2892 Check If Array Is Good

Check if Array is Good

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

impl Solution {
    pub fn is_good(mut nums: Vec<i32>) -> bool {
        let mut temp = 0;
        nums.sort();
        for k in &nums[..nums.len()-1] {
            if *k == temp + 1 {
                temp = *k;
            } else {
                return false;
            }
        }
        //print!("{:?} {}" , nums, temp);
        if nums[nums.len()-1] == temp {
            return true;
        }
        return false;
    }
}