#### 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 for-block.

• Space Complexity: , the space used by left, right, and count.

Analysis written by: @awice.