Maximum Value of a String in an Array - Practice Coding Problems

Maximum Value of a String in an Array - Problem

Array String Easy

Imagine you're a data analyst working with mixed alphanumeric datasets. You need to evaluate each string based on a unique scoring system:

Numeric strings (containing only digits): Score equals their numeric value in base 10
Alphanumeric strings (containing letters or mixed characters): Score equals their length

Given an array strs of alphanumeric strings, your task is to find and return the maximum score among all strings.

Example:

"123" → Score: 123 (numeric value)
"abc" → Score: 3 (length)
"12a" → Score: 3 (length, since it contains a letter)

Input & Output

example_1.py — Mixed String Types

$ Input: strs = ["alic3", "bob", "3", "4", "00000"]

› Output: 5

💡 Note: "alic3" contains non-digit characters, so its value is its length (5). "bob" has length 3. "3" is numeric with value 3. "4" is numeric with value 4. "00000" is numeric with value 0. Maximum is 5.

example_2.py — All Numeric Strings

$ Input: strs = ["1", "01", "001", "0001"]

› Output: 1

💡 Note: All strings contain only digits. "1" = 1, "01" = 1, "001" = 1, "0001" = 1. All have the same numeric value, so maximum is 1.

example_3.py — Large Numeric Value

$ Input: strs = ["999", "a", "bb", "ccc"]

› Output: 999

💡 Note: "999" is numeric with value 999. "a" has length 1, "bb" has length 2, "ccc" has length 3. The numeric value 999 is much larger than any length.

Constraints

1 ≤ strs.length ≤ 100
1 ≤ strs[i].length ≤ 9
strs[i] consists of only lowercase English letters and digits
All strings are non-empty

Visualization

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Understanding the Visualization

Input Analysis

Receive array of mixed alphanumeric strings

String Classification

For each string, determine if it's purely numeric or contains letters

Value Calculation

Numeric strings → convert to integer, Mixed strings → count characters

Maximum Tracking

Continuously update the highest value found so far

Key Takeaway

🎯 Key Insight: This problem requires distinguishing between purely numeric strings (evaluated by their integer value) and mixed alphanumeric strings (evaluated by their length). A single pass through the array with proper string classification gives us the optimal O(n*m) solution.

Asked in

a Amazon 25 G Google 18 ⊞ Microsoft 15 f Meta 12

The optimal approach uses a single linear scan through the array. For each string, we check if it contains only digits using built-in functions like isdigit(). If numeric, we convert to integer; otherwise, we use the string length. We track the maximum value throughout the iteration. This solution has O(n*m) time complexity where n is the number of strings and m is the average string length, and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Linear Scan)	O(n*m)	O(1)	Check each string individually and track the maximum value

Brute Force (Linear Scan) — Algorithm Steps

Initialize maxValue to 0
For each string in the array:
Check if the string contains only digits
If numeric: convert to integer and use as value
If alphanumeric: use string length as value
Update maxValue if current value is larger
Return maxValue

Visualization

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Step-by-Step Walkthrough

Initialize

Set maxValue = 0

Check String

Examine if string is numeric or alphanumeric

Calculate Value

Convert to int if numeric, else use length

Update Maximum

Keep track of the highest value seen

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

int isNumeric(char* s) {
    int len = strlen(s);
    if (len == 0) return 0;
    
    for (int i = 0; i < len; i++) {
        if (!isdigit(s[i])) return 0;
    }
    return 1;
}

int maximumValue(char** strs, int strsSize) {
    int maxValue = 0;
    
    for (int i = 0; i < strsSize; i++) {
        int value;
        if (isNumeric(strs[i])) {
            value = atoi(strs[i]);
        } else {
            value = strlen(strs[i]);
        }
        
        if (value > maxValue) {
            maxValue = value;
        }
    }
    
    return maxValue;
}

int main() {
    char* strs[] = {"alic3", "bob", "3", "4", "00000"};
    printf("%d\n", maximumValue(strs, 5));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n*m)

n strings, each requiring m characters to check if numeric

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for tracking

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Linear Scan) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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