CheckIfArrayIsGood.cpp

/*
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.
*/

/// Solution 1

class Solution {
public:
    bool isGood(vector<int>& nums) {
        int mx=*max_element(nums.begin(), nums.end());
        int n = nums.size();
        if (n != mx + 1) {
            return false;
        }
        unordered_map<int, int> count;
        for (int num : nums) {
            count[num]++;
        }
        for (int i = 1; i <= mx - 1; ++i) {
            if (count[i] != 1) {
                return false;
            }
        }
        if (count[mx] != 2) {
            return false;
        }
    return true;
    }
};

/// Solution 2

class Solution {
public:
    bool isGood(vector<int>& nums) {
        int n=nums.size();
        sort(nums.begin(),nums.end());
        for(int i=0;i<n-1;i++){
            if(nums[i]!=(i+1)) return false;
        }
        if(nums[n-1]!=n-1) return false;
    return true;
    }
};