leetCode-Degree of an Array

Posted 2020-10-15 kevincong

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: [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.

My Solution:

class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer,Integer>map = new HashMap<Integer,Integer>();
        int len = nums.length;
        for(int i = 0;i < len;i++){
            map.put(nums[i],(map.get(nums[i]) == null)?0:map.get(nums[i])+1);
        }
        int maxCount = 0;
        List<Integer>list = new ArrayList<Integer>();
        for(Integer key : map.keySet()){
            if(map.get(key) > maxCount){
                maxCount = map.get(key);
            }
        }
        for(Integer key : map.keySet()){
            if(map.get(key) == maxCount){
                list.add(key);
            }
        }
        List<Integer> temp = new ArrayList<Integer>();
        for(int i:nums){
            temp.add(Integer.valueOf(i));
        }
        int shortest = temp.lastIndexOf(list.get(0)) - temp.indexOf(list.get(0)) + 1;
        for(int i = 0;i < list.size();i++){
            int first = temp.indexOf(list.get(i));
            int last = temp.lastIndexOf(list.get(i));
            int length = last - first +1;
            if(length < shortest){
                shortest = length;
            }
        }
        return shortest;
    }
}

Better Solution:
class Solution {
    public int findShortestSubArray(int[] nums) {
           if (nums.length == 0 || nums == null) return 0;
        Map<Integer, int[]> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++){
           if (!map.containsKey(nums[i])){
               map.put(nums[i], new int[]{1, i, i});  // the first element in array is degree, second is first index of this key, third is last index of this key
           } else {
               int[] temp = map.get(nums[i]);
               temp[0]++;
               temp[2] = i;
           }
        }
        int degree = Integer.MIN_VALUE, res = Integer.MAX_VALUE;
        for (int[] value : map.values()){
            if (value[0] > degree){
                degree = value[0];
                res = value[2] - value[1] + 1;
            } else if (value[0] == degree){
                res = Math.min( value[2] - value[1] + 1, res);
            } 
        }
        return res;
    }
}

Best Solution:

class Solution {
    public int findShortestSubArray(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        int maxNum = 0;
        for (int n : nums){
            maxNum = Math.max(n, maxNum);
        }
        //由于数组的元素都为正整数,因此int[maxNum + 1]的下标可以包含数组中所有元素
        //start以nums的元素为下标，保存对应的起始点
        int[] start = new int[maxNum + 1];
        //end以nums的元素为下标，保存对应的终止点
        int[] end = new int[maxNum + 1];
        //que以nums的元素为下标，保存对应的次数
        int[] que = new int[maxNum + 1];

        for (int i = 0; i < nums.length; i++){
        //第一次出现nums[i]时，将i存入start,即起始点
            if (que[nums[i]] == 0)
                start[nums[i]] = i;
            //每次出现同一个nums元素，更新end数组,即终止点
            end[nums[i]] = i;
            //每次出现同一个nums元素，更新que数组，计数
            que[nums[i]]++;
        }

        int max = 0;
        //找出最大degree
        for (int n : que)
            max = Math.max(max, n);

        List<Integer> maxNums = new ArrayList<>();
        #找出que中拥有最大计数的下标
        for (int i = 0; i < que.length; i++){
            if (que[i] == max)
                maxNums.add(i);
        }

        int res = nums.length;
        #遍历maxNums,找出满足degree的最小子集长度
        for (int n : maxNums){
            int r = end[n] - start[n] + 1;
            res = Math.min(r, res);
        }

        return res;
    }

 

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