0245. Shortest Word Distance III
https://leetcode.com/problems/shortest-word-distance-iii
Description
Given an array of strings wordsDict
and two 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 = "makes", word2 = "coding"
**Output:** 1
Example 2:
**Input:** wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "makes"
**Output:** 3
Constraints:
1 <= wordsDict.length <= 105
1 <= wordsDict[i].length <= 10
wordsDict[i]
consists of lowercase English letters.word1
andword2
are inwordsDict
.
ac2
this one is stupid, why you need to have a map? a waste time and space.
class Solution {
public int shortestWordDistance(String[] words, String word1, String word2) {
// edge cases
if (words.length < 2) return 0;
Map<String, List<Integer>> map = new HashMap<String, List<Integer>>();
// walk array, get map
for (int i = 0; i < words.length; i++) {
if (map.containsKey(words[i])) {
map.get(words[i]).add(i);
} else {
List<Integer> tmp = new ArrayList<Integer>();
tmp.add(i);
map.put(words[i], tmp);
}
}
return minDiff(map.get(word1), map.get(word2));
}
private int minDiff(List<Integer> l1, List<Integer> l2) {
int diff = Integer.MAX_VALUE;
// same list, same word
if (l1 == l2) {
for (int i = 1; i < l1.size(); i++) {
diff = Math.min(diff,
Math.abs(l1.get(i) - l1.get(i-1)));
}
return diff;
}
// binary search longer one
List<Integer> s = l1, l = l2;
if (l1.size() > l2.size()) {
l = l1; s = l2;
}
for (int i = 0; i < s.size(); i++) {
int target = s.get(i);
int left = 0, right = l.size()-1;
while (left + 1 < right) {
int mid = left + (right - left) / 2;
if (l.get(mid) > target) right = mid;
else left = mid;
}
diff = Math.min(diff,
Math.min(Math.abs(l.get(left) - target), Math.abs(l.get(right) - target)));
}
return diff;
}
}
ac2
class Solution {
public int shortestWordDistance(String[] words, String word1, String word2) {
// edge cases
if (words.length < 2) return 0;
int prev = -1;
int diff = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1) || words[i].equals(word2)) {
if (prev != -1) {
if (word1.equals(word2)) {
diff = Math.min(diff, i - prev);
} else if(!words[i].equals(words[prev])) {
diff = Math.min(diff, i - prev);
}
}
prev = i;
}
}
return diff;
}
}
ac3
// intuition: 1) time complexity can be optimized? 2) space can be optimized?
class Solution {
public int shortestWordDistance(String[] words, String word1, String word2) {
// edge cases
if (words.length < 2) return 0;
int i1 = -1, i2 = -1;
int diff = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) {
if (word1.equals(word2)) {
i2 = i1;
i1 = i;
} else {
i1 = i;
}
} else if (words[i].equals(word2)) {
i2 = i;
}
if (i1 >= 0 && i2 >= 0) diff = Math.min(diff,
Math.abs(i1-i2));
}
return diff;
}
}
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