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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);
    }
    
    

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