Valid Word Square

class Solution {
public:
    bool isValidWordSquare(vector<string>& words) {
        for (int i = 0; i < words.size(); i++) {
            for (int j = 0; j < words[i].length(); j++) {
                if (j >= words.size() || 
                    i >= words[j].length() || 
                    words[i][j] != words[j][i]) {
                    return false;
                }
            }
        }
        return true;
    }
};

This solution checks whether a given list of words forms a valid word square in C++. A word square is a matrix of words that read the same horizontally and vertically. The implementation follows these steps:

1. Iterate over each word in the list.
2. For each character in a word, verify if the corresponding position vertically across the words holds the matching character.
3. If a mismatch occurs or the indices go out of the valid range, return false, indicating the words do not form a valid word square.
4. If all conditions are satisfied, return true.

The primary checks involve ensuring that for each character words[i][j], the corresponding vertical character words[j][i] exists and is identical. If either the row or column extends beyond the size of the words list or the characters do not match, the function immediately returns false. Otherwise, after checking all the words and characters without finding any issues, the function returns true, confirming the list of words forms a valid word square.

class Solution {
    public boolean isWordSquare(List<String> wordsList) {
        for (int rowIndex = 0; rowIndex < wordsList.size(); ++rowIndex) {
            for (int colIndex = 0; colIndex < wordsList.get(rowIndex).length(); ++colIndex) {
                if (colIndex >= wordsList.size() || 
                    rowIndex >= wordsList.get(colIndex).length() || 
                    wordsList.get(rowIndex).charAt(colIndex) != wordsList.get(colIndex).charAt(rowIndex)) {
                    return false;
                }
            }
        }
        return true;
    }
}

The presented Java solution implements a method to validate whether a list of strings forms a valid word square. A word square is a sequence of words arranged into a square grid, where the words run the same both horizontally and vertically.

• Begin by iterating over each word in the wordsList with a nested loop structure. The outer loop’s variable rowIndex represents the current row of words, and the inner loop’s variable colIndex represents the corresponding character in the current word.
• Check for three conditions during the character matching:
  • Verify if colIndex is less than the total number of words. This ensures the column exists for the corresponding row character.
  • Check if rowIndex is less than the length of the word at the current column. This confirms the row character exists in the current column word.
  • Compare characters: the character at the current row and column (wordsList.get(rowIndex).charAt(colIndex)) should match the character at the transposed position (wordsList.get(colIndex).charAt(rowIndex)).
• If any condition fails, the method immediately returns false, indicating the words do not form a valid word square.
• If all iterations pass the checks without returning false, then after the loops complete, return true as the list represents a valid word square.

This solution efficiently checks for the validity of a word square by ensuring every character position aligns correctly across both rows and columns, leveraging a direct indexing approach that avoids any transformations or additional data structures.

var isWordSquare = function(wordList) {
    for (let index1 = 0; index1 < wordList.length; ++index1) {
        for (let index2 = 0; index2 < wordList[index1].length; ++index2) {
            if (index2 >= wordList.length || 
                index1 >= wordList[index2].length || 
                wordList[index1][index2] != wordList[index2][index1]){
                return false;
            }
        }
    }
    return true;
};

The problem "Valid Word Square" can be solved using the JavaScript function isWordSquare. This function checks if a given list of words forms a valid word square. A word square is a square grid of words where the kth row and column read the exact same string.

Here's a concise breakdown of the solution:

• The function isWordSquare takes wordList as an argument, which is an array of strings.
• Iterate through each word in wordList using a double nested loop where index1 represents the row index and index2 represents the column index.
• For every character position (index1, index2):
  • Check if the current column index2 exceeds the number of words in the list or if the current row index1 exceeds the length of the word at the current column. This helps in ensuring that the matrix is symmetric.
  • Check if the character at the (index1, index2) position matches the character at the (index2, index1) position.
  • If any of the above checks fail, return false indicating that the words do not form a valid word square.
• If all comparisons are successfully symmetric, return true indicating a valid word square.

This JavaScript solution efficiently determines the validity of a word arrangement as a square by leveraging column and row symmetry checks. The function uses straightforward indexing and comparison to ensure each row matches its corresponding column across the entire matrix.

class Solution:
    def isValidWordSquare(self, words: List[str]) -> bool:
        for i in range(len(words)):
            for j in range(len(words[i])):
                if j >= len(words) or \
                   i >= len(words[j]) or \
                   words[i][j] != words[j][i]:
                    return False
        return True

This Python3 solution checks whether a given list of strings forms a valid word square. A word square is valid if the words line along the rows and columns consistently, meaning the character at the i-th row and j-th column must match the character at the j-th row and i-th column. Here’s how the function operates:

1. Iterate over each word and character in the words list using two nested loops, where i represents the current word index and j represents the character index in the word.
2. Check if the current column j exceeds the number of words or if the current word length at word j is less than i. This step ensures that characters align properly without exceeding their respective string lengths.
3. Compare the character at the i-th row and j-th column with the character at the j-th row and i-th column. If they don’t match, return False.
4. If all checks pass through without breaking, return True after exiting the loops, confirming the words form a valid word square.

This method uses direct index access and simple conditional checks to efficiently validate the word square without the need for additional data structures, providing a solution with clear and understandable logic.