2517. Maximum Tastiness of Candy Basket ( 類似2616

類似

T: O(nlogn)
S: O(1)

```java
class Solution {
    public int maximumTastiness(int[] price, int k) {
        Arrays.sort(price);
        int left = 0;
        int right = price[price.length-1] - price[0];

        int result = 0;
        while (left <= right) {
            int mid = left + (right - left)/2;

            int elementsUseNumber = 1;
            int curNumber = price[0]; // choose candies
            for (int i = 1; i < price.length;i++) {
                if (price[i] - curNumber >= mid) { // if candidates can >= max_diff
                    elementsUseNumber++;
                    curNumber = price[i];
                }
            }

            if (elementsUseNumber >= k) {
                result = Math.max(result, mid);
                left = mid + 1; // male k samller, so pick larger number
            } else {
                right = mid - 1;
            }
        }
        return result;
    }
}
/*
[13,5,1,8,21,2]

1 2 5 8 13 21
 1 3.3 5  8
         我叫這個 max_diff (8)

 8 13 21
  5. 8
  13
  want larger diff (choose pair)

 5 13 21
  8. 8
  16 


  so how about to search the diff range (0 ~ price[last] - price[0]) and k?

  max_diff = 8 k = 3
             7
             6 k = 4...

             if we want to choose from more k cadidates, the max_diff will be lower

             所以可以發現 max_diff 和 k 是有一個相對關係的....
             所以可以用 binary search

T: O(nlogn)
S: O(1)
*/
```