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.Intuition and Algorithm
\nAn 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
\nclass Solution(object):\n def findShortestSubArray(self, nums):\n left, right, count = {}, {}, {}\n for i, x in enumerate(nums):\n if x not in left: left[x] = i\n right[x] = i\n count[x] = count.get(x, 0) + 1\n\n ans = len(nums)\n degree = max(count.values())\n for x in count:\n if count[x] == degree:\n ans = min(ans, right[x] - left[x] + 1)\n\n return ans\n
Java
\nclass Solution {\n public int findShortestSubArray(int[] nums) {\n Map<Integer, Integer> left = new HashMap(),\n right = new HashMap(), count = new HashMap();\n\n for (int i = 0; i < nums.length; i++) {\n int x = nums[i];\n if (left.get(x) == null) left.put(x, i);\n right.put(x, i);\n count.put(x, count.getOrDefault(x, 0) + 1);\n }\n\n int ans = nums.length;\n int degree = Collections.max(count.values());\n for (int x: count.keySet()) {\n if (count.get(x) == degree) {\n ans = Math.min(ans, right.get(x) - left.get(x) + 1);\n }\n }\n return ans;\n }\n}\n
Complexity Analysis
\nTime 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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