697. Degree of an Array


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.


    • 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.

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