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.class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
int n = nums.size();
unordered_map<int,int> count,first;
int degree = 0, res = INT_MAX;
for(int i=0;i<n;i++)
{
count[nums[i]]++;
if(first.find(nums[i])==first.end())
first[nums[i]] = i;
if(count[nums[i]]>degree)
{
degree = count[nums[i]];
res = i-first[nums[i]]+1;
}
if(count[nums[i]]==degree)
{
res = min(res,i-first[nums[i]]+1);
}
}
return res;
}
};