LeetCode #3799 — MEDIUM

Word Squares II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a string array words, consisting of distinct 4-letter strings, each containing lowercase English letters.

A word square consists of 4 distinct words: top, left, right and bottom, arranged as follows:

top forms the top row.
bottom forms the bottom row.
left forms the left column (top to bottom).
right forms the right column (top to bottom).

It must satisfy:

top[0] == left[0], top[3] == right[0]
bottom[0] == left[3], bottom[3] == right[3]

Return all valid distinct word squares, sorted in ascending lexicographic order by the 4-tuple (top, left, right, bottom).

Example 1:

Input: words = ["able","area","echo","also"]

Output: [["able","area","echo","also"],["area","able","also","echo"]]

Explanation:

There are exactly two valid 4-word squares that satisfy all corner constraints:

"able" (top), "area" (left), "echo" (right), "also" (bottom)
- top[0] == left[0] == 'a'
- top[3] == right[0] == 'e'
- bottom[0] == left[3] == 'a'
- bottom[3] == right[3] == 'o'
"area" (top), "able" (left), "also" (right), "echo" (bottom)
- All corner constraints are satisfied.

Thus, the answer is [["able","area","echo","also"],["area","able","also","echo"]].

Example 2:

Input: words = ["code","cafe","eden","edge"]

Output: []

Explanation:

No combination of four words satisfies all four corner constraints. Thus, the answer is empty array [].

Constraints:

4 <= words.length <= 15
words[i].length == 4
words[i] consists of only lowercase English letters.
All words[i] are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given a string array words, consisting of distinct 4-letter strings, each containing lowercase English letters. A word square consists of 4 distinct words: top, left, right and bottom, arranged as follows: top forms the top row. bottom forms the bottom row. left forms the left column (top to bottom). right forms the right column (top to bottom). It must satisfy: top[0] == left[0], top[3] == right[0] bottom[0] == left[3], bottom[3] == right[3] Return all valid distinct word squares, sorted in ascending lexicographic order by the 4-tuple (top, left, right, bottom).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking

Example 1

["able","area","echo","also"]

Example 2

["code","cafe","eden","edge"]

Step 02

Core Insight

What unlocks the optimal approach

Use bruteforce

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3799: Word Squares II
class Solution {
    public List<List<String>> wordSquares(String[] words) {
        Arrays.sort(words);
        int n = words.length;
        List<List<String>> ans = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            String top = words[i];
            for (int j = 0; j < n; j++) {
                if (j != i) {
                    String left = words[j];
                    for (int k = 0; k < n; k++) {
                        if (k != j && k != i) {
                            String right = words[k];
                            for (int h = 0; h < n; h++) {
                                if (h != k && h != j && h != i) {
                                    String bottom = words[h];
                                    if (top.charAt(0) == left.charAt(0)
                                        && top.charAt(3) == right.charAt(0)
                                        && bottom.charAt(0) == left.charAt(3)
                                        && bottom.charAt(3) == right.charAt(3)) {
                                        ans.add(List.of(top, left, right, bottom));
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3799: Word Squares II
func wordSquares(words []string) [][]string {
	sort.Strings(words)
	n := len(words)
	ans := [][]string{}

	for i := 0; i < n; i++ {
		top := words[i]
		for j := 0; j < n; j++ {
			if j != i {
				left := words[j]
				for k := 0; k < n; k++ {
					if k != j && k != i {
						right := words[k]
						for h := 0; h < n; h++ {
							if h != k && h != j && h != i {
								bottom := words[h]
								if top[0] == left[0] &&
									top[3] == right[0] &&
									bottom[0] == left[3] &&
									bottom[3] == right[3] {
									ans = append(ans, []string{top, left, right, bottom})
								}
							}
						}
					}
				}
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #3799: Word Squares II
class Solution:
    def wordSquares(self, words: List[str]) -> List[List[str]]:
        words.sort()
        n = len(words)
        ans = []
        for i in range(n):
            top = words[i]
            for j in range(n):
                if j != i:
                    left = words[j]
                    for k in range(n):
                        if k != j and k != i:
                            right = words[k]
                            for h in range(n):
                                if h != k and h != j and h != i:
                                    bottom = words[h]
                                    if (
                                        top[0] == left[0]
                                        and top[3] == right[0]
                                        and bottom[0] == left[3]
                                        and bottom[3] == right[3]
                                    ):
                                        ans.append([top, left, right, bottom])
        return ans

// Accepted solution for LeetCode #3799: Word Squares II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3799: Word Squares II
// class Solution {
//     public List<List<String>> wordSquares(String[] words) {
//         Arrays.sort(words);
//         int n = words.length;
//         List<List<String>> ans = new ArrayList<>();
//         for (int i = 0; i < n; i++) {
//             String top = words[i];
//             for (int j = 0; j < n; j++) {
//                 if (j != i) {
//                     String left = words[j];
//                     for (int k = 0; k < n; k++) {
//                         if (k != j && k != i) {
//                             String right = words[k];
//                             for (int h = 0; h < n; h++) {
//                                 if (h != k && h != j && h != i) {
//                                     String bottom = words[h];
//                                     if (top.charAt(0) == left.charAt(0)
//                                         && top.charAt(3) == right.charAt(0)
//                                         && bottom.charAt(0) == left.charAt(3)
//                                         && bottom.charAt(3) == right.charAt(3)) {
//                                         ans.add(List.of(top, left, right, bottom));
//                                     }
//                                 }
//                             }
//                         }
//                     }
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3799: Word Squares II
function wordSquares(words: string[]): string[][] {
    words.sort();
    const n = words.length;
    const ans: string[][] = [];

    for (let i = 0; i < n; i++) {
        const top = words[i];
        for (let j = 0; j < n; j++) {
            if (j !== i) {
                const left = words[j];
                for (let k = 0; k < n; k++) {
                    if (k !== j && k !== i) {
                        const right = words[k];
                        for (let h = 0; h < n; h++) {
                            if (h !== k && h !== j && h !== i) {
                                const bottom = words[h];
                                if (
                                    top[0] === left[0] &&
                                    top[3] === right[0] &&
                                    bottom[0] === left[3] &&
                                    bottom[3] === right[3]
                                ) {
                                    ans.push([top, left, right, bottom]);
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.