Solutions

Prerequisites

Before attempting this problem, you should be comfortable with:

Hash Maps - Counting frequency of elements efficiently in O(1) per operation
Sorting - Sorting arrays and iterating in sorted order to find elements with specific properties
Frequency Counting - Tracking how many times each element appears in an array

1. Sorting

Intuition

After sorting in descending order, we can scan from largest to smallest. A number is unique if it differs from its neighbors. We skip over groups of duplicates and return the first number that appears exactly once.

Algorithm

Handle the edge case: if there is only one element, return it.
Sort the array in descending order.
Iterate through the sorted array:
- If the current element differs from the next (or is the last element), it is unique. Return it.
- Otherwise, skip all consecutive duplicates.
If no unique number is found, return -1.

class Solution:
    def largestUniqueNumber(self, nums: List[int]) -> int:
        n = len(nums)

        # If there's only one element, it's unique by default
        if n == 1:
            return nums[0]

        nums.sort(reverse=True)

        # Start from the beginning (largest numbers)
        currentIndex = 0

        while currentIndex < n:
            # If it's the first element or different from the next one, it's unique
            if (
                currentIndex == n - 1
                or nums[currentIndex] != nums[currentIndex + 1]
            ):
                return nums[currentIndex]
            # Skip duplicates
            while (
                currentIndex < n - 1
                and nums[currentIndex] == nums[currentIndex + 1]
            ):
                currentIndex += 1
            # Move to the next unique number
            currentIndex += 1

        return -1

class Solution {
    public int largestUniqueNumber(int[] nums) {
        int n = nums.length;

        // If there's only one element, it's unique by default
        if (n == 1) {
            return nums[0];
        }

        Arrays.sort(nums);

        // Start from the end (largest numbers)
        int currentIndex = n - 1;

        while (currentIndex >= 0) {
            // If it's the first element or different from the previous one, it's unique
            if (
                currentIndex == 0 ||
                nums[currentIndex] != nums[currentIndex - 1]
            ) {
                return nums[currentIndex];
            }

            // Skip duplicates
            while (
                currentIndex > 0 && nums[currentIndex] == nums[currentIndex - 1]
            ) {
                currentIndex--;
            }

            // Move to the next unique number
            currentIndex--;
        }

        return -1;
    }
}

class Solution {
public:
    int largestUniqueNumber(vector<int>& nums) {
        int n = nums.size();

        // If there's only one element, it's unique by default
        if (n == 1) {
            return nums[0];
        }

        sort(nums.begin(), nums.end(), greater<int>());

        // Start from the beginning (largest numbers)
        int currentIndex = 0;

        while (currentIndex < n) {
            // If it's the last element or different from the next one,
            // it's unique
            if (currentIndex == n - 1 ||
                nums[currentIndex] != nums[currentIndex + 1]) {
                return nums[currentIndex];
            }

            // Skip duplicates
            while (currentIndex < n - 1 &&
                   nums[currentIndex] == nums[currentIndex + 1]) {
                currentIndex++;
            }

            // Move to the next unique number
            currentIndex++;
        }

        return -1;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number}
     */
    largestUniqueNumber(nums) {
        const n = nums.length;

        // If there's only one element, it's unique by default
        if (n === 1) {
            return nums[0];
        }

        nums.sort((a, b) => b - a);

        // Start from the beginning (largest numbers)
        let currentIndex = 0;

        while (currentIndex < n) {
            // If it's the first element or different from the next one, it's unique
            if (
                currentIndex === n - 1
                || nums[currentIndex] !== nums[currentIndex + 1]
            ) {
                return nums[currentIndex];
            }

            // Skip duplicates
            while (
                currentIndex < n - 1
                && nums[currentIndex] === nums[currentIndex + 1]
            ) {
                currentIndex++;
            }

            // Move to the next unique number
            currentIndex++;
        }

        return -1;
    }
}

public class Solution {
    public int LargestUniqueNumber(int[] nums) {
        int n = nums.Length;

        // If there's only one element, it's unique by default
        if (n == 1) {
            return nums[0];
        }

        Array.Sort(nums);

        // Start from the end (largest numbers)
        int currentIndex = n - 1;

        while (currentIndex >= 0) {
            // If it's the first element or different from the previous one, it's unique
            if (currentIndex == 0 || nums[currentIndex] != nums[currentIndex - 1]) {
                return nums[currentIndex];
            }

            // Skip duplicates
            while (currentIndex > 0 && nums[currentIndex] == nums[currentIndex - 1]) {
                currentIndex--;
            }

            // Move to the next unique number
            currentIndex--;
        }

        return -1;
    }
}

func largestUniqueNumber(nums []int) int {
    n := len(nums)

    // If there's only one element, it's unique by default
    if n == 1 {
        return nums[0]
    }

    sort.Sort(sort.Reverse(sort.IntSlice(nums)))

    // Start from the beginning (largest numbers)
    currentIndex := 0

    for currentIndex < n {
        // If it's the last element or different from the next one, it's unique
        if currentIndex == n-1 || nums[currentIndex] != nums[currentIndex+1] {
            return nums[currentIndex]
        }

        // Skip duplicates
        for currentIndex < n-1 && nums[currentIndex] == nums[currentIndex+1] {
            currentIndex++
        }

        // Move to the next unique number
        currentIndex++
    }

    return -1
}

class Solution {
    fun largestUniqueNumber(nums: IntArray): Int {
        val n = nums.size

        // If there's only one element, it's unique by default
        if (n == 1) {
            return nums[0]
        }

        nums.sortDescending()

        // Start from the beginning (largest numbers)
        var currentIndex = 0

        while (currentIndex < n) {
            // If it's the last element or different from the next one, it's unique
            if (currentIndex == n - 1 || nums[currentIndex] != nums[currentIndex + 1]) {
                return nums[currentIndex]
            }

            // Skip duplicates
            while (currentIndex < n - 1 && nums[currentIndex] == nums[currentIndex + 1]) {
                currentIndex++
            }

            // Move to the next unique number
            currentIndex++
        }

        return -1
    }
}

class Solution {
    func largestUniqueNumber(_ nums: [Int]) -> Int {
        let n = nums.count

        // If there's only one element, it's unique by default
        if n == 1 {
            return nums[0]
        }

        let sorted = nums.sorted(by: >)

        // Start from the beginning (largest numbers)
        var currentIndex = 0

        while currentIndex < n {
            // If it's the last element or different from the next one, it's unique
            if currentIndex == n - 1 || sorted[currentIndex] != sorted[currentIndex + 1] {
                return sorted[currentIndex]
            }

            // Skip duplicates
            while currentIndex < n - 1 && sorted[currentIndex] == sorted[currentIndex + 1] {
                currentIndex += 1
            }

            // Move to the next unique number
            currentIndex += 1
        }

        return -1
    }
}

Time & Space Complexity

Time complexity: $O(n \cdot \log n)$
Space complexity: $O(S)$ Depends on the language of implementation

Where $n$ is the length of the nums array and $S$ is the sorting algorthm

2. Sorted Map

Intuition

We can count the frequency of each number using a sorted map (tree map). Since the map maintains keys in sorted order, we iterate from the largest key downward and return the first key with frequency 1.

Algorithm

Build a frequency map counting occurrences of each number.
Use a sorted map structure that keeps keys in order.
Iterate through the keys in descending order.
Return the first key with a frequency of 1, or -1 if none exists.

class Solution:
    def largestUniqueNumber(self, nums: List[int]) -> int:
        # Create a frequency map
        frequency_map = {}
        for num in nums:
            frequency_map[num] = frequency_map.get(num, 0) + 1

        # Create a sorted OrderedDict
        sorted_map = OrderedDict(sorted(frequency_map.items(), reverse=True))

        # Find the largest unique number
        for num, freq in sorted_map.items():
            if freq == 1:
                return num

        return -1

class Solution {
    public int largestUniqueNumber(int[] nums) {
        // Use a TreeMap to store numbers and their frequencies
        TreeMap<Integer, Integer> frequencyMap = new TreeMap<>();

        // Populate the frequencyMap
        for (int num : nums) {
            frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);
        }

        // Initialize the result to -1 (default if no unique number is found)
        int largestUnique = -1;

        // Iterate through the map in reverse order (largest to smallest)
        for (Integer num : frequencyMap.descendingKeySet()) {
            // If the frequency is 1, we've found our largest unique number
            if (frequencyMap.get(num) == 1) {
                largestUnique = num;
                break;
            }
        }

        return largestUnique;
    }
}

class Solution {
public:
    int largestUniqueNumber(vector<int>& nums) {
        // Use a map to store numbers and their frequencies
        map<int, int> frequencyMap;

        // Populate the frequencyMap
        for (int num : nums) {
            frequencyMap[num]++;
        }

        // Initialize the result to -1 (default if no unique number is found)
        int largestUnique = -1;

        // Iterate through the map in reverse order (largest to smallest)
        for (auto it = frequencyMap.rbegin(); it != frequencyMap.rend(); ++it) {
            // If the frequency is 1, we've found our largest unique number
            if (it->second == 1) {
                largestUnique = it->first;
                break;
            }
        }

        return largestUnique;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number}
     */
    largestUniqueNumber(nums) {
        // Create a frequency map
        const frequencyMap = {};
        for (const num of nums) {
            frequencyMap[num] = (frequencyMap[num] || 0) + 1;
        }

        // Create a sorted OrderedDict
        const sortedMap = new Map(
            Object.entries(frequencyMap).sort((a, b) => b[0] - a[0])
        );

        // Find the largest unique number
        for (const [num, freq] of sortedMap) {
            if (freq === 1) {
                return Number(num);
            }
        }

        return -1;
    }
}

public class Solution {
    public int LargestUniqueNumber(int[] nums) {
        // Use a SortedDictionary to store numbers and their frequencies
        SortedDictionary<int, int> frequencyMap = new SortedDictionary<int, int>();

        // Populate the frequencyMap
        foreach (int num in nums) {
            if (frequencyMap.ContainsKey(num)) {
                frequencyMap[num]++;
            } else {
                frequencyMap[num] = 1;
            }
        }

        // Initialize the result to -1 (default if no unique number is found)
        int largestUnique = -1;

        // Iterate through the map in reverse order (largest to smallest)
        foreach (int num in frequencyMap.Keys.Reverse()) {
            // If the frequency is 1, we've found our largest unique number
            if (frequencyMap[num] == 1) {
                largestUnique = num;
                break;
            }
        }

        return largestUnique;
    }
}

func largestUniqueNumber(nums []int) int {
    // Create a frequency map
    frequencyMap := make(map[int]int)
    for _, num := range nums {
        frequencyMap[num]++
    }

    // Get all keys and sort them in descending order
    keys := make([]int, 0, len(frequencyMap))
    for k := range frequencyMap {
        keys = append(keys, k)
    }
    sort.Sort(sort.Reverse(sort.IntSlice(keys)))

    // Find the largest unique number
    for _, num := range keys {
        if frequencyMap[num] == 1 {
            return num
        }
    }

    return -1
}

class Solution {
    fun largestUniqueNumber(nums: IntArray): Int {
        // Use a TreeMap to store numbers and their frequencies
        val frequencyMap = sortedMapOf<Int, Int>()

        // Populate the frequencyMap
        for (num in nums) {
            frequencyMap[num] = frequencyMap.getOrDefault(num, 0) + 1
        }

        // Initialize the result to -1 (default if no unique number is found)
        var largestUnique = -1

        // Iterate through the map in reverse order (largest to smallest)
        for (num in frequencyMap.keys.reversed()) {
            // If the frequency is 1, we've found our largest unique number
            if (frequencyMap[num] == 1) {
                largestUnique = num
                break
            }
        }

        return largestUnique
    }
}

class Solution {
    func largestUniqueNumber(_ nums: [Int]) -> Int {
        // Create a frequency map
        var frequencyMap = [Int: Int]()
        for num in nums {
            frequencyMap[num, default: 0] += 1
        }

        // Sort keys in descending order
        let sortedKeys = frequencyMap.keys.sorted(by: >)

        // Find the largest unique number
        for num in sortedKeys {
            if frequencyMap[num] == 1 {
                return num
            }
        }

        return -1
    }
}

Time & Space Complexity

Time complexity: $O(n \cdot \log n)$
Space complexity: $O(n)$

Where $n$ is the length of the nums array.

3. Map

Intuition

Using a regular hash map, we count frequencies in linear time. Then we scan all entries and track the maximum number with frequency 1. This avoids the overhead of maintaining sorted order.

Algorithm

Build a frequency map counting occurrences of each number.
Initialize the result to -1.
Iterate through all entries in the map:
- If frequency equals 1 and the number is larger than the current result, update the result.
Return the result.

class Solution:
    def largestUniqueNumber(self, nums: List[int]) -> int:
        # Use Counter to count frequencies of numbers
        frequency_map = Counter(nums)

        # Find the largest number with frequency 1, or -1 if none found
        return max(
            (num for num, freq in frequency_map.items() if freq == 1),
            default=-1,
        )

class Solution {
    public int largestUniqueNumber(int[] nums) {
        // Create a HashMap to store the frequency of each number
        Map<Integer, Integer> frequencyMap = new HashMap<>();

        // Populate the frequencyMap
        for (int num : nums) {
            frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);
        }

        // Initialize the result to -1 (default if no unique number is found)
        int largestUnique = -1;

        for (int num : frequencyMap.keySet()) {
            // Check if the number appears only once and is larger than the current largestUnique
            if (frequencyMap.get(num) == 1 && num > largestUnique) {
                largestUnique = num;
            }
        }

        return largestUnique;
    }
}

class Solution {
public:
    int largestUniqueNumber(vector<int>& nums) {
        // Create an unordered_map to store the frequency of each number
        unordered_map<int, int> frequencyMap;

        // Populate the frequencyMap
        for (int num : nums) {
            frequencyMap[num]++;
        }

        // Initialize the result to -1 (default if no unique number is found)
        int largestUnique = -1;

        for (auto& pair : frequencyMap) {
            // Check if the number appears only once and is larger than the
            // current largestUnique
            if (pair.second == 1 && pair.first > largestUnique) {
                largestUnique = pair.first;
            }
        }

        return largestUnique;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number}
     */
    largestUniqueNumber(nums) {
        // Create a HashMap to store the frequency of each number
        const frequencyMap = new Map();

        // Populate the frequencyMap
        for (const num of nums) {
            frequencyMap.set(num, (frequencyMap.get(num) || 0) + 1);
        }

        // Initialize the result to -1 (default if no unique number is found)
        let largestUnique = -1;
        for (const num of frequencyMap.keys()) {
            // Check if the number appears only once and is larger than the current largestUnique
            if (frequencyMap.get(num) === 1 && num > largestUnique) {
                largestUnique = num;
            }
        }
        return largestUnique;
    }
}

public class Solution {
    public int LargestUniqueNumber(int[] nums) {
        // Create a Dictionary to store the frequency of each number
        Dictionary<int, int> frequencyMap = new Dictionary<int, int>();

        // Populate the frequencyMap
        foreach (int num in nums) {
            if (frequencyMap.ContainsKey(num)) {
                frequencyMap[num]++;
            } else {
                frequencyMap[num] = 1;
            }
        }

        // Initialize the result to -1 (default if no unique number is found)
        int largestUnique = -1;

        foreach (int num in frequencyMap.Keys) {
            // Check if the number appears only once and is larger than the current largestUnique
            if (frequencyMap[num] == 1 && num > largestUnique) {
                largestUnique = num;
            }
        }

        return largestUnique;
    }
}

func largestUniqueNumber(nums []int) int {
    // Create a map to store the frequency of each number
    frequencyMap := make(map[int]int)

    // Populate the frequencyMap
    for _, num := range nums {
        frequencyMap[num]++
    }

    // Initialize the result to -1 (default if no unique number is found)
    largestUnique := -1

    for num, freq := range frequencyMap {
        // Check if the number appears only once and is larger than the current largestUnique
        if freq == 1 && num > largestUnique {
            largestUnique = num
        }
    }

    return largestUnique
}

class Solution {
    fun largestUniqueNumber(nums: IntArray): Int {
        // Create a HashMap to store the frequency of each number
        val frequencyMap = mutableMapOf<Int, Int>()

        // Populate the frequencyMap
        for (num in nums) {
            frequencyMap[num] = frequencyMap.getOrDefault(num, 0) + 1
        }

        // Initialize the result to -1 (default if no unique number is found)
        var largestUnique = -1

        for ((num, freq) in frequencyMap) {
            // Check if the number appears only once and is larger than the current largestUnique
            if (freq == 1 && num > largestUnique) {
                largestUnique = num
            }
        }

        return largestUnique
    }
}

class Solution {
    func largestUniqueNumber(_ nums: [Int]) -> Int {
        // Create a Dictionary to store the frequency of each number
        var frequencyMap = [Int: Int]()

        // Populate the frequencyMap
        for num in nums {
            frequencyMap[num, default: 0] += 1
        }

        // Initialize the result to -1 (default if no unique number is found)
        var largestUnique = -1

        for (num, freq) in frequencyMap {
            // Check if the number appears only once and is larger than the current largestUnique
            if freq == 1 && num > largestUnique {
                largestUnique = num
            }
        }

        return largestUnique
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

Where $n$ is the length of the nums array.

Common Pitfalls

Returning the First Unique Instead of the Largest

When iterating through numbers, it is tempting to return immediately upon finding a unique number. However, the problem asks for the largest unique number. Without sorting or tracking the maximum, returning early may yield a smaller unique value when a larger one exists.

Confusing Unique with Distinct

A number is unique if it appears exactly once in the array, not if it is different from its neighbors. Using logic that checks for distinct consecutive elements after sorting misses duplicates that are separated by other values in the original array.