LeetCode in Kotlin
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.lengthwill be between 1 and 50,000.nums[i]will be an integer between 0 and 49,999.
Solution
class Solution {
private class Value(var count: Int, var start: Int, var end: Int)
fun findShortestSubArray(nums: IntArray): Int {
var max = 1
val map: MutableMap<Int, Value> = HashMap()
for (i in nums.indices) {
val j = nums[i]
if (map.containsKey(j)) {
val v = map[j]
v!!.count++
max = Math.max(max, v.count)
v.end = i
} else {
map[j] = Value(1, i, i)
}
}
var min = Int.MAX_VALUE
for (entry in map.entries.iterator()) {
val v: Value = entry.value
if (v.count == max) {
min = min.coerceAtMost(v.end - v.start)
}
}
return min + 1
}
}