243. Shortest Word Distance - Detailed Explanation

Free Coding Questions Catalog
Boost your coding skills with our essential coding questions catalog. Take a step towards a better tech career now!

Problem Statement

You’re given an array of strings words and two distinct strings word1 and word2 that both appear in words. Return the shortest distance (in indices) between any occurrence of word1 and any occurrence of word2. Formally, find the minimum value of |i – j| such that words[i] == word1 and words[j] == word2.

Examples

Example 1

Input  
 words = ["practice","makes","perfect","coding","makes"]  
 word1 = "coding", word2 = "practice"  
Output  
 3  
Explanation  
 "coding" is at index 3 and "practice" at index 0 → |3 − 0| = 3  

Example 2

Input  
 words = ["practice","makes","perfect","coding","makes"]  
 word1 = "makes", word2 = "coding"  
Output  
 1  
Explanation  
 Closest "makes" at index 4 and "coding" at index 3 → |4 − 3| = 1  

Constraints

  • 2 ≤ words.length ≤ 3·10⁴
  • 1 ≤ words[i].length ≤ 10
  • word1 != word2
  • Both word1 and word2 are guaranteed to appear in words.

Hints

  1. Brute force checking every pair of positions is O(n²) and too slow for large n.
  2. You only need to remember the last seen index of each target word as you scan.
  3. Update the minimum distance whenever you encounter either target and you have seen the other at least once.

Approach

One‑Pass with Two Pointers

  1. Initialize idx1 = -1, idx2 = -1, and minDist = +∞.
  2. Loop i from 0 to words.length-1:
    • If words[i] == word1, set idx1 = i.
    • Else if words[i] == word2, set idx2 = i.
    • If both idx1 and idx2 are ≥ 0, update
      minDist = min(minDist, |idx1 - idx2|)
      
  3. Return minDist at the end.

This keeps it O(n) time and O(1) extra space.

Step‑by‑Step (Example 2)

words = ["practice","makes","perfect","coding","makes"]
word1="makes", word2="coding"

i=0: "practice" → no change
i=1: "makes"   → idx1=1, idx2=-1 → can't update yet
i=2: "perfect" → no change
i=3: "coding"  → idx2=3, idx1=1 → minDist = |3-1|=2
i=4: "makes"   → idx1=4, idx2=3 → minDist = min(2,|4-3|=1) = 1
Return 1

Complexity Analysis

  • Time: O(n), a single pass through words.
  • Space: O(1), only a few integer variables.

Python Code

Python3
Python3

. . . .

Java Code

Java
Java

. . . .

Common Mistakes

  • Forgetting to check distance right after updating an index.
  • Using two separate loops instead of a single scan (still correct but less concise).
  • Initializing minDist to 0 instead of a large value, causing incorrect early minima.

Edge Cases

  • The first occurrence of one word appears at index 0 and the other at the very end.
  • Multiple interleaved occurrences—ensure you always compare to the most recent other index.
TAGS
leetcode
CONTRIBUTOR
Design Gurus Team
-

GET YOUR FREE

Coding Questions Catalog

Design Gurus Newsletter - Latest from our Blog
Boost your coding skills with our essential coding questions catalog.
Take a step towards a better tech career now!
Explore Answers
Related Courses
Grokking the Coding Interview: Patterns for Coding Questions
Grokking the Coding Interview Patterns in Java, Python, JS, C++, C#, and Go. The most comprehensive course with 476 Lessons.
Grokking Modern AI Fundamentals
Master the fundamentals of AI today to lead the tech revolution of tomorrow.
Grokking Data Structures & Algorithms for Coding Interviews
Unlock Coding Interview Success: Dive Deep into Data Structures and Algorithms.
Image
One-Stop Portal For Tech Interviews.
Copyright © 2025 Design Gurus, LLC. All rights reserved.
;