# 1796. Second Largest Digit in a String

## Description

Given an alphanumeric string s, return the second largest numerical digit that appears in s, or -1 if it does not exist.

An alphanumeric string is a string consisting of lowercase English letters and digits.

Example 1:

Input: s = "dfa12321afd"
Output: 2
Explanation: The digits that appear in s are [1, 2, 3]. The second largest digit is 2.


Example 2:

Input: s = "abc1111"
Output: -1
Explanation: The digits that appear in s are [1]. There is no second largest digit.


Constraints:

• 1 <= s.length <= 500
• s consists of only lowercase English letters and/or digits.

## Solutions

Solution 1: One Pass

We define $a$ and $b$ to represent the largest and second largest numbers in the string, initially $a = b = -1$.

We traverse the string $s$. If the current character is a digit, we convert it to a number $v$. If $v > a$, it means that $v$ is the largest number currently appearing, we update $b$ to $a$, and update $a$ to $v$; if $v < a$, it means that $v$ is the second largest number currently appearing, we update $b$ to $v$.

After the traversal, we return $b$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

Solution 2: Bit Manipulation

We can use an integer $mask$ to mark the numbers that appear in the string, where the $i$-th bit of $mask$ indicates whether the number $i$ has appeared.

We traverse the string $s$. If the current character is a digit, we convert it to a number $v$, and set the $v$-th bit of $mask$ to $1$.

Finally, we traverse $mask$ from high to low, find the second bit that is $1$, and the corresponding number is the second largest number. If there is no second largest number, return $-1$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

• class Solution {
public int secondHighest(String s) {
int a = -1, b = -1;
for (int i = 0; i < s.length(); ++i) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
int v = c - '0';
if (v > a) {
b = a;
a = v;
} else if (v > b && v < a) {
b = v;
}
}
}
return b;
}
}

• class Solution {
public:
int secondHighest(string s) {
int a = -1, b = -1;
for (char& c : s) {
if (isdigit(c)) {
int v = c - '0';
if (v > a) {
b = a, a = v;
} else if (v > b && v < a) {
b = v;
}
}
}
return b;
}
};

• class Solution:
def secondHighest(self, s: str) -> int:
a = b = -1
for c in s:
if c.isdigit():
v = int(c)
if v > a:
a, b = v, a
elif b < v < a:
b = v
return b


• func secondHighest(s string) int {
a, b := -1, -1
for _, c := range s {
if c >= '0' && c <= '9' {
v := int(c - '0')
if v > a {
b, a = a, v
} else if v > b && v < a {
b = v
}
}
}
return b
}

• function secondHighest(s: string): number {
let first = -1;
let second = -1;
for (const c of s) {
if (c >= '0' && c <= '9') {
const num = c.charCodeAt(0) - '0'.charCodeAt(0);
if (first < num) {
[first, second] = [num, first];
} else if (first !== num && second < num) {
second = num;
}
}
}
return second;
}


• impl Solution {
pub fn second_highest(s: String) -> i32 {
let mut first = -1;
let mut second = -1;
for c in s.as_bytes() {
if char::is_digit(*c as char, 10) {
let num = (c - b'0') as i32;
if first < num {
second = first;
first = num;
} else if num < first && second < num {
second = num;
}
}
}
second
}
}