Given an array
arr of positive integers sorted in a strictly increasing order, and an integer
kth positive integer that is missing from this array.
Input: arr = [2,3,4,7,11], k = 5 Output: 9 Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.
Input: arr = [1,2,3,4], k = 2 Output: 6 Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6.
1 <= arr.length <= 1000
1 <= arr[i] <= 1000
1 <= k <= 1000
arr[i] < arr[j]for
1 <= i < j <= arr.length
We can first find a list of missing positives from the list and then we have to handle different scenarios to find kth missing positive. If the list of missing positives has a length greater than k, returns the kth positive from the list. If the list of missing positives is empty, the k th missing positive would be the largest number in the list + k. If the missing positives’ length is shorter than k, the kth missing positive would be the largest number in the original list + k – len(missing_positives).
class Solution: def findKthPositive(self, arr: List[int], k: int) -> int: missing_positives =  n = max(arr) arr_set = set(arr) for i in range(1, n + 1): if i not in arr_set: missing_positives.append(i) if not missing_positives: return arr[len(arr) - 1] + k if k > len(missing_positives): return arr[len(arr) - 1] + (k - len(missing_positives)) return missing_positives[k - 1]
- Time Complexity: O(N).
- Space Complexity: O(N).