Problem

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.

Pre analysis

Will apply heuristic approach and keep track of longest subarray and its length. Will use dictionary to keep track of count of each element.

Another solution

var findShortestSubArray = function (nums) {
  let map = {};
  let max = 0;
  let min = nums.length;
  let maxNums = [];
  for (let i = 0; i < nums.length; i++) {
    if (map[nums[i]] === undefined) {
      map[nums[i]] = 1;
    } else {
      map[nums[i]]++;
    }
    if (map[nums[i]] > max) {
      max = map[nums[i]];
      maxNums = [nums[i]];
    } else if (map[nums[i]] === max) {
      maxNums.push(nums[i]);
    }
  }
  for (let i = 0; i < maxNums.length; i++) {
    let num = maxNums[i];
    let start = nums.indexOf(num);
    let end = nums.lastIndexOf(num);
    if (end - start + 1 < min) {
      min = end - start + 1;
    }
  }
  return min;
};