697. Degree of an Array

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: [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: [1,2,2,3,1,4,2]
Output: 6

Note:

nums.length will be between 1 and 50,000.

nums[i] will be an integer between 0 and 49,999.

用兩個hashmap分別紀錄每個數字對應到的次數還有第一次出現的位置。

// One Pass with Two Hashmaps
int findShortestSubArray(vector<int>& nums) { // time: O(n); space: O(n)
    unordered_map<int, int> counter, first;
    int degree = 0, len = INT_MAX;
    for (int i = 0; i < (int)nums.size(); ++i) {
        if (!first.count(nums[i])) first[nums[i]] = i;
        if (++counter[nums[i]] > degree) {
            degree = counter[nums[i]];
            len = i - first[nums[i]] + 1;
        } else if (counter[nums[i]] == degree) {
            len = min(len, i - first[nums[i]] + 1);
        }
    }
    return len;
}

也可以把出現次數和出現位置合併在一個hashmap的value裡，這邊value用vector<int>表示，vector size為3，紀錄{出現次數, 第一次出現位置, 最後一次出現位置}。第一個for loop完可以建構出這個hashmap，第二次loop整個hashmap中的所有entry，然後根據value的vector來判斷。

// Two Pass with One Hashmap
int findShortestSubArray(vector<int>& nums) { // time: O(n); space: O(n)
    unordered_map<int, vector<int> > mp; // num in the array -> <times, first idx, last idx>
    for (int i = 0; i < (int)nums.size(); ++i) {
        if (!mp.count(nums[i])) {
            mp[nums[i]] = {1, i, i};
        } else {
            ++mp[nums[i]][0];
            mp[nums[i]][2] = i;
        }
    }
    int degree = INT_MIN, res = INT_MAX;
    for (auto& e : mp) {
        vector<int> v = e.second;
        if (v[0] > degree) {
            degree = v[0];
            res = v[2] - v[1] + 1;
        } else if (v[0] == degree) {
            res = min(res, v[2] - v[1] + 1);
        }
    }
    return res;
}