243. Shortest Word Distance - Detailed Explanation
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
andword2
are guaranteed to appear inwords
.
Hints
- Brute force checking every pair of positions is O(n²) and too slow for large n.
- You only need to remember the last seen index of each target word as you scan.
- Update the minimum distance whenever you encounter either target and you have seen the other at least once.
Approach
One‑Pass with Two Pointers
- Initialize
idx1 = -1
,idx2 = -1
, andminDist = +∞
. - Loop
i
from0
towords.length-1
:- If
words[i] == word1
, setidx1 = i
. - Else if
words[i] == word2
, setidx2 = i
. - If both
idx1
andidx2
are ≥ 0, updateminDist = min(minDist, |idx1 - idx2|)
- If
- 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.
Related Problems
TAGS
leetcode
CONTRIBUTOR
Design Gurus Team
-
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