Problem
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 - 1exactly once, plus two occurrences of n). For example, base[1] = [1, 1] andbase[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 <= 1001 <= num[i] <= 200
Solution (Java)
class Solution {
public boolean isGood(int[] nums) {
int n = nums.length;
int arrLen = n-1;
int[] arr = new int[arrLen];
for(int num: nums){
int index = num-1;
if(index > arrLen -1 || index<= -1) return false;
arr[index]++;
}
if(arr[arrLen -1] != 2) return false;
for(int i=0;i < arrLen-1; i++){
if(arr[i]!=1) return false;
}
return true;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).