LeetCode #2496 — EASY

Maximum Value of a String in an Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Array

Example 1

["alic3","bob","3","4","00000"]

Example 2

["1","01","001","0001"]

Core Insight

What unlocks the optimal approach

For strings comprising only of digits, convert them into integers.
For all other strings, calculate their length.

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 #2496: Maximum Value of a String in an Array
class Solution {
    public int maximumValue(String[] strs) {
        int ans = 0;
        for (var s : strs) {
            ans = Math.max(ans, f(s));
        }
        return ans;
    }

    private int f(String s) {
        int x = 0;
        for (int i = 0, n = s.length(); i < n; ++i) {
            char c = s.charAt(i);
            if (Character.isLetter(c)) {
                return n;
            }
            x = x * 10 + (c - '0');
        }
        return x;
    }
}

// Accepted solution for LeetCode #2496: Maximum Value of a String in an Array
func maximumValue(strs []string) (ans int) {
	f := func(s string) (x int) {
		for _, c := range s {
			if c >= 'a' && c <= 'z' {
				return len(s)
			}
			x = x*10 + int(c-'0')
		}
		return
	}
	for _, s := range strs {
		if x := f(s); ans < x {
			ans = x
		}
	}
	return
}

# Accepted solution for LeetCode #2496: Maximum Value of a String in an Array
class Solution:
    def maximumValue(self, strs: List[str]) -> int:
        def f(s: str) -> int:
            return int(s) if all(c.isdigit() for c in s) else len(s)

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

// Accepted solution for LeetCode #2496: Maximum Value of a String in an Array
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
    }
}

// Accepted solution for LeetCode #2496: Maximum Value of a String in an Array
function maximumValue(strs: string[]): number {
    const f = (s: string) => (Number.isNaN(Number(s)) ? s.length : Number(s));
    return Math.max(...strs.map(f));
}

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(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.