
• 0 - (-1) = 1.
• 2 - 0 = 2.
• Min Distance: 1 (between -1 and 0).
• Output: 1.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def shortestDistance(self, points: List[List[int]]) -> int:
        # Step 1: Extract x-values and sort
        x_coords = [point[0] for point in points]
        x_coords.sort()
        n = len(x_coords)

        # Step 2: Handle edge case
        if n < 2:
            return 0  # Shouldn’t happen per constraints

        # Step 3: Find minimum difference
        min_dist = float('inf')
        for i in range(1, n):
            dist = x_coords[i] - x_coords[i-1]
            min_dist = min(min_dist, dist)

        # Step 4: Return result
        return min_dist

• Line 4: Extract x-values into a list and sort.
• Lines 7-9: Edge case (not typically needed due to constraints).
• Lines 12-14: Compute differences:
• Subtract adjacent elements, update min_dist.
• Time Complexity: O(n log n)—dominated by sorting.
• Space Complexity: O(1)—excluding input list.

This is like a gap measurer—fast and neat!

Alternative Solution: Brute-Force Pairwise Comparison

Why an Alternative Approach?

The brute-force pairwise approach compares every pair of points—O(n²) time, O(1) space. It’s simpler to understand and implement, requiring no sorting, but it’s less efficient for large inputs. It’s like measuring every possible gap with a tape measure!

How It Works

Picture this as a point-by-point check:

• Step 1: Extract x-values.
• Step 2: Compare all pairs:
• Use nested loops to calculate |x1 - x2|.
• Step 3: Track minimum distance.
• Step 4: Return the smallest gap.

It’s like a thorough distance sweep—check everything!

Step-by-Step Example

Example:

Copy codex
-1
0
2

• Pairs:
• |-1 - 0| = 1.
• |-1 - 2| = 3.
• |0 - 2| = 2.
• Min Distance: 1 (between -1 and 0).
• Output: 1.

Code for Brute-Force Approach

Copy codeclass Solution:
    def shortestDistance(self, points: List[List[int]]) -> int:
        # Step 1: Extract x-values
        x_coords = [point[0] for point in points]
        n = len(x_coords)

        # Step 2: Handle edge case
        if n < 2:
            return 0

        # Step 3: Compare all pairs
        min_dist = float('inf')
        for i in range(n):
            for j in range(i + 1, n):
                dist = abs(x_coords[i] - x_coords[j])
                min_dist = min(min_dist, dist)

        # Step 4: Return result
        return min_dist

• Line 4: Extract x-values.
• Lines 12-15: Nested loops:
• Compute absolute difference, update min_dist.
• Time Complexity: O(n²)—all pairs checked.
• Space Complexity: O(1)—minimal space.

It’s a full distance scan—simple but slow!

Comparing the Two Solutions

• Sorting (Best):
• Pros: O(n log n) time, O(1) space, efficient for large n.
• Cons: Requires sorting intuition.
• Brute-Force (Alternative):
• Pros: O(n²) time, O(1) space, easy to grasp.
• Cons: Slow for large inputs.

Sorting wins for efficiency.

Additional Examples and Edge Cases

• Input: [[0], [1]]
• Output: 1.
• Input: [[2], [2]]
• Output: 0 (if duplicates allowed, but typically distinct).

Sorting handles better with optimization.

Complexity Breakdown

• Sorting: Time O(n log n), Space O(1).
• Brute-Force: Time O(n²), Space O(1).

Sorting is optimal.

Key Takeaways

• Sorting: Adjacent gaps—smart!
• Brute-Force: Full check—clear!
• 1D: Lines are fun.
• Python Tip: Sorting rocks—see Python Basics.

Final Thoughts: Measure That Gap

LeetCode 613: Shortest Distance in a Line in Python is a neat 1D distance challenge. Sorting with adjacent differences offers speed and elegance, while brute-force keeps it simple. Want more? Try LeetCode 414: Third Maximum Number or LeetCode 628: Maximum Product of Three Numbers. Ready to survey? Head to Solve LeetCode 613 on LeetCode and find that shortest distance today!