# 1133. Largest Unique Number

## Description

Given an integer array nums, return the largest integer that only occurs once. If no integer occurs once, return -1.

Example 1:

Input: nums = [5,7,3,9,4,9,8,3,1]
Output: 8
Explanation: The maximum integer in the array is 9 but it is repeated. The number 8 occurs only once, so it is the answer.

Example 2:

Input: nums = [9,9,8,8]
Output: -1
Explanation: There is no number that occurs only once.


Constraints:

• 1 <= nums.length <= 2000
• 0 <= nums[i] <= 1000

## Solutions

Solution 1: Counting + Reverse Traversal

Given the data range in the problem, we can use an array of length $1001$ to count the occurrence of each number. Then, we traverse the array in reverse order to find the first number that appears only once. If no such number is found, we return $-1$.

The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array, and $M$ is the maximum number that appears in the array. In this problem, $M \leq 1000$.

• class Solution {
public int largestUniqueNumber(int[] nums) {
int[] cnt = new int[1001];
for (int x : nums) {
++cnt[x];
}
for (int x = 1000; x >= 0; --x) {
if (cnt[x] == 1) {
return x;
}
}
return -1;
}
}

• class Solution {
public:
int largestUniqueNumber(vector<int>& nums) {
int cnt[1001]{};
for (int& x : nums) {
++cnt[x];
}
for (int x = 1000; ~x; --x) {
if (cnt[x] == 1) {
return x;
}
}
return -1;
}
};

• class Solution:
def largestUniqueNumber(self, nums: List[int]) -> int:
cnt = Counter(nums)
return next((x for x in range(1000, -1, -1) if cnt[x] == 1), -1)


• func largestUniqueNumber(nums []int) int {
cnt := [1001]int{}
for _, x := range nums {
cnt[x]++
}
for x := 1000; x >= 0; x-- {
if cnt[x] == 1 {
return x
}
}
return -1
}

• function largestUniqueNumber(nums: number[]): number {
const cnt = new Array(1001).fill(0);
for (const x of nums) {
++cnt[x];
}
for (let x = 1000; x >= 0; --x) {
if (cnt[x] == 1) {
return x;
}
}
return -1;
}


• /**
* @param {number[]} nums
* @return {number}
*/
var largestUniqueNumber = function (nums) {
const cnt = new Array(1001).fill(0);
for (const x of nums) {
++cnt[x];
}
for (let x = 1000; x >= 0; --x) {
if (cnt[x] == 1) {
return x;
}
}
return -1;
};