697. Degree of an Array
package LeetCode_697 /** * 697. Degree of an Array * https://leetcode.com/problems/degree-of-an-array/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: 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: 1. nums.length will be between 1 and 50,000. 2. nums[i] will be an integer between 0 and 49,999. * */ class Solution { /* * solution: HashMap, Time:O(n), Space:O(n); * 1. save each num's index of appear by List, * 2. find out most frequency number (degree), * 3. find out the smallest length of sub-array that has length as same as degree, * */ fun findShortestSubArray(nums: IntArray): Int? { val map = HashMap<Int, ArrayList<Int>>() for (i in nums.indices) { if (!map.contains(nums[i])) { val list = ArrayList<Int>() list.add(i) map.put(nums[i], list) } else { map.get(nums[i])!!.add(i) } } var degree = 0 for (item in map) { degree = Math.max(degree, item.value.size) } var result = Int.MAX_VALUE for (item in map) { //find out the smallest length of sub-array that has length as same as degree, if (item.value.size == degree) { //calculate the length by index, result = Math.min(result, item.value.get(item.value.lastIndex) - item.value.get(0) + 1) } } return result } }