LeetCode 619: Biggest Single Number Solution in Python – A Step-by-Step Guide

What Is LeetCode 619: Biggest Single Number?

In LeetCode 619: Biggest Single Number, you’re given a table my_numbers with a single column num, containing integers. Your task is to find the largest number that appears exactly once in the table, returning null (None in Python) if no such number exists. For example, in [5, 7, 3, 7], 5 and 3 appear once, so 5 is the biggest single number; in [1, 1, 2, 2], all numbers repeat, so you return null. This problem tests your ability to count frequencies and filter data, typically posed as an SQL query with GROUP BY and HAVING, but we’ll solve it in Python for a coding twist.

Problem Statement

• Input: A list of records (or table rows) with [num].
• Output: An integer (the largest number appearing once) or None if none exist.
• Rules:
• Find numbers with frequency 1.
• Return the maximum among them, or null if none.

Constraints

• Table has at least 1 row.
• 0 ≤ num ≤ 1000.
• Number of rows ≤ 10⁴.

Examples

• Input:
``` num 5 7 3 7 ```
• Output: 5 (5 and 3 appear once, 5 > 3)
• Input:
``` num 1 1 2 2 ```
• Output: None (No number appears once)
• Input:
``` num 8 ```
• Output: 8 (Single number)

These examples show the goal—let’s solve it!

Understanding the Problem: Finding the Lone Giant

To solve LeetCode 619: Biggest Single Number in Python, we need to count how often each number appears in the list, filter for those appearing exactly once, and find the largest among them—or return None if none qualify. A naive approach might check each number against all others, but we can optimize with frequency counting. We’ll use:

• Best Solution (Dictionary-Based Frequency Counting): O(n) time, O(n) space—fast and efficient (n = number of rows).
• Alternative Solution (Sorting with Iteration): O(n log n) time, O(n) space—visual but slower.

Let’s start with the dictionary-based solution, breaking it down for beginners.

Best Solution: Dictionary-Based Frequency Counting

Why This Is the Best Solution

The dictionary-based frequency counting approach is the top choice for LeetCode 619 because it’s efficient—O(n) time with O(n) space—and uses a dictionary to count occurrences in one pass, then quickly identifies the largest single number. It’s straightforward, scalable within constraints (≤ 10⁴ rows), and mimics SQL’s HAVING COUNT = 1 logic. It’s like tallying votes and picking the biggest solo winner!

How It Works

Think of this as counting attendance:

• Step 1: Build Frequency Dictionary:
• Map each number to its count in the list.
• Step 2: Filter Singles:
• Find numbers with count 1.
• Step 3: Find Maximum:
• Get the largest number from the singles, or None if empty.
• Step 4: Return Result:
• Output the result.

It’s like a roll call—count and spot the lone star!

Step-by-Step Example

Example:

Copy codenum
5
7
3
7

• Step 1: Frequency dict:
• {5: 1, 7: 2, 3: 1}.
• Step 2: Singles:
• 5: 1, 3: 1.
• Step 3: Max:
• max(5, 3) = 5.
• Output: 5.

Example:

Copy codenum
1
1
2
2

• Step 1: {1: 2, 2: 2}.
• Step 2: No singles.
• Step 3: None.
• Output: None.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codefrom collections import defaultdict

class Solution:
    def biggestSingleNumber(self, my_numbers: List[List[int]]) -> int:
        # Step 1: Build frequency dictionary
        freq = defaultdict(int)
        for row in my_numbers:
            num = row[0]
            freq[num] += 1

        # Step 2: Find biggest single number
        max_single = None
        for num, count in freq.items():
            if count == 1:
                if max_single is None or num > max_single:
                    max_single = num

        # Step 3: Return result
        return max_single

• Line 1: Import defaultdict for ease.
• Lines 6-9: Count frequencies:
• Increment count for each num.
• Lines 12-16: Filter and find max:
• Track largest number with count 1.
• Time Complexity: O(n)—one pass to build dict, one to find max.
• Space Complexity: O(n)—store frequencies.

This is like a number counter—fast and simple!

Alternative Solution: Sorting with Iteration

Why an Alternative Approach?

The sorting-based approach sorts the numbers first—O(n log n) time, O(n) space. It’s more visual, allowing you to spot singles by checking adjacent duplicates, but it’s less efficient due to sorting overhead. It’s like lining up numbers and scanning for loners!

How It Works

Picture this as a number lineup:

• Step 1: Extract and sort numbers.
• Step 2: Check for singles:
• Iterate, comparing with neighbors.
• Step 3: Track max single number.
• Step 4: Return result.

It’s like a sorted roll call—spot the solos!

Step-by-Step Example

Example:

Copy codenum
5
7
3
7

• Step 1: Sort: [3, 5, 7, 7].
• Step 2: Check:
• 3: No left, right=5, single, max=3.
• 5: Left=3, right=7, single, max=5.
• 7: Left=5, right=7, not single.
• 7: Left=7, no right, not single.
• Step 3: Max single = 5.
• Output: 5.

Code for Sorting Approach

Copy codeclass Solution:
    def biggestSingleNumber(self, my_numbers: List[List[int]]) -> int:
        # Step 1: Extract and sort numbers
        nums = [row[0] for row in my_numbers]
        nums.sort()
        n = len(nums)

        # Step 2: Handle edge cases
        if n == 0:
            return None
        if n == 1:
            return nums[0]

        # Step 3: Find biggest single number
        max_single = None
        for i in range(n):
            # Check if single (not equal to neighbors)
            is_single = True
            if i > 0 and nums[i] == nums[i-1]:
                is_single = False
            if i < n-1 and nums[i] == nums[i+1]:
                is_single = False
            if is_single:
                if max_single is None or nums[i] > max_single:
                    max_single = nums[i]

        # Step 4: Return result
        return max_single

• Line 4: Extract and sort numbers.
• Lines 7-11: Handle edge cases.
• Lines 14-23: Check for singles:
• Compare with left and right neighbors.
• Time Complexity: O(n log n)—sorting dominates.
• Space Complexity: O(n)—store sorted list.

It’s a sorted number scan—clear but slower!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n) time, O(n) space, fast and clean.
• Cons: Less visual.
• Sorting (Alternative):
• Pros: O(n log n) time, O(n) space, intuitive with sorted view.
• Cons: Slower due to sort.

Dictionary wins for efficiency.

Additional Examples and Edge Cases

• Input: [[8]]
• Output: 8.
• Input: [[1], [1]]
• Output: None.

Both handle these well.

Complexity Breakdown

• Dictionary: Time O(n), Space O(n).
• Sorting: Time O(n log n), Space O(n).

Dictionary is optimal.

Key Takeaways

• Dictionary: Quick counting—smart!
• Sorting: Visual scan—clear!
• Data: Frequencies are fun.
• Python Tip: Dicts rock—see Python Basics.

Final Thoughts: Spot That Number

LeetCode 619: Biggest Single Number in Python is a neat frequency challenge. Dictionary-based counting offers speed and simplicity, while sorting provides a visual twist. Want more? Try LeetCode 136: Single Number or LeetCode 389: Find the Difference. Ready to hunt? Head to Solve LeetCode 619 on LeetCode and find that biggest single number today!