DevExCode

LeetCode 245: Shortest Word Distance III

ArrayString

LeetCode 245 Solution Explanation

Explanation

To solve this problem, we can iterate through the input array of words and keep track of the indices of the two target words. We update the indices as we encounter the target words. We then calculate the minimum distance between the two indices. If the words can be the same, we need to handle the case where the two indices are the same.

Algorithm:

  1. Initialize variables index1, index2, and minDistance to track the indices and minimum distance.
  2. Iterate through the input words array:
    • If the current word is equal to word1, update index1.
    • If the current word is equal to word2, update index2.
    • If both indices are valid, calculate the distance and update minDistance.
    • If word1 and word2 are the same, handle the case where index1 and index2 are the same.
  3. Finally, return minDistance.

Time Complexity: O(n) where n is the number of words in the input array.

Space Complexity: O(1)

View LeetCode 245 on LeetCode

LeetCode 245 Solutions in Java, C++, Python

class Solution {
    public int shortestWordDistance(String[] words, String word1, String word2) {
        int index1 = -1, index2 = -1;
        int minDistance = Integer.MAX_VALUE;
        
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) {
                index1 = i;
            }
            if (words[i].equals(word2)) {
                if (word1.equals(word2)) {
                    index1 = index2;
                }
                index2 = i;
            }
            if (index1 != -1 && index2 != -1) {
                minDistance = Math.min(minDistance, Math.abs(index1 - index2));
            }
        }
        
        return minDistance;
    }
}

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