LeetCode #1324 — MEDIUM

Print Words Vertically

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

The Problem

Problem Statement

Given a string s. Return all the words vertically in the same order in which they appear in s.
Words are returned as a list of strings, complete with spaces when is necessary. (Trailing spaces are not allowed).
Each word would be put on only one column and that in one column there will be only one word.

Example 1:

Input: s = "HOW ARE YOU"
Output: ["HAY","ORO","WEU"]
Explanation: Each word is printed vertically. 
 "HAY"
 "ORO"
 "WEU"

Example 2:

Input: s = "TO BE OR NOT TO BE"
Output: ["TBONTB","OEROOE","   T"]
Explanation: Trailing spaces is not allowed. 
"TBONTB"
"OEROOE"
"   T"

Example 3:

Input: s = "CONTEST IS COMING"
Output: ["CIC","OSO","N M","T I","E N","S G","T"]

Constraints:

1 <= s.length <= 200
s contains only upper case English letters.
It's guaranteed that there is only one space between 2 words.

Step 01

Brute Force Baseline

Problem summary: Given a string s. Return all the words vertically in the same order in which they appear in s. Words are returned as a list of strings, complete with spaces when is necessary. (Trailing spaces are not allowed). Each word would be put on only one column and that in one column there will be only one word.

Baseline thinking

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

Pattern signal: Array

Example 1

"HOW ARE YOU"

Example 2

"TO BE OR NOT TO BE"

Example 3

"CONTEST IS COMING"

Step 02

Core Insight

What unlocks the optimal approach

Use the maximum length of words to determine the length of the returned answer. However, don't forget to remove trailing spaces.

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 #1324: Print Words Vertically
class Solution {
    public List<String> printVertically(String s) {
        String[] words = s.split(" ");
        int n = 0;
        for (var w : words) {
            n = Math.max(n, w.length());
        }
        List<String> ans = new ArrayList<>();
        for (int j = 0; j < n; ++j) {
            StringBuilder t = new StringBuilder();
            for (var w : words) {
                t.append(j < w.length() ? w.charAt(j) : ' ');
            }
            while (t.length() > 0 && t.charAt(t.length() - 1) == ' ') {
                t.deleteCharAt(t.length() - 1);
            }
            ans.add(t.toString());
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1324: Print Words Vertically
func printVertically(s string) (ans []string) {
	words := strings.Split(s, " ")
	n := 0
	for _, w := range words {
		n = max(n, len(w))
	}
	for j := 0; j < n; j++ {
		t := []byte{}
		for _, w := range words {
			if j < len(w) {
				t = append(t, w[j])
			} else {
				t = append(t, ' ')
			}
		}
		for len(t) > 0 && t[len(t)-1] == ' ' {
			t = t[:len(t)-1]
		}
		ans = append(ans, string(t))
	}
	return
}

# Accepted solution for LeetCode #1324: Print Words Vertically
class Solution:
    def printVertically(self, s: str) -> List[str]:
        words = s.split()
        n = max(len(w) for w in words)
        ans = []
        for j in range(n):
            t = [w[j] if j < len(w) else ' ' for w in words]
            while t[-1] == ' ':
                t.pop()
            ans.append(''.join(t))
        return ans

// Accepted solution for LeetCode #1324: Print Words Vertically
struct Solution;

impl Solution {
    fn print_vertically(s: String) -> Vec<String> {
        let v: Vec<&str> = s.split_whitespace().collect();
        let mut res = vec![];
        for (j, s) in v.iter().enumerate() {
            for (i, c) in s.char_indices() {
                if i == res.len() {
                    res.push("".to_string());
                }
                while res[i].len() < j {
                    res[i].push(' ');
                }
                res[i].push(c);
            }
        }
        res
    }
}
#[test]
fn test() {
    let s = "HOW ARE YOU".to_string();
    let res = vec_string!["HAY", "ORO", "WEU"];
    assert_eq!(Solution::print_vertically(s), res);
    let s = "TO BE OR NOT TO BE".to_string();
    let res = vec_string!["TBONTB", "OEROOE", "   T"];
    assert_eq!(Solution::print_vertically(s), res);
    let s = "CONTEST IS COMING".to_string();
    let res = vec_string!["CIC", "OSO", "N M", "T I", "E N", "S G", "T"];
    assert_eq!(Solution::print_vertically(s), res);
}

// Accepted solution for LeetCode #1324: Print Words Vertically
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1324: Print Words Vertically
// class Solution {
//     public List<String> printVertically(String s) {
//         String[] words = s.split(" ");
//         int n = 0;
//         for (var w : words) {
//             n = Math.max(n, w.length());
//         }
//         List<String> ans = new ArrayList<>();
//         for (int j = 0; j < n; ++j) {
//             StringBuilder t = new StringBuilder();
//             for (var w : words) {
//                 t.append(j < w.length() ? w.charAt(j) : ' ');
//             }
//             while (t.length() > 0 && t.charAt(t.length() - 1) == ' ') {
//                 t.deleteCharAt(t.length() - 1);
//             }
//             ans.add(t.toString());
//         }
//         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(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.