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.word1 and word2 are in wordsDict.word1 != word2
Solutions
Solution 1
PythonJavaC++Go
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
