697.degree-of-an-array.py

#
# @lc app=leetcode id=697 lang=python3
#
# [697] Degree of an Array
#
# https://leetcode.com/problems/degree-of-an-array/description/
#
# algorithms
# Easy (55.93%)
# Likes:    2644
# Dislikes: 1546
# Total Accepted:    179.3K
# Total Submissions: 320.5K
# Testcase Example:  '[1,2,2,3,1]'
#
# 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: nums = [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: nums = [1,2,2,3,1,4,2]
# Output: 6
# Explanation: 
# The degree is 3 because the element 2 is repeated 3 times.
# So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
# 
# 
# 
# Constraints:
# 
# 
# nums.length will be between 1 and 50,000.
# nums[i] will be an integer between 0 and 49,999.
# 
# 
#

# @lc code=start
class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        count = defaultdict(int)
        left = {}
        right = {}
        degree = 0
        
        for i, num in enumerate(nums):
            count[num] += 1
            if num not in left:
                left[num] = i
            right[num] = i
            degree = max(degree, count[num])
        
        ans = len(nums)
        for num in count:
            if count[num] == degree:
                ans = min(ans, right[num] - left[num] + 1)
        
        return ans
        
# @lc code=end