LeetCode 2784. Check if Array is Good Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Counting

We can use a hash table or array cnt to record the number of occurrences of each element in the array nums. Then we determine whether the following conditions are met:

1. cnt[n] = 2, i.e., the largest element in the array appears twice;
2. For 1 ≤ i ≤ n-1, it holds that cnt[i] = 1, i.e., except for the largest element, all other elements appear only once.

If the above two conditions are met, then the array nums is a good array, otherwise it is not.

The time complexity is O(n), and the space complexity is O(n). Where n is the length of the array nums.

PythonJavaC++GoTypeScriptC#

class Solution: def isGood(self, nums: List[int]) -> bool: cnt = Counter(nums) n = len(nums) - 1 return cnt[n] == 2 and all(cnt[i] for i in range(1, n))(code-box)

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

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

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

function isGood(nums: number[]): boolean { const n = nums.length - 1; const cnt: number[] = Array(201).fill(0); for (const x of nums) { ++cnt[x]; } if (cnt[n] !== 2) { return false; } for (let i = 1; i < n; ++i) { if (cnt[i] !== 1) { return false; } } return true; }(code-box)

public class Solution { public bool IsGood(int[] nums) { int n = nums.Length - 1; int[] cnt = new int[201]; foreach (int x in nums) { ++cnt[x]; } if (cnt[n] != 2) { return false; } for (int i = 1; i < n; ++i) { if (cnt[i] != 1) { return false; } } return true; } }(code-box)