697. Degree of an Array

Posted 2020-10-13 lvbbg

697. Degree of an Array

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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.

 1 class Solution {
 2 public:
 3     int findShortestSubArray(vector<int>& nums) {
 4         unordered_map<int, pair<int, int>> m1;
 5         unordered_map<int, int> m2;
 6         int count = 0,res = nums.size();
 7         for (int i = 0; i < nums.size(); i++)
 8         {
 9             m2[nums[i]]++;
10             if (m1.find(nums[i]) == m1.end()) m1[nums[i]] = pair<int, int>(i, i);
11             else m1[nums[i]].second = i;
12             if (m2[nums[i]] > count) count = m2[nums[i]];
13         }
14         for (auto num : m2)
15         {
16             if (count == num.second)
17             {
18                 res = min(res, m1[num.first].second - m1[num.first].first + 1);
19             }
20         }
21         return res;
22     }
23 };