LeetCode #2624 — MEDIUM

Snail Traversal

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Write code that enhances all arrays such that you can call the snail(rowsCount, colsCount) method that transforms the 1D array into a 2D array organised in the pattern known as snail traversal order. Invalid input values should output an empty array. If rowsCount * colsCount !== nums.length, the input is considered invalid.

Snail traversal order starts at the top left cell with the first value of the current array. It then moves through the entire first column from top to bottom, followed by moving to the next column on the right and traversing it from bottom to top. This pattern continues, alternating the direction of traversal with each column, until the entire current array is covered. For example, when given the input array [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15] with rowsCount = 5 and colsCount = 4, the desired output matrix is shown below. Note that iterating the matrix following the arrows corresponds to the order of numbers in the original array.

Example 1:

Input: 
nums = [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]
rowsCount = 5
colsCount = 4
Output: 
[
 [19,17,16,15],
 [10,1,14,4],
 [3,2,12,20],
 [7,5,18,11],
 [9,8,6,13]
]

Example 2:

Input: 
nums = [1,2,3,4]
rowsCount = 1
colsCount = 4
Output: [[1, 2, 3, 4]]

Example 3:

Input: 
nums = [1,3]
rowsCount = 2
colsCount = 2
Output: []
Explanation: 2 multiplied by 2 is 4, and the original array [1,3] has a length of 2; therefore, the input is invalid.

Constraints:

0 <= nums.length <= 250
1 <= nums[i] <= 1000
1 <= rowsCount <= 250
1 <= colsCount <= 250

Step 01

Brute Force Baseline

Problem summary: Write code that enhances all arrays such that you can call the snail(rowsCount, colsCount) method that transforms the 1D array into a 2D array organised in the pattern known as snail traversal order. Invalid input values should output an empty array. If rowsCount * colsCount !== nums.length, the input is considered invalid. Snail traversal order starts at the top left cell with the first value of the current array. It then moves through the entire first column from top to bottom, followed by moving to the next column on the right and traversing it from bottom to top. This pattern continues, alternating the direction of traversal with each column, until the entire current array is covered. For example, when given the input array [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15] with rowsCount = 5 and colsCount = 4, the desired output matrix is shown below. Note that

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

[19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]
5
4

Example 2

[1,2,3,4]
1
4

Example 3

[1,3]
2
2

Core Insight

What unlocks the optimal approach

Different ways to approach this problem. Perhaps store a boolean if you are moving up or down and a current column. Reverse the direction and increment the column every time you hits a wall.
Is there a way way to do this without storing state - by just using math?

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 #2624: Snail Traversal
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2624: Snail Traversal
// declare global {
//     interface Array<T> {
//         snail(rowsCount: number, colsCount: number): number[][];
//     }
// }
// 
// Array.prototype.snail = function (rowsCount: number, colsCount: number): number[][] {
//     if (rowsCount * colsCount !== this.length) {
//         return [];
//     }
//     const ans: number[][] = Array.from({ length: rowsCount }, () => Array(colsCount));
//     for (let h = 0, i = 0, j = 0, k = 1; h < this.length; ++h) {
//         ans[i][j] = this[h];
//         i += k;
//         if (i === rowsCount || i === -1) {
//             i -= k;
//             k = -k;
//             ++j;
//         }
//     }
//     return ans;
// };
// 
// /**
//  * const arr = [1,2,3,4];
//  * arr.snail(1,4); // [[1,2,3,4]]
//  */

// Accepted solution for LeetCode #2624: Snail Traversal
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2624: Snail Traversal
// declare global {
//     interface Array<T> {
//         snail(rowsCount: number, colsCount: number): number[][];
//     }
// }
// 
// Array.prototype.snail = function (rowsCount: number, colsCount: number): number[][] {
//     if (rowsCount * colsCount !== this.length) {
//         return [];
//     }
//     const ans: number[][] = Array.from({ length: rowsCount }, () => Array(colsCount));
//     for (let h = 0, i = 0, j = 0, k = 1; h < this.length; ++h) {
//         ans[i][j] = this[h];
//         i += k;
//         if (i === rowsCount || i === -1) {
//             i -= k;
//             k = -k;
//             ++j;
//         }
//     }
//     return ans;
// };
// 
// /**
//  * const arr = [1,2,3,4];
//  * arr.snail(1,4); // [[1,2,3,4]]
//  */

# Accepted solution for LeetCode #2624: Snail Traversal
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2624: Snail Traversal
# declare global {
#     interface Array<T> {
#         snail(rowsCount: number, colsCount: number): number[][];
#     }
# }
# 
# Array.prototype.snail = function (rowsCount: number, colsCount: number): number[][] {
#     if (rowsCount * colsCount !== this.length) {
#         return [];
#     }
#     const ans: number[][] = Array.from({ length: rowsCount }, () => Array(colsCount));
#     for (let h = 0, i = 0, j = 0, k = 1; h < this.length; ++h) {
#         ans[i][j] = this[h];
#         i += k;
#         if (i === rowsCount || i === -1) {
#             i -= k;
#             k = -k;
#             ++j;
#         }
#     }
#     return ans;
# };
# 
# /**
#  * const arr = [1,2,3,4];
#  * arr.snail(1,4); // [[1,2,3,4]]
#  */

// Accepted solution for LeetCode #2624: Snail Traversal
// Rust example auto-generated from ts 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 (ts):
// // Accepted solution for LeetCode #2624: Snail Traversal
// declare global {
//     interface Array<T> {
//         snail(rowsCount: number, colsCount: number): number[][];
//     }
// }
// 
// Array.prototype.snail = function (rowsCount: number, colsCount: number): number[][] {
//     if (rowsCount * colsCount !== this.length) {
//         return [];
//     }
//     const ans: number[][] = Array.from({ length: rowsCount }, () => Array(colsCount));
//     for (let h = 0, i = 0, j = 0, k = 1; h < this.length; ++h) {
//         ans[i][j] = this[h];
//         i += k;
//         if (i === rowsCount || i === -1) {
//             i -= k;
//             k = -k;
//             ++j;
//         }
//     }
//     return ans;
// };
// 
// /**
//  * const arr = [1,2,3,4];
//  * arr.snail(1,4); // [[1,2,3,4]]
//  */

// Accepted solution for LeetCode #2624: Snail Traversal
declare global {
    interface Array<T> {
        snail(rowsCount: number, colsCount: number): number[][];
    }
}

Array.prototype.snail = function (rowsCount: number, colsCount: number): number[][] {
    if (rowsCount * colsCount !== this.length) {
        return [];
    }
    const ans: number[][] = Array.from({ length: rowsCount }, () => Array(colsCount));
    for (let h = 0, i = 0, j = 0, k = 1; h < this.length; ++h) {
        ans[i][j] = this[h];
        i += k;
        if (i === rowsCount || i === -1) {
            i -= k;
            k = -k;
            ++j;
        }
    }
    return ans;
};

/**
 * const arr = [1,2,3,4];
 * arr.snail(1,4); // [[1,2,3,4]]
 */

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(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.