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

Solution 1

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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())

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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;
    }
}

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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;
    }
};

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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
}

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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);
}