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.
Build confidence with an intuition-first walkthrough focused on hash map fundamentals.
You are given a string word. A letter is called special if it appears both in lowercase and uppercase in word.
Return the number of special letters in word.
Example 1:
Input: word = "aaAbcBC"
Output: 3
Explanation:
The special characters in word are 'a', 'b', and 'c'.
Example 2:
Input: word = "abc"
Output: 0
Explanation:
No character in word appears in uppercase.
Example 3:
Input: word = "abBCab"
Output: 1
Explanation:
The only special character in word is 'b'.
Constraints:
1 <= word.length <= 50word consists of only lowercase and uppercase English letters.Problem summary: You are given a string word. A letter is called special if it appears both in lowercase and uppercase in word. Return the number of special letters in word.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Hash Map
"aaAbcBC"
"abc"
"abBCab"
detect-capital)greatest-english-letter-in-upper-and-lower-case)count-the-number-of-special-characters-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3120: Count the Number of Special Characters I
class Solution {
public int numberOfSpecialChars(String word) {
boolean[] s = new boolean['z' + 1];
for (int i = 0; i < word.length(); ++i) {
s[word.charAt(i)] = true;
}
int ans = 0;
for (int i = 0; i < 26; ++i) {
if (s['a' + i] && s['A' + i]) {
++ans;
}
}
return ans;
}
}
// Accepted solution for LeetCode #3120: Count the Number of Special Characters I
func numberOfSpecialChars(word string) (ans int) {
s := make([]bool, 'z'+1)
for _, c := range word {
s[c] = true
}
for i := 0; i < 26; i++ {
if s['a'+i] && s['A'+i] {
ans++
}
}
return
}
# Accepted solution for LeetCode #3120: Count the Number of Special Characters I
class Solution:
def numberOfSpecialChars(self, word: str) -> int:
s = set(word)
return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))
// Accepted solution for LeetCode #3120: Count the Number of Special Characters I
// 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 #3120: Count the Number of Special Characters I
// class Solution {
// public int numberOfSpecialChars(String word) {
// boolean[] s = new boolean['z' + 1];
// for (int i = 0; i < word.length(); ++i) {
// s[word.charAt(i)] = true;
// }
// int ans = 0;
// for (int i = 0; i < 26; ++i) {
// if (s['a' + i] && s['A' + i]) {
// ++ans;
// }
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #3120: Count the Number of Special Characters I
function numberOfSpecialChars(word: string): number {
const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
for (let i = 0; i < word.length; ++i) {
s[word.charCodeAt(i)] = true;
}
let ans: number = 0;
for (let i = 0; i < 26; ++i) {
if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
++ans;
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.