Question

1133. Largest Unique Number

中文文档

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:
  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)

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;
  }
};

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;
};

Solution 2

class Solution:
  def largestUniqueNumber(self, nums: List[int]) -> int:
    cnt = Counter(nums)
    return max((x for x, v in cnt.items() if v == 1), default=-1)