LeetCode #3324 — MEDIUM

Find the Sequence of Strings Appeared on the Screen

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

Solve on LeetCode

The Problem

Problem Statement

You are given a string target.

Alice is going to type target on her computer using a special keyboard that has only two keys:

Key 1 appends the character "a" to the string on the screen.
Key 2 changes the last character of the string on the screen to its next character in the English alphabet. For example, "c" changes to "d" and "z" changes to "a".

Note that initially there is an empty string "" on the screen, so she can only press key 1.

Return a list of all strings that appear on the screen as Alice types target, in the order they appear, using the minimum key presses.

Example 1:

Input: target = "abc"

Output: ["a","aa","ab","aba","abb","abc"]

Explanation:

The sequence of key presses done by Alice are:

Press key 1, and the string on the screen becomes "a".
Press key 1, and the string on the screen becomes "aa".
Press key 2, and the string on the screen becomes "ab".
Press key 1, and the string on the screen becomes "aba".
Press key 2, and the string on the screen becomes "abb".
Press key 2, and the string on the screen becomes "abc".

Example 2:

Input: target = "he"

Output: ["a","b","c","d","e","f","g","h","ha","hb","hc","hd","he"]

Constraints:

1 <= target.length <= 400
target consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string target. Alice is going to type target on her computer using a special keyboard that has only two keys: Key 1 appends the character "a" to the string on the screen. Key 2 changes the last character of the string on the screen to its next character in the English alphabet. For example, "c" changes to "d" and "z" changes to "a". Note that initially there is an empty string "" on the screen, so she can only press key 1. Return a list of all strings that appear on the screen as Alice types target, in the order they appear, using the minimum key presses.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"abc"

Example 2

"he"

Core Insight

What unlocks the optimal approach

Append the character <code>'a'</code> using key 1.
Convert it to the required character using key 2.

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 #3324: Find the Sequence of Strings Appeared on the Screen
class Solution {
    public List<String> stringSequence(String target) {
        List<String> ans = new ArrayList<>();
        for (char c : target.toCharArray()) {
            String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
            for (char a = 'a'; a <= c; ++a) {
                String t = s + a;
                ans.add(t);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3324: Find the Sequence of Strings Appeared on the Screen
func stringSequence(target string) (ans []string) {
	for _, c := range target {
		s := ""
		if len(ans) > 0 {
			s = ans[len(ans)-1]
		}
		for a := 'a'; a <= c; a++ {
			t := s + string(a)
			ans = append(ans, t)
		}
	}
	return
}

# Accepted solution for LeetCode #3324: Find the Sequence of Strings Appeared on the Screen
class Solution:
    def stringSequence(self, target: str) -> List[str]:
        ans = []
        for c in target:
            s = ans[-1] if ans else ""
            for a in ascii_lowercase:
                t = s + a
                ans.append(t)
                if a == c:
                    break
        return ans

// Accepted solution for LeetCode #3324: Find the Sequence of Strings Appeared on the Screen
// 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 #3324: Find the Sequence of Strings Appeared on the Screen
// class Solution {
//     public List<String> stringSequence(String target) {
//         List<String> ans = new ArrayList<>();
//         for (char c : target.toCharArray()) {
//             String s = ans.isEmpty() ? "" : ans.get(ans.size() - 1);
//             for (char a = 'a'; a <= c; ++a) {
//                 String t = s + a;
//                 ans.add(t);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3324: Find the Sequence of Strings Appeared on the Screen
function stringSequence(target: string): string[] {
    const ans: string[] = [];
    for (const c of target) {
        let s = ans.length > 0 ? ans[ans.length - 1] : '';
        for (let a = 'a'.charCodeAt(0); a <= c.charCodeAt(0); a++) {
            const t = s + String.fromCharCode(a);
            ans.push(t);
        }
    }
    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.