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.
Solutions
class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
unordered_map<int, int> freqs;
unordered_map<int, pair<int, int>> ranges;
for(int i = 0; i < nums.size(); ++i){
if(++freqs[nums[i]] == 1){
ranges[nums[i]].first = i;
}
ranges[nums[i]].second = i;
}
int max_freq = 0;
for(auto it = freqs.begin(); it != freqs.end(); ++it){
max_freq = max(max_freq, it->second);
}
int min_len = nums.size();
for(auto it = freqs.begin(); it != freqs.end(); ++it){
if(it->second == max_freq){
auto p = ranges[it->first];
min_len = min(min_len, p.second - p.first + 1);
}
}
return min_len;
}
};
class Solution {
public:
int findShortestSubArray(vector<int>& nums) {
unordered_map<int, vector<int>> hash;
for(int i = 0; i < nums.size(); ++i){
hash[nums[i]].push_back(i);
}
int max_freq = 0;
int min_len = nums.size();
for(auto it = hash.begin(); it != hash.end(); ++it){
if(it->second.size() == max_freq){
min_len = min(min_len, it->second.back() - it->second.front() + 1);
}else if(it->second.size() > max_freq){
max_freq = it->second.size();
min_len = it->second.back() - it->second.front() + 1;
}
}
return min_len;
}
};
