Monday, September 14, 2020

LeetCode [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: 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.
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class Solution {
    public int findShortestSubArray(int[] nums) {
        int n = nums.length;
        int[][] map = new int[50000][3];//startIndex, endIndex, count
        int ret = n, degree = 0;
        for(int i=0; i<n; ++i){
            int v = nums[i];
            if(map[v][2]==0) map[v][0] = i;//startIndex
            map[v][1] = i;//endIndex
            map[v][2]++;//count
        
            if(map[v][2]>degree){
                degree = map[v][2];
                ret = map[v][1]-map[v][0]+1;
            }else if(map[v][2]==degree && map[v][1]-map[v][0]+1<ret){
                ret = map[v][1]-map[v][0]+1;
            }
        }
        return ret;
    }
}

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