697. Degree of an Array

697. Degree of an Array (Easy)

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.

Companies:
Mathworks, VMware, Visa

Related Topics:
Array

Similar Questions:

• Maximum Subarray (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/degree-of-an-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int, int> cnt, left, right;
        int deg = 0, ans = INT_MAX;
        for (int i = 0; i < nums.size(); ++i) {
            int n = nums[i];
            cnt[n]++;
            if (cnt[n] > deg) deg = cnt[n];
            if (cnt[n] == 1) left[n] = i;
            right[n] = i;
        }
        for (auto p : cnt) {
            if (p.second != deg) continue;
            ans = min(ans, right[p.first] - left[p.first] + 1);
        }
        return ans;
    }
};