Question

Formatted question description: https://leetcode.ca/all/245.html

Given an array of strings wordsDict and two strings that already exist in the array word1 and word2, return the shortest distance between the occurrence of 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 and word2 are in wordsDict.

Algorithm

A condition is added here, that is, two words may be the same.

When word1 and word2 are equal, use p1 to save the result of p2, and p2 is assigned to the current position i, so that the result can be updated.

If word1 and word2 are not equal, the same logic is still valid.

Code

public class Shortest_Word_Distance_III {

    public static void main(String[] args) {
        Shortest_Word_Distance_III out = new Shortest_Word_Distance_III();
        Solution s = out.new Solution();

        // output: 2
        System.out.println(
            s.shortestWordDistance(new String[]{"makes", "practice", "makes", "perfect", "coding", "makes"}, "makes", "makes"));
    }


    public class Solution {
        public int shortestWordDistance(String[] words, String word1, String word2) {
            int posA = -1;
            int posB = -1;
            int minDistance = Integer.MAX_VALUE;

            for (int i = 0; i < words.length; i++) {
                String word = words[i];

                if (word.equals(word1)) {
                    posA = i;
                } else if (word.equals(word2)) {
                    posB = i; // this is covering normal cases, word1 not same as word2
                }

                if (posA != -1 && posB != -1 && posA != posB) { // @note: update before reset posB
                    minDistance = Math.min(minDistance, Math.abs(posA - posB));
                }

                if (word1.equals(word2)) {
                    // always updating posA in above line, so after checking update posB
                    // so that, posB is always the most recent word's index
                    posB = posA;
                }
            }

            return minDistance;
        }
    }
}

############

class Solution {
    public int shortestWordDistance(String[] wordsDict, String word1, String word2) {
        int ans = wordsDict.length;
        if (word1.equals(word2)) {
            for (int i = 0, j = -1; i < wordsDict.length; ++i) {
                if (wordsDict[i].equals(word1)) {
                    if (j != -1) {
                        ans = Math.min(ans, i - j);
                    }
                    j = i;
                }
            }
        } else {
            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;
    }
}

class Solution {
public:
    int shortestWordDistance(vector<string>& wordsDict, string word1, string word2) {
        int n = wordsDict.size();
        int ans = n;
        if (word1 == word2) {
            for (int i = 0, j = -1; i < n; ++i) {
                if (wordsDict[i] == word1) {
                    if (j != -1) {
                        ans = min(ans, i - j);
                    }
                    j = i;
                }
            }
        } else {
            for (int k = 0, i = -1, j = -1; k < n; ++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;
    }
};

class Solution:
    def shortestWordDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
        ans = len(wordsDict)
        if word1 == word2:
            j = -1
            for i, w in enumerate(wordsDict):
                if w == word1:
                    if j != -1: # i != -1 too, so both words found
                        ans = min(ans, i - j)
                    j = i
        else:
            i = j = -1
            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


class Solution: # combine above if-else
    def shortestWordDistance(self, words: List[str], word1: str, word2: str) -> int:
        posA = -1
        posB = -1
        minDistance = float("inf")

        for i in range(len(words)):
            word = words[i]

            if word == word1:
                posA = i
            elif word == word2:
                posB = i

            if posA != -1 and posB != -1 and posA != posB:
                minDistance = min(minDistance, abs(posA - posB))

            if word1 == word2:
                posB = posA

        return minDistance

############

class Solution(object):
  def shortestWordDistance(self, words, word1, word2):
    """
    :type words: List[str]
    :type word1: str
    :type word2: str
    :rtype: int
    """
    ans = float("inf")
    idx1 = idx2 = -1
    for i in range(0, len(words)):
      word = words[i]
      if word in (word1, word2):
        if word == word1:
          idx1 = i
          if idx2 != -1 and idx1 != idx2:
            ans = min(ans, abs(idx2 - idx1))
        if word == word2:
          idx2 = i
          if idx1 != -1 and idx1 != idx2:
            ans = min(ans, abs(idx2 - idx1))
    return ans

func shortestWordDistance(wordsDict []string, word1 string, word2 string) int {
	ans := len(wordsDict)
	if word1 == word2 {
		j := -1
		for i, w := range wordsDict {
			if w == word1 {
				if j != -1 {
					ans = min(ans, i-j)
				}
				j = i
			}
		}
	} else {
		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 min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

245 - Shortest Word Distance III

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