LeetCode #2405 — MEDIUM

Optimal Partition of String

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

The Problem

Problem Statement

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = "abacaba"
Output: 4
Explanation:
Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
It can be shown that 4 is the minimum number of substrings needed.

Example 2:

Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").

Constraints:

1 <= s.length <= 10⁵
s consists of only English lowercase letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once. Return the minimum number of substrings in such a partition. Note that each character should belong to exactly one substring in a partition.

Baseline thinking

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

Pattern signal: Hash Map · Greedy

Example 1

"abacaba"

Example 2

"ssssss"

Core Insight

What unlocks the optimal approach

Try to come up with a greedy approach.
From left to right, extend every substring in the partition as much as possible.

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 #2405: Optimal Partition of String
class Solution {
    public int partitionString(String s) {
        int ans = 1, mask = 0;
        for (int i = 0; i < s.length(); ++i) {
            int x = s.charAt(i) - 'a';
            if ((mask >> x & 1) == 1) {
                ++ans;
                mask = 0;
            }
            mask |= 1 << x;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2405: Optimal Partition of String
func partitionString(s string) int {
	ans, mask := 1, 0
	for _, c := range s {
		x := int(c - 'a')
		if mask>>x&1 == 1 {
			ans++
			mask = 0
		}
		mask |= 1 << x
	}
	return ans
}

# Accepted solution for LeetCode #2405: Optimal Partition of String
class Solution:
    def partitionString(self, s: str) -> int:
        ans, mask = 1, 0
        for x in map(lambda c: ord(c) - ord("a"), s):
            if mask >> x & 1:
                ans += 1
                mask = 0
            mask |= 1 << x
        return ans

// Accepted solution for LeetCode #2405: Optimal Partition of String
impl Solution {
    pub fn partition_string(s: String) -> i32 {
        let mut ans = 1;
        let mut mask = 0;
        for x in s.chars().map(|c| (c as u8 - b'a') as u32) {
            if mask >> x & 1 == 1 {
                ans += 1;
                mask = 0;
            }
            mask |= 1 << x;
        }
        ans
    }
}

// Accepted solution for LeetCode #2405: Optimal Partition of String
function partitionString(s: string): number {
    let [ans, mask] = [1, 0];
    for (const c of s) {
        const x = c.charCodeAt(0) - 97;
        if ((mask >> x) & 1) {
            ++ans;
            mask = 0;
        }
        mask |= 1 << x;
    }
    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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.