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2784. Check if Array is Good
Description
You are given an integer array nums
. We consider an array good if it is a permutation of an array base[n]
.
base[n] = [1, 2, ..., n  1, n, n]
(in other words, it is an array of length n + 1
which contains 1
to n  1
exactly once, plus two occurrences of n
). For example, base[1] = [1, 1]
and base[3] = [1, 2, 3, 3]
.
Return true
if the given array is good, otherwise return false
.
Note: A permutation of integers represents an arrangement of these numbers.
Example 1:
Input: nums = [2, 1, 3] Output: false Explanation: Since the maximum element of the array is 3, the only candidate n for which this array could be a permutation of base[n], is n = 3. However, base[3] has four elements but array nums has three. Therefore, it can not be a permutation of base[3] = [1, 2, 3, 3]. So the answer is false.
Example 2:
Input: nums = [1, 3, 3, 2] Output: true Explanation: Since the maximum element of the array is 3, the only candidate n for which this array could be a permutation of base[n], is n = 3. It can be seen that nums is a permutation of base[3] = [1, 2, 3, 3] (by swapping the second and fourth elements in nums, we reach base[3]). Therefore, the answer is true.
Example 3:
Input: nums = [1, 1] Output: true Explanation: Since the maximum element of the array is 1, the only candidate n for which this array could be a permutation of base[n], is n = 1. It can be seen that nums is a permutation of base[1] = [1, 1]. Therefore, the answer is true.
Example 4:
Input: nums = [3, 4, 4, 1, 2, 1] Output: false Explanation: Since the maximum element of the array is 4, the only candidate n for which this array could be a permutation of base[n], is n = 4. However, base[4] has five elements but array nums has six. Therefore, it can not be a permutation of base[4] = [1, 2, 3, 4, 4]. So the answer is false.
Constraints:
1 <= nums.length <= 100
1 <= num[i] <= 200
Solutions

class Solution { public boolean isGood(int[] nums) { int n = nums.length  1; int[] cnt = new int[201]; for (int x : nums) { ++cnt[x]; } cnt[n] = 2; for (int i = 1; i < n; ++i) { cnt[i] = 1; } for (int x : cnt) { if (x != 0) { return false; } } return true; } }

class Solution { public: bool isGood(vector<int>& nums) { int n = nums.size()  1; vector<int> cnt(201); for (int x : nums) { ++cnt[x]; } cnt[n] = 2; for (int i = 1; i < n; ++i) { cnt[i]; } for (int x : cnt) { if (x) { return false; } } return true; } };

class Solution: def isGood(self, nums: List[int]) > bool: n = len(nums)  1 cnt = Counter(nums) cnt[n] = 2 for i in range(1, n): cnt[i] = 1 return all(v == 0 for v in cnt.values())

func isGood(nums []int) bool { n := len(nums)  1 cnt := [201]int{} for _, x := range nums { cnt[x]++ } cnt[n] = 2 for i := 1; i < n; i++ { cnt[i] } for _, x := range cnt { if x != 0 { return false } } return true }

function isGood(nums: number[]): boolean { const n = nums.length  1; const cnt: number[] = new Array(201).fill(0); for (const x of nums) { ++cnt[x]; } cnt[n] = 2; for (let i = 1; i < n; ++i) { cnt[i]; } return cnt.every(x => x >= 0); }