LeetCode #3121 — MEDIUM

Count the Number of Special Characters II

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

The Problem

Problem Statement

You are given a string word. A letter c is called special if it appears both in lowercase and uppercase in word, and every lowercase occurrence of c appears before the first uppercase occurrence of c.

Return the number of special letters in word.

Example 1:

Input: word = "aaAbcBC"

Output: 3

Explanation:

The special characters are 'a', 'b', and 'c'.

Example 2:

Input: word = "abc"

Output: 0

Explanation:

There are no special characters in word.

Example 3:

Input: word = "AbBCab"

Output: 0

Explanation:

There are no special characters in word.

Constraints:

1 <= word.length <= 2 * 10⁵
word consists of only lowercase and uppercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string word. A letter c is called special if it appears both in lowercase and uppercase in word, and every lowercase occurrence of c appears before the first uppercase occurrence of c. Return the number of special letters in word.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"aaAbcBC"

Example 2

"abc"

Example 3

"AbBCab"

Core Insight

What unlocks the optimal approach

For each character <code>c</code>, store the first occurrence of its uppercase and the last occurrence of its lowercase.

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 #3121: Count the Number of Special Characters II
class Solution {
    public int numberOfSpecialChars(String word) {
        int[] first = new int['z' + 1];
        int[] last = new int['z' + 1];
        for (int i = 1; i <= word.length(); ++i) {
            int j = word.charAt(i - 1);
            if (first[j] == 0) {
                first[j] = i;
            }
            last[j] = i;
        }
        int ans = 0;
        for (int i = 0; i < 26; ++i) {
            if (last['a' + i] > 0 && first['A' + i] > 0 && last['a' + i] < first['A' + i]) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3121: Count the Number of Special Characters II
func numberOfSpecialChars(word string) (ans int) {
	first := make([]int, 'z'+1)
	last := make([]int, 'z'+1)
	for i, c := range word {
		if first[c] == 0 {
			first[c] = i + 1
		}
		last[c] = i + 1
	}
	for i := 0; i < 26; i++ {
		if last['a'+i] > 0 && first['A'+i] > 0 && last['a'+i] < first['A'+i] {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3121: Count the Number of Special Characters II
class Solution:
    def numberOfSpecialChars(self, word: str) -> int:
        first, last = {}, {}
        for i, c in enumerate(word):
            if c not in first:
                first[c] = i
            last[c] = i
        return sum(
            a in last and b in first and last[a] < first[b]
            for a, b in zip(ascii_lowercase, ascii_uppercase)
        )

// Accepted solution for LeetCode #3121: Count the Number of Special Characters II
// 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 #3121: Count the Number of Special Characters II
// class Solution {
//     public int numberOfSpecialChars(String word) {
//         int[] first = new int['z' + 1];
//         int[] last = new int['z' + 1];
//         for (int i = 1; i <= word.length(); ++i) {
//             int j = word.charAt(i - 1);
//             if (first[j] == 0) {
//                 first[j] = i;
//             }
//             last[j] = i;
//         }
//         int ans = 0;
//         for (int i = 0; i < 26; ++i) {
//             if (last['a' + i] > 0 && first['A' + i] > 0 && last['a' + i] < first['A' + i]) {
//                 ++ans;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3121: Count the Number of Special Characters II
function numberOfSpecialChars(word: string): number {
    const first: number[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => 0);
    const last: number[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => 0);
    for (let i = 0; i < word.length; ++i) {
        const j = word.charCodeAt(i);
        if (first[j] === 0) {
            first[j] = i + 1;
        }
        last[j] = i + 1;
    }
    let ans: number = 0;
    for (let i = 0; i < 26; ++i) {
        if (
            last['a'.charCodeAt(0) + i] &&
            first['A'.charCodeAt(0) + i] &&
            last['a'.charCodeAt(0) + i] < first['A'.charCodeAt(0) + i]
        ) {
            ++ans;
        }
    }
    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 + |\Sigma|)

Space

O(|\Sigma|)

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.

Count the Number of Special Characters II

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Mutating counts without cleanup