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Formatted question description: https://leetcode.ca/all/2496.html

# 2496. Maximum Value of a String in an Array

• Difficulty: Easy.
• Related Topics: .
• Similar Questions: Maximum Subarray.

## Problem

The value of an alphanumeric string can be defined as:

• The numeric representation of the string in base 10, if it comprises of digits only.

• The length of the string, otherwise.

Given an array strs of alphanumeric strings, return the **maximum value of any string in **strs.

Example 1:

Input: strs = ["alic3","bob","3","4","00000"]
Output: 5
Explanation:
- "alic3" consists of both letters and digits, so its value is its length, i.e. 5.
- "bob" consists only of letters, so its value is also its length, i.e. 3.
- "3" consists only of digits, so its value is its numeric equivalent, i.e. 3.
- "4" also consists only of digits, so its value is 4.
- "00000" consists only of digits, so its value is 0.
Hence, the maximum value is 5, of "alic3".


Example 2:

Input: strs = ["1","01","001","0001"]
Output: 1
Explanation:
Each string in the array has value 1. Hence, we return 1.


Constraints:

• 1 <= strs.length <= 100

• 1 <= strs[i].length <= 9

• strs[i] consists of only lowercase English letters and digits.

## Solution (Java, C++, Python)

• class Solution {
public int maximumValue(String[] strs) {
int ans = 0;
for (String s : strs) {
ans = Math.max(ans, f(s));
}
return ans;
}

private int f(String s) {
for (int i = 0; i < s.length(); ++i) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z') {
return s.length();
}
}
return Integer.parseInt(s);
}
}

• class Solution {
public:
int maximumValue(vector<string>& strs) {
auto f = [](string& s) {
int n = s.size(), m = 0;
for (char& c : s) {
if (!isdigit(c)) return n;
m = m * 10 + (c - '0');
}
return m;
};
int ans = 0;
for (auto& s : strs) ans = max(ans, f(s));
return ans;
}
};

• class Solution:
def maximumValue(self, strs: List[str]) -> int:
def f(s):
return int(s) if all(c.isdigit() for c in s) else len(s)

return max(f(s) for s in strs)


• func maximumValue(strs []string) (ans int) {
f := func(s string) int {
n, m := len(s), 0
for _, c := range s {
if c >= 'a' && c <= 'z' {
return n
}
m = m*10 + int(c-'0')
}
return m
}
for _, s := range strs {
if t := f(s); ans < t {
ans = t
}
}
return
}

• function maximumValue(strs: string[]): number {
let ans = 0;
for (const s of strs) {
const num = Number(s);
ans = Math.max(ans, Number.isNaN(num) ? s.length : num);
}
return ans;
}


• impl Solution {
pub fn maximum_value(strs: Vec<String>) -> i32 {
let mut ans = 0;
for s in strs.iter() {
let num = s.parse().unwrap_or(s.len());
ans = ans.max(num);
}
ans as i32
}
}


• public class Solution {
public int MaximumValue(string[] strs) {
return strs.Max(f);
}

private int f(string s) {
int x = 0;
foreach (var c in s) {
if (c >= 'a') {
return s.Length;
}
x = x * 10 + (c - '0');
}
return x;
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).