243. Shortest Word Distance 🔒
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.
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:
2 <= wordsDict.length <= 3 * 1041 <= wordsDict[i].length <= 10wordsDict[i]consists of lowercase English letters.word1andword2are inwordsDict.word1 != word2
Solutions
Solution 1
Python3
class Solution:
def shortestDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
i = j = -1
ans = inf
for k, w in enumerate(wordsDict):
if w == word1:
i = k
if w == word2:
j = k
if i != -1 and j != -1:
ans = min(ans, abs(i - j))
return ans
Java
class Solution {
public int shortestDistance(String[] wordsDict, String word1, String word2) {
int ans = 0x3f3f3f3f;
for (int k = 0, i = -1, j = -1; k < wordsDict.length; ++k) {
if (wordsDict[k].equals(word1)) {
i = k;
}
if (wordsDict[k].equals(word2)) {
j = k;
}
if (i != -1 && j != -1) {
ans = Math.min(ans, Math.abs(i - j));
}
}
return ans;
}
}
C++
class Solution {
public:
int shortestDistance(vector<string>& wordsDict, string word1, string word2) {
int ans = INT_MAX;
for (int k = 0, i = -1, j = -1; k < wordsDict.size(); ++k) {
if (wordsDict[k] == word1) {
i = k;
}
if (wordsDict[k] == word2) {
j = k;
}
if (i != -1 && j != -1) {
ans = min(ans, abs(i - j));
}
}
return ans;
}
};
Go
func shortestDistance(wordsDict []string, word1 string, word2 string) int {
ans := 0x3f3f3f3f
i, j := -1, -1
for k, w := range wordsDict {
if w == word1 {
i = k
}
if w == word2 {
j = k
}
if i != -1 && j != -1 {
ans = min(ans, abs(i-j))
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}