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1608. Special Array With X Elements Greater Than or Equal X.md

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1608. Special Array With X Elements Greater Than or Equal X

the first one in Weekly Contest 209.

Difficulty : Easy

Related Topics : Array

You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.

Notice that x does not have to be an element in nums.

Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.

Example 1:

Input: nums = [3,5]
Output: 2
Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.

Example 2:

Input: nums = [0,0]
Output: -1
Explanation: No numbers fit the criteria for x.
If x = 0, there should be 0 numbers >= x, but there are 2.
If x = 1, there should be 1 number >= x, but there are 0.
If x = 2, there should be 2 numbers >= x, but there are 0.
x cannot be greater since there are only 2 numbers in nums.

Example 3:

Input: nums = [0,4,3,0,4]
Output: 3
Explanation: There are 3 values that are greater than or equal to 3.

Example 4:

Input: nums = [3,6,7,7,0]
Output: -1

Constraints:

  • 1 <= nums.length <=100
  • 0 <= nums[i] <= 1000

Solution

  • mine
    • Java
      • Runtime: 2 ms, faster than 44.13%, Memory Usage: 36.3 MB, less than 39.89% of Java online submissions 
        // O(N * Max(nums))time
// O(1)space
public int specialArray(int[] nums) {
    Arrays.sort(nums);
    int max = nums[nums.length - 1];
    int res = -1;
    for(int i = 0; i <= max; i++){
        int t = 0;
        for(int j = 0; j < nums.length; j++){
            if(nums[j] >= i){
                t++;
            }
            if(t > i) break;
        }
        if(t == i) return i;
    }
    return res;
}
  • the most votes
  • Runtime: 1 ms, faster than 86.04%, Memory Usage: 36.6 MB, less than 39.89% of Java online submissions 
    //O(N*logN)time
//O(1)space
public int specialArray(int[] nums) {
    Arrays.sort(nums);
    for (int i = 0; i < nums.length; i++) {
        int n = nums.length - i;
        boolean cond1 = n <= nums[i];
        boolean cond2 = (i - 1 < 0) || (n > nums[i - 1]);
        if (cond1 && cond2) return n;
    }
    return -1;
}