Largest Unique Number

Easy · Arrays & Hashing · Explanation and code example

Map

How it works

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.

Steps

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

Complexity

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

Where n is the length of the nums array.

Things to watch for

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.

Python code example

from typing import List
from collections import Counter

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,
        )

C++ code example

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;
    }
};
Source and license

Adapted from NeetCode’s source · MIT license. Node types are supplied by LeetCode where needed.

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