Valid Word Square

Valid Word Square - Problem

Array Matrix Easy

Given an array of strings words, return true if it forms a valid word square.

A sequence of strings forms a valid word square if the kth row and column read the same string, where 0 <= k < max(numRows, numColumns).

In other words, for each position (i, j), the character at words[i][j] should equal the character at words[j][i] (if both positions exist).

Input & Output

Example 1 — Valid Word Square

$ Input: words = ["abcd","bnrt","crmy","dtye"]

› Output: true

💡 Note: Row 0: "abcd" matches Column 0: "abcd", Row 1: "bnrt" matches Column 1: "bnrt", etc. All rows equal their corresponding columns.

Example 2 — Invalid Word Square

$ Input: words = ["abcd","bnrt","crm","dt"]

› Output: false

💡 Note: Row 2 is "crm" but Column 2 would be "mrd" (reading vertically). They don't match, so it's not a valid word square.

Example 3 — Single Word

$ Input: words = ["a"]

› Output: true

💡 Note: Single character forms a valid 1×1 word square where row 0 equals column 0.

Constraints

1 ≤ words.length ≤ 500
1 ≤ words[i].length ≤ 500
words[i] consists of lowercase English letters

Visualization

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Asked in

G Google 25 f Facebook 18

The key insight is that a valid word square must be symmetric: row i must equal column i. The optimized approach checks only the upper triangle to avoid redundant comparisons. Time: O(n²), Space: O(1).

Common Approaches

✓ Brute Force Character-by-Character

⏱️ Time: O(n²) Space: O(1)

For each position (i,j) in the grid, verify that words[i][j] equals words[j][i]. Handle out-of-bounds cases by ensuring both positions exist or both don't exist.

Optimized Single Pass

⏱️ Time: O(n²) Space: O(1)

Since word squares are symmetric, we only need to check positions where i <= j. This eliminates redundant comparisons and makes the algorithm more efficient while maintaining correctness.

Brute Force Character-by-Character — Algorithm Steps

Iterate through all possible positions
For each (i,j), check if words[i][j] == words[j][i]
Handle cases where strings have different lengths

Visualization

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Step-by-Step Walkthrough

Input Grid

Array of strings representing word square

Compare Positions

Check words[i][j] == words[j][i] for all i,j

Result

Return true if all positions match

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char words[][100], int n) {
    if (n == 0) return true;
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            int lenI = strlen(words[i]);
            int lenJ = strlen(words[j]);
            
            // Check if position (i,j) exists
            bool hasIJ = j < lenI;
            // Check if position (j,i) exists
            bool hasJI = i < lenJ;
            
            if (hasIJ != hasJI) {
                return false;
            }
            
            if (hasIJ && words[i][j] != words[j][i]) {
                return false;
            }
        }
    }
    
    return true;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for JSON array
    char words[50][100];
    int n = 0;
    
    char *start = strchr(line, '[');
    if (start) {
        start++;
        char *token = strtok(start, ",]");
        while (token && n < 50) {
            // Remove quotes and whitespace
            while (*token == ' ' || *token == '"') token++;
            char *end = token + strlen(token) - 1;
            while (end > token && (*end == ' ' || *end == '"' || *end == ']')) {
                *end = '\0';
                end--;
            }
            strcpy(words[n], token);
            n++;
            token = strtok(NULL, ",]");
        }
    }
    
    bool result = solution(words, n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check each position (i,j) in an n×n grid

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for indices and comparison

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

428 Likes

Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Character-by-Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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