Largest Unique Number

easy array hash table counting

Problem

You are given an integer array nums. Return the largest integer that appears exactly once in the array. If no integer appears exactly once, return -1.

Inputnums = [5, 7, 3, 9, 4, 9, 8, 3, 1]

Output8

The values appearing exactly once are 5, 7, 4, 8, and 1; the largest of these is 8.

nums (comma-separated)

def largest_unique_number(nums):
    count = {}
    for x in nums:
        count[x] = count.get(x, 0) + 1
    ans = -1
    for x, c in count.items():
        if c == 1 and x > ans:
            ans = x
    return ans

function largestUniqueNumber(nums) {
  const count = new Map();
  for (const x of nums) count.set(x, (count.get(x) || 0) + 1);
  let ans = -1;
  for (const [x, c] of count) {
    if (c === 1 && x > ans) ans = x;
  }
  return ans;
}

class Solution {
    public int largestUniqueNumber(int[] nums) {
        Map<Integer, Integer> count = new HashMap<>();
        for (int x : nums) count.put(x, count.getOrDefault(x, 0) + 1);
        int ans = -1;
        for (Map.Entry<Integer, Integer> e : count.entrySet()) {
            if (e.getValue() == 1 && e.getKey() > ans) ans = e.getKey();
        }
        return ans;
    }
}

int largestUniqueNumber(vector<int>& nums) {
    unordered_map<int, int> count;
    for (int x : nums) count[x]++;
    int ans = -1;
    for (auto& e : count) {
        if (e.second == 1 && e.first > ans) ans = e.first;
    }
    return ans;
}

Explanation

We want the biggest number that appears exactly once. The two things we care about for each value are: how often does it appear, and how big is it. A frequency map handles the first part perfectly.

First pass: walk the array and tally counts into count, so count[x] tells us how many times x shows up.

Second pass: start with ans = -1 (the answer when nothing qualifies). Look at each distinct value x with count c; if c == 1 it is unique, and if it is also bigger than the best we have, we update ans = x.

Example: nums = [5,7,3,9,4,9,8,3,1]. The 9s and 3s repeat, so they are out. The values appearing once are 5, 7, 4, 8, 1, and the largest among them is 8.

If no value has a count of 1, ans never changes and we correctly return -1.

Time: O(n) Space: O(n)