leetcode

Second Largest Digit in a String - Link

Question Description

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

A numerical digit is a character in the range '0' to '9'.

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Constraints

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

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Approach

Use a single pass through the string to track the maximum and second maximum digits found.

Algorithm:

1. Initialize max = -1 and secondMax = -1
2. Iterate through each character in the string
3. If the character is a digit, convert it to integer value
4. Update tracking variables:
   • If current digit > max: update secondMax = max, then max = current digit
   • If current digit < max but > secondMax: update secondMax = current digit
5. Return secondMax

This approach works because:

• Single pass ensures O(n) time complexity
• Only two variables needed for constant space
• Handles duplicates correctly (ignores them after first occurrence)
• Returns -1 if no second largest digit exists

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Dry Run

Example 1: s = "dfa12321afd"

Step-by-step execution:

• ch=’d’: not digit, skip
• ch=’f’: not digit, skip
• ch=’a’: not digit, skip
• ch=’1’: digit, num=1, 1 > -1, so secondMax=-1, max=1
• ch=’2’: digit, num=2, 2 > 1, so secondMax=1, max=2
• ch=’3’: digit, num=3, 3 > 2, so secondMax=2, max=3
• ch=’2’: digit, num=2, 2 < 3 && 2 == 2, no update (duplicate)
• ch=’1’: digit, num=1, 1 < 3 && 1 < 2, no update
• ch=’a’: not digit, skip
• ch=’f’: not digit, skip
• ch=’d’: not digit, skip

Result: secondMax = 2

Example 2: s = "abc"

Step-by-step execution:

• No digits found
• max = -1, secondMax = -1

Result: -1

Example 3: s = "a1b"

Step-by-step execution:

• ch=’a’: not digit, skip
• ch=’1’: digit, num=1, 1 > -1, so secondMax=-1, max=1
• ch=’b’: not digit, skip

Result: -1 (only one unique digit)

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Solution

class Solution {
    public int secondHighest(String s) {
        int max = -1;
        int sMax = -1;

        for (char ch : s.toCharArray()) {
            if (Character.isDigit(ch)) {
                int num = ch - '0';
                if (num > max) {
                    sMax = max;
                    max = num;
                } else if (num < max && num > sMax) {
                    sMax = num;
                }
            }
        }
        return sMax;
    }
}

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Time and Space Complexity

• Time Complexity: O(n) - single pass through the string
• Space Complexity: O(1) - only using constant extra space for variables

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Alternative Approaches

Using Set Approach (Handles Duplicates Explicitly)

class Solution {
    public int secondHighest(String s) {
        Set<Integer> digits = new HashSet<>();

        for (char ch : s.toCharArray()) {
            if (Character.isDigit(ch)) {
                digits.add(ch - '0');
            }
        }

        if (digits.size() < 2) {
            return -1;
        }

        // Find the largest digit
        int max = Collections.max(digits);

        // Remove the largest and find the new maximum
        digits.remove(max);

        return Collections.max(digits);
    }
}

Explanation: Use a Set to collect unique digits, then find the maximum twice.

Time Complexity: O(n) - single pass plus set operations Space Complexity: O(k) - where k is number of unique digits (≤ 10)

Stream API Approach (Java 8+)

class Solution {
    public int secondHighest(String s) {
        int max = s.chars()
                  .filter(Character::isDigit)
                  .map(c -> c - '0')
                  .distinct()
                  .max()
                  .orElse(-1);

        int secondMax = s.chars()
                        .filter(Character::isDigit)
                        .map(c -> c - '0')
                        .distinct()
                        .filter(d -> d < max)
                        .max()
                        .orElse(-1);

        return secondMax;
    }
}

Explanation: Use Java streams to process digits functionally.

Time Complexity: O(n) - stream processing Space Complexity: O(k) - for distinct elements

ASCII-based Approach

class Solution {
    public int secondHighest(String s) {
        int max = -1;
        int secondMax = -1;

        for (char ch : s.toCharArray()) {
            if (ch >= '0' && ch <= '9') {
                int num = ch - '0';
                if (num > max) {
                    secondMax = max;
                    max = num;
                } else if (num > secondMax && num < max) {
                    secondMax = num;
                }
            }
        }
        return secondMax;
    }
}

Explanation: Same logic but using ASCII range check instead of Character.isDigit().

Time Complexity: O(n) - single pass Space Complexity: O(1) - constant space

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