LeetCode #3223 — MEDIUM

Minimum Length of String After Operations

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

Solve on LeetCode

The Problem

Problem Statement

You are given a string s.

You can perform the following process on s any number of times:

Choose an index i in the string such that there is at least one character to the left of index i that is equal to s[i], and at least one character to the right that is also equal to s[i].
Delete the closest occurrence of s[i] located to the left of i.
Delete the closest occurrence of s[i] located to the right of i.

Return the minimum length of the final string s that you can achieve.

Example 1:

Input: s = "abaacbcbb"

Output: 5

Explanation:
We do the following operations:

Choose index 2, then remove the characters at indices 0 and 3. The resulting string is s = "bacbcbb".
Choose index 3, then remove the characters at indices 0 and 5. The resulting string is s = "acbcb".

Example 2:

Input: s = "aa"

Output: 2

Explanation:
We cannot perform any operations, so we return the length of the original string.

Constraints:

1 <= s.length <= 2 * 10⁵
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s. You can perform the following process on s any number of times: Choose an index i in the string such that there is at least one character to the left of index i that is equal to s[i], and at least one character to the right that is also equal to s[i]. Delete the closest occurrence of s[i] located to the left of i. Delete the closest occurrence of s[i] located to the right of i. Return the minimum length of the final string s that you can achieve.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"abaacbcbb"

Example 2

"aa"

Step 02

Core Insight

What unlocks the optimal approach

Only the frequency of each character matters in finding the final answer.
If a character occurs less than 3 times, we cannot perform any process with it.
Suppose there is a character that occurs at least 3 times in the string, we can repeatedly delete two of these characters until there are at most 2 occurrences left of it.

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 #3223: Minimum Length of String After Operations
class Solution {
    public int minimumLength(String s) {
        int[] cnt = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        int ans = 0;
        for (int x : cnt) {
            if (x > 0) {
                ans += x % 2 == 1 ? 1 : 2;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3223: Minimum Length of String After Operations
func minimumLength(s string) (ans int) {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	for _, x := range cnt {
		if x > 0 {
			if x&1 == 1 {
				ans += 1
			} else {
				ans += 2
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3223: Minimum Length of String After Operations
class Solution:
    def minimumLength(self, s: str) -> int:
        cnt = Counter(s)
        return sum(1 if x & 1 else 2 for x in cnt.values())

// Accepted solution for LeetCode #3223: Minimum Length of String After Operations
// 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 #3223: Minimum Length of String After Operations
// class Solution {
//     public int minimumLength(String s) {
//         int[] cnt = new int[26];
//         for (int i = 0; i < s.length(); ++i) {
//             ++cnt[s.charAt(i) - 'a'];
//         }
//         int ans = 0;
//         for (int x : cnt) {
//             if (x > 0) {
//                 ans += x % 2 == 1 ? 1 : 2;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3223: Minimum Length of String After Operations
function minimumLength(s: string): number {
    const cnt = new Map<string, number>();
    for (const c of s) {
        cnt.set(c, (cnt.get(c) || 0) + 1);
    }
    let ans = 0;
    for (const x of cnt.values()) {
        ans += x & 1 ? 1 : 2;
    }
    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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

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.