# 245. Shortest Word Distance III

## LeetCode [245. Shortest Word Distance III](https://leetcode-cn.com/problems/shortest-word-distance-iii/)

### Description

Given an array of strings `wordsDict` and two different strings that already exist in the array `word1` and `word2`, return *the shortest distance between these two words in the list*.

**Note** that `word1` and `word2` may be the same. It is guaranteed that they represent **two individual words** in the list.

**Example 1:**

```
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "coding", word2 = "practice"
Output: 3
```

**Example 2:**

```
Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
Output: 1
```

**Constraints:**

* `1 <= wordsDict.length <= 3 * 104`
* `1 <= wordsDict[i].length <= 10`
* `wordsDict[i]` consists of lowercase English letters.
* `word1` and `word2` are in `wordsDict`.

### Tags

Array

### Solution

Based on the answer of LeetCode [243. Shortest Word Distance](https://txfs19260817.gitbook.io/leetcode-go-notes/solutions/243.-shortest-word-distance), we add a statement for the case when `word1` and `word2` are identical.

```go
if word1 == word2 && p1 != -1 && i-p1 < ans {
    ans = i - p1
}
```

### Complexity

* Time complexity: $$O(n)$$
* Space complexity: $$O(1)$$

### Code

```go
func shortestWordDistance(wordsDict []string, word1 string, word2 string) int {
	ans, p1, p2 := 1<<31, -1, -1
	for i, w := range wordsDict {
		if w == word1 {
			if word1 == word2 && p1 != -1 && i-p1 < ans {
				ans = i - p1
			}
			p1 = i
		} else if w == word2 {
			p2 = i
		}
		if p1 != -1 && p2 != -1 {
			if n := abs(p1 - p2); n < ans {
				ans = n
			}
		}
	}
	return ans
}

func abs(a int) int {
	if a < 0 {
		return -1 * a
	}
	return a
}
```