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697. Degree of an Array

Description

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.

Solutions

  • class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        Map<Integer, Integer> left = new HashMap<>();
        Map<Integer, Integer> right = new HashMap<>();
        int degree = 0;
        for (int i = 0; i < nums.length; ++i) {
            int v = nums[i];
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
            degree = Math.max(degree, cnt.get(v));
            if (!left.containsKey(v)) {
                left.put(v, i);
            }
            right.put(v, i);
        }
        int ans = 1000000;
        for (int v : nums) {
            if (cnt.get(v) == degree) {
                int t = right.get(v) - left.get(v) + 1;
                if (ans > t) {
                    ans = t;
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int, int> cnt;
        unordered_map<int, int> left;
        unordered_map<int, int> right;
        int degree = 0;
        for (int i = 0; i < nums.size(); ++i) {
            int v = nums[i];
            degree = max(degree, ++cnt[v]);
            if (!left.count(v)) {
                left[v] = i;
            }
            right[v] = i;
        }
        int ans = 1e6;
        for (int v : nums) {
            if (cnt[v] == degree) {
                int t = right[v] - left[v] + 1;
                if (ans > t) {
                    ans = t;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        degree = cnt.most_common()[0][1]
        left, right = {}, {}
        for i, v in enumerate(nums):
            if v not in left:
                left[v] = i
            right[v] = i
        ans = inf
        for v in nums:
            if cnt[v] == degree:
                t = right[v] - left[v] + 1
                if ans > t:
                    ans = t
        return ans


  • func findShortestSubArray(nums []int) (ans int) {
	ans = 50000
	numsMap := make(map[int]int, len(nums))
	for _, num := range nums {
		numsMap[num]++
	}
	var maxDegree int
	for _, num := range numsMap {
		maxDegree = max(num, maxDegree)
	}
	degreeNums := getMaxDegreeElem(maxDegree, numsMap)
	for _, num := range degreeNums {
		f := findSubArray(num, nums)
		ans = min(ans, f)
	}
	return
}

func findSubArray(target int, nums []int) int {
	start := getStartIdx(target, nums)
	end := getEndIdx(target, nums)
	return (end - start) + 1
}

func getStartIdx(target int, nums []int) (start int) {
	for idx, num := range nums {
		if num == target {
			start = idx
			break
		}
	}
	return start
}

func getEndIdx(target int, nums []int) (end int) {
	for i := len(nums) - 1; i > 0; i-- {
		if nums[i] == target {
			end = i
			break
		}
	}
	return
}

func getMaxDegreeElem(maxDegree int, numsMap map[int]int) []int {
	var ans []int
	for key, value := range numsMap {
		if value == maxDegree {
			ans = append(ans, key)
		}
	}
	return ans
}

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