LeetCode #3816 — HARD

Lexicographically Smallest String After Deleting Duplicate Characters

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a string s that consists of lowercase English letters.

You can perform the following operation any number of times (possibly zero times):

Choose any letter that appears at least twice in the current string s and delete any one occurrence.

Return the lexicographically smallest resulting string that can be formed this way.

Example 1:

Input: s = "aaccb"

Output: "aacb"

Explanation:

We can form the strings "acb", "aacb", "accb", and "aaccb". "aacb" is the lexicographically smallest one.

For example, we can obtain "aacb" by choosing 'c' and deleting its first occurrence.

Example 2:

Input: s = "z"

Output: "z"

Explanation:

We cannot perform any operations. The only string we can form is "z".

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a string s that consists of lowercase English letters. You can perform the following operation any number of times (possibly zero times): Choose any letter that appears at least twice in the current string s and delete any one occurrence. Return the lexicographically smallest resulting string that can be formed this way.

Baseline thinking

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

Pattern signal: Hash Map · Stack · Greedy

Example 1

"aaccb"

Example 2

"z"

Core Insight

What unlocks the optimal approach

Solve greedily.
Each distinct letter must appear at least once in the final string.
For each letter, maintain a deque of its positions.
At each step, try letters from <code>'a'</code> to <code>'z'</code> and pick the smallest letter whose earliest position lies within a safe window.
Do not pick an occurrence if choosing it would make some other letter impossible to keep.
Mark positions as used and repeat, always minimizing the next chosen character.

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

Largest constraint values

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 #3816: Lexicographically Smallest String After Deleting Duplicate Characters
class Solution {
    public String lexSmallestAfterDeletion(String s) {
        int[] cnt = new int[26];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        StringBuilder stk = new StringBuilder();
        for (int i = 0; i < n; ++i) {
            char c = s.charAt(i);
            while (stk.length() > 0 && stk.charAt(stk.length() - 1) > c
                && cnt[stk.charAt(stk.length() - 1) - 'a'] > 1) {
                --cnt[stk.charAt(stk.length() - 1) - 'a'];
                stk.setLength(stk.length() - 1);
            }
            stk.append(c);
        }
        while (cnt[stk.charAt(stk.length() - 1) - 'a'] > 1) {
            --cnt[stk.charAt(stk.length() - 1) - 'a'];
            stk.setLength(stk.length() - 1);
        }
        return stk.toString();
    }
}

// Accepted solution for LeetCode #3816: Lexicographically Smallest String After Deleting Duplicate Characters
func lexSmallestAfterDeletion(s string) string {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	stk := []byte{}
	for _, c := range s {
		for len(stk) > 0 && stk[len(stk)-1] > byte(c) && cnt[stk[len(stk)-1]-'a'] > 1 {
			cnt[stk[len(stk)-1]-'a']--
			stk = stk[:len(stk)-1]
		}
		stk = append(stk, byte(c))
	}
	for cnt[stk[len(stk)-1]-'a'] > 1 {
		cnt[stk[len(stk)-1]-'a']--
		stk = stk[:len(stk)-1]
	}
	return string(stk)
}

# Accepted solution for LeetCode #3816: Lexicographically Smallest String After Deleting Duplicate Characters
class Solution:
    def lexSmallestAfterDeletion(self, s: str) -> str:
        cnt = Counter(s)
        stk = []
        for c in s:
            while stk and stk[-1] > c and cnt[stk[-1]] > 1:
                cnt[stk.pop()] -= 1
            stk.append(c)
        while stk and cnt[stk[-1]] > 1:
            cnt[stk.pop()] -= 1
        return "".join(stk)

// Accepted solution for LeetCode #3816: Lexicographically Smallest String After Deleting Duplicate Characters
// 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 #3816: Lexicographically Smallest String After Deleting Duplicate Characters
// class Solution {
//     public String lexSmallestAfterDeletion(String s) {
//         int[] cnt = new int[26];
//         int n = s.length();
//         for (int i = 0; i < n; ++i) {
//             ++cnt[s.charAt(i) - 'a'];
//         }
//         StringBuilder stk = new StringBuilder();
//         for (int i = 0; i < n; ++i) {
//             char c = s.charAt(i);
//             while (stk.length() > 0 && stk.charAt(stk.length() - 1) > c
//                 && cnt[stk.charAt(stk.length() - 1) - 'a'] > 1) {
//                 --cnt[stk.charAt(stk.length() - 1) - 'a'];
//                 stk.setLength(stk.length() - 1);
//             }
//             stk.append(c);
//         }
//         while (cnt[stk.charAt(stk.length() - 1) - 'a'] > 1) {
//             --cnt[stk.charAt(stk.length() - 1) - 'a'];
//             stk.setLength(stk.length() - 1);
//         }
//         return stk.toString();
//     }
// }

// Accepted solution for LeetCode #3816: Lexicographically Smallest String After Deleting Duplicate Characters
function lexSmallestAfterDeletion(s: string): string {
    const cnt: number[] = new Array(26).fill(0);
    const n = s.length;
    const a = 'a'.charCodeAt(0);
    for (let i = 0; i < n; ++i) {
        ++cnt[s.charCodeAt(i) - a];
    }
    const stk: string[] = [];
    for (let i = 0; i < n; ++i) {
        const c = s[i];
        while (
            stk.length > 0 &&
            stk[stk.length - 1] > c &&
            cnt[stk[stk.length - 1].charCodeAt(0) - a] > 1
        ) {
            --cnt[stk.pop()!.charCodeAt(0) - a];
        }
        stk.push(c);
    }
    while (cnt[stk[stk.length - 1].charCodeAt(0) - a] > 1) {
        --cnt[stk.pop()!.charCodeAt(0) - a];
    }
    return stk.join('');
}

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 left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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.