Algorithm


Problem Name: 697. Degree of an Array

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.

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming


class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map < int, int>idx, cnt;
        int degree = 0, minlen = nums.size();
        for(int i = 0; i  <  nums.size(); i++){
            if(!idx.count(nums[i])) idx[nums[i]] = i;
            if(++cnt[nums[i]] == degree) minlen = min(minlen, i - idx[nums[i]] + 1);
            if(cnt[nums[i]] > degree){
                degree = cnt[nums[i]];
                minlen = i - idx[nums[i]] + 1;
            }
        }
        return minlen;
    }
};
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Input

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nums = [1,2,2,3,1]

Output

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2

#2 Code Example with Java Programming

Code - Java Programming


class Solution {
  public int findShortestSubArray(int[] nums) {
    Map map = new HashMap<>();
    int maxCount = 0;
    for (int num : nums) {
      map.put(num, map.getOrDefault(num, 0) + 1);
      maxCount = Math.max(maxCount, map.get(num));
    }
    int start = 0;
    int end = 0;
    map.clear();
    int minLength = Integer.MAX_VALUE;
    while (end  <  nums.length) {
      map.put(nums[end], map.getOrDefault(nums[end], 0) + 1);
      while (start <= end && map.get(nums[end]) == maxCount) {
        minLength = Math.min(minLength, end - start + 1);
        map.put(nums[start], map.get(nums[start]) - 1);
        start++;
      }
      end++;
    }
    return minLength;
  }
}
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Input

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nums = [1,2,2,3,1]

Output

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2

#3 Code Example with Javascript Programming

Code - Javascript Programming


const findShortestSubArray = function(nums) {
  const left = {};
  const right = {};
  const count = {};

  for (let i = 0; i  <  nums.length; i++) {
    if (!left.hasOwnProperty(nums[i])) {
      left[nums[i]] = i;
    }
    right[nums[i]] = i;
    count[nums[i]] = count[nums[i]] ? count[nums[i]] + 1 : 1;
  }
  const degree = Math.max(...Object.keys(count).map(el => count[el]));
  let res = nums.length;
  for (let el in count) {
    if (count.hasOwnProperty(el) && count[el] === degree) {
      res = Math.min(res, right[el] - left[el] + 1);
    }
  }

  return res;
};
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Input

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nums = [1,2,2,3,1,4,2]

Output

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6

#4 Code Example with Python Programming

Code - Python Programming


class Solution:
    def findShortestSubArray(self, nums):
        cnt, seen = collections.Counter(nums), collections.defaultdict(list)
        degree = max(cnt.values())
        for i, v in enumerate(nums): seen[v].append(i)
        return min(seen[v][-1] - seen[v][0] + 1 for v in cnt if cnt[v] == degree)
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Input

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nums = [1,2,2,3,1,4,2]

Output

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6

#5 Code Example with C# Programming

Code - C# Programming


using System;
using System.Collections.Generic;
using System.Linq;

namespace LeetCode
{
    public class _0697_DegreeOfAnArray
    {
        public int FindShortestSubArray(int[] nums)
        {
            var counts = new Dictionary < int, int>();
            var left = new Dictionary<int, int>();
            var right = new Dictionary < int, int>();

            for (int i = 0; i  <  nums.Length; i++)
            {
                if (!left.ContainsKey(nums[i]))
                    left[nums[i]] = i;
                right[nums[i]] = i;
                if (!counts.ContainsKey(nums[i]))
                    counts[nums[i]] = 1;
                else
                    counts[nums[i]]++;
            }

            var max = counts.Values.Max();
            var answer = nums.Length;
            foreach (var key in counts.Keys)
            {
                if (counts[key] == max)
                {
                    answer = Math.Min(answer, right[key] - left[key] + 1);
                }
            }

            return answer;
        }
    }
}
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Input

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nums = [1,2,2,3,1]

Output

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2
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