697. Degree of an Array

📋 Đề Bài

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:

Input: nums = [1,2,2,3,1]
Output: 2
Explanation: 
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.

Example 2:

Input: nums = [1,2,2,3,1,4,2]
Output: 6
Explanation: 
The degree is 3 because the element 2 is repeated 3 times.
So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.

Constraints:

nums.length will be between 1 and 50,000.
nums[i] will be an integer between 0 and 49,999.

🧠 Thuật Toán & Kỹ Thuật

Hash Table (Bảng băm)

⏱️ Thời gian O(n)

💾 Không gian O(n)

💻 Lời Giải

C++ 0697-degree-of-an-array.cpp

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int maxFreq = 0;
        const int n = nums.size();
        int cnt[50000] = {};
        unordered_map<int, int> indexs;
        int res = INT_MAX;
        
        for (int i = 0; i < n; ++i) {
            cnt[nums[i]]++;
            if (!indexs.count(nums[i])) {
                indexs[nums[i]] = i;
            }
            if (maxFreq == cnt[nums[i]]) {
                res = min(res, i - indexs[nums[i]] + 1);
            }
            else if (maxFreq < cnt[nums[i]]) {
                maxFreq = cnt[nums[i]];
                res = i - indexs[nums[i]] + 1;
            }
        }
        
        return res;
    }
};