LeetCode 1554: Strings Differ by One Character Solution

Master LeetCode problem 1554 (Strings Differ by One Character), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1554. Strings Differ by One Character

Medium

Hash Table String Rolling Hash Hash Function

Problem Explanation

Explanation:

To solve this problem, we can iterate through each pair of strings and check if they differ by exactly one character. We can do this by comparing the characters at each index of the strings. If we find more than one differing character, we move on to the next pair. If we find exactly one differing character, we return true.

Algorithm:

Iterate through each pair of strings in the input list.
Compare the characters at each index of the two strings.
If the characters differ, increment a counter.
If the counter exceeds 1, move to the next pair of strings.
If the counter is exactly 1, return true.

Time Complexity:

The time complexity of this approach is O(n*m), where n is the length of the input list and m is the average length of the strings in the list.

Space Complexity:

The space complexity of this approach is O(1) as we are not using any extra space.

: :

Solution Code

class Solution {
    public boolean differByOne(String[] dict) {
        for (int i = 0; i < dict.length; i++) {
            for (int j = 0; j < dict.length; j++) {
                if (i != j) {
                    int count = 0;
                    for (int k = 0; k < dict[i].length(); k++) {
                        if (dict[i].charAt(k) != dict[j].charAt(k)) {
                            count++;
                        }
                        if (count > 1) break;
                    }
                    if (count == 1) return true;
                }
            }
        }
        return false;
    }
}

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