LeetCode #3597 — MEDIUM

Partition String

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

Solve on LeetCode

The Problem

Problem Statement

Given a string s, partition it into unique segments according to the following procedure:

Start building a segment beginning at index 0.
Continue extending the current segment character by character until the current segment has not been seen before.
Once the segment is unique, add it to your list of segments, mark it as seen, and begin a new segment from the next index.
Repeat until you reach the end of s.

Return an array of strings segments, where segments[i] is the i^th segment created.

Example 1:

Input: s = "abbccccd"

Output: ["a","b","bc","c","cc","d"]

Explanation:

Index	Segment After Adding	Seen Segments	Current Segment Seen Before?	New Segment	Updated Seen Segments
0	"a"	[]	No	""	["a"]
1	"b"	["a"]	No	""	["a", "b"]
2	"b"	["a", "b"]	Yes	"b"	["a", "b"]
3	"bc"	["a", "b"]	No	""	["a", "b", "bc"]
4	"c"	["a", "b", "bc"]	No	""	["a", "b", "bc", "c"]
5	"c"	["a", "b", "bc", "c"]	Yes	"c"	["a", "b", "bc", "c"]
6	"cc"	["a", "b", "bc", "c"]	No	""	["a", "b", "bc", "c", "cc"]
7	"d"	["a", "b", "bc", "c", "cc"]	No	""	["a", "b", "bc", "c", "cc", "d"]

Hence, the final output is ["a", "b", "bc", "c", "cc", "d"].

Example 2:

Input: s = "aaaa"

Output: ["a","aa"]

Explanation:

Index	Segment After Adding	Seen Segments	Current Segment Seen Before?	New Segment	Updated Seen Segments
0	"a"	[]	No	""	["a"]
1	"a"	["a"]	Yes	"a"	["a"]
2	"aa"	["a"]	No	""	["a", "aa"]
3	"a"	["a", "aa"]	Yes	"a"	["a", "aa"]

Hence, the final output is ["a", "aa"].

Constraints:

1 <= s.length <= 10⁵
s contains only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, partition it into unique segments according to the following procedure: Start building a segment beginning at index 0. Continue extending the current segment character by character until the current segment has not been seen before. Once the segment is unique, add it to your list of segments, mark it as seen, and begin a new segment from the next index. Repeat until you reach the end of s. Return an array of strings segments, where segments[i] is the ith segment created.

Baseline thinking

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

Pattern signal: Hash Map · Trie

Example 1

"abbccccd"

Example 2

"aaaa"

Step 02

Core Insight

What unlocks the optimal approach

Simulate as described

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 #3597: Partition String 
class Solution {
    public List<String> partitionString(String s) {
        Set<String> vis = new HashSet<>();
        List<String> ans = new ArrayList<>();
        String t = "";
        for (char c : s.toCharArray()) {
            t += c;
            if (vis.add(t)) {
                ans.add(t);
                t = "";
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3597: Partition String 
func partitionString(s string) (ans []string) {
	vis := make(map[string]bool)
	t := ""
	for _, c := range s {
		t += string(c)
		if !vis[t] {
			vis[t] = true
			ans = append(ans, t)
			t = ""
		}
	}
	return
}

# Accepted solution for LeetCode #3597: Partition String 
class Solution:
    def partitionString(self, s: str) -> List[str]:
        vis = set()
        ans = []
        t = ""
        for c in s:
            t += c
            if t not in vis:
                vis.add(t)
                ans.append(t)
                t = ""
        return ans

// Accepted solution for LeetCode #3597: Partition String 
// 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 #3597: Partition String 
// class Solution {
//     public List<String> partitionString(String s) {
//         Set<String> vis = new HashSet<>();
//         List<String> ans = new ArrayList<>();
//         String t = "";
//         for (char c : s.toCharArray()) {
//             t += c;
//             if (vis.add(t)) {
//                 ans.add(t);
//                 t = "";
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3597: Partition String 
function partitionString(s: string): string[] {
    const vis = new Set<string>();
    const ans: string[] = [];
    let t = '';
    for (const c of s) {
        t += c;
        if (!vis.has(t)) {
            vis.add(t);
            ans.push(t);
            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 × sqrtn)

Space

O(n)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.