LeetCode #3417 — EASY

Zigzag Grid Traversal With Skip

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given an m x n 2D array grid of positive integers.

Your task is to traverse grid in a zigzag pattern while skipping every alternate cell.

Zigzag pattern traversal is defined as following the below actions:

  • Start at the top-left cell (0, 0).
  • Move right within a row until the end of the row is reached.
  • Drop down to the next row, then traverse left until the beginning of the row is reached.
  • Continue alternating between right and left traversal until every row has been traversed.

Note that you must skip every alternate cell during the traversal.

Return an array of integers result containing, in order, the value of the cells visited during the zigzag traversal with skips.

Example 1:

Input: grid = [[1,2],[3,4]]

Output: [1,4]

Explanation:

Example 2:

Input: grid = [[2,1],[2,1],[2,1]]

Output: [2,1,2]

Explanation:

Example 3:

Input: grid = [[1,2,3],[4,5,6],[7,8,9]]

Output: [1,3,5,7,9]

Explanation:

Constraints:

  • 2 <= n == grid.length <= 50
  • 2 <= m == grid[i].length <= 50
  • 1 <= grid[i][j] <= 2500

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an m x n 2D array grid of positive integers. Your task is to traverse grid in a zigzag pattern while skipping every alternate cell. Zigzag pattern traversal is defined as following the below actions: Start at the top-left cell (0, 0). Move right within a row until the end of the row is reached. Drop down to the next row, then traverse left until the beginning of the row is reached. Continue alternating between right and left traversal until every row has been traversed. Note that you must skip every alternate cell during the traversal. Return an array of integers result containing, in order, the value of the cells visited during the zigzag traversal with skips.

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,2],[3,4]]

Example 2

[[2,1],[2,1],[2,1]]

Example 3

[[1,2,3],[4,5,6],[7,8,9]]

Related Problems

  • Binary Tree Zigzag Level Order Traversal (binary-tree-zigzag-level-order-traversal)
  • Longest ZigZag Path in a Binary Tree (longest-zigzag-path-in-a-binary-tree)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3417: Zigzag Grid Traversal With Skip
class Solution {
    public List<Integer> zigzagTraversal(int[][] grid) {
        boolean ok = true;
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < grid.length; ++i) {
            if (i % 2 == 1) {
                reverse(grid[i]);
            }
            for (int x : grid[i]) {
                if (ok) {
                    ans.add(x);
                }
                ok = !ok;
            }
        }
        return ans;
    }

    private void reverse(int[] nums) {
        for (int i = 0, j = nums.length - 1; i < j; ++i, --j) {
            int t = nums[i];
            nums[i] = nums[j];
            nums[j] = t;
        }
    }
}
Step 06

Interactive Study Demo

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

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.

Related Problems
#1Two Sumeasy→#6Zigzag Conversionmedium→#14Longest Common Prefixeasy→#31Next Permutationmedium→#36Valid Sudokumedium→