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.
class Solution:
def findShortestSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
first, counter, res, degree = {}, {}, 0, 0
for i, v in enumerate(nums):
first.setdefault(v, i)
counter[v] = counter.get(v, 0) + 1
if counter[v] > degree:
degree = counter[v]
res = i - first[v] + 1
elif counter[v] == degree:
res = min(res, i - first[v]+1)
return res
my answer(Wrong):
temp = sorted(nums)
i = 0
j = 0
freq={}
count = 0
while(j <len(temp)):
if temp[i] == temp[j]:
count += 1
freq[str(temp[i])] = count
j += 1
else:
i = j
count = 1
freq[str(temp[i])] = count
j += 1
degree = max(freq.items(), key = lambda x : x[1])[1]
# element = int(max(freq.items(), key = lambda x : x[1])[0])
elements = []
for item in freq:
if freq[item] == degree:
elements.append(int(item))
flag = []
min = 50000
for element in elements:
for k in range(len(nums)):
if nums[k] == element:
flag.append(k)
length = flag[-1] - flag[0] +1
flag.clear()
if min > length:
min = length
return min(run time error)