Approach #1: Left and Right Index [Accepted]
Intuition and Algorithm
An array that has degree d
, must have some element x
occur d
times. If some subarray has the same degree, then some element x
(that occured d
times), still occurs d
times. The shortest such subarray would be from the first occurrence of x
until the last occurrence.
For each element in the given array, let's know left
, the index of its first occurrence; and right
, the index of its last occurrence. For example, with nums = [1,2,3,2,5]
we have left[2] = 1
and right[2] = 3
.
Then, for each element x
that occurs the maximum number of times, right[x]  left[x] + 1
will be our candidate answer, and we'll take the minimum of those candidates.
Python
class Solution(object): def findShortestSubArray(self, nums): left, right, count = {}, {}, {} for i, x in enumerate(nums): if x not in left: left[x] = i right[x] = i count[x] = count.get(x, 0) + 1 ans = len(nums) degree = max(count.values()) for x in count: if count[x] == degree: ans = min(ans, right[x]  left[x] + 1) return ans
Java
class Solution { public int findShortestSubArray(int[] nums) { Map<Integer, Integer> left = new HashMap(), right = new HashMap(), count = new HashMap(); for (int i = 0; i < nums.length; i++) { int x = nums[i]; if (left.get(x) == null) left.put(x, i); right.put(x, i); count.put(x, count.getOrDefault(x, 0) + 1); } int ans = nums.length; int degree = Collections.max(count.values()); for (int x: count.keySet()) { if (count.get(x) == degree) { ans = Math.min(ans, right.get(x)  left.get(x) + 1); } } return ans; } }
Complexity Analysis

Time Complexity: , where is the length of
nums
. Every loop is through items with work inside the forblock. 
Space Complexity: , the space used by
left
,right
, andcount
.
Analysis written by: @awice.