1552_Magnetic force between two balls_Binary_Search

Title description

Ideas

The question intends to maximize the minimum, similar to this, maximize the minimum, minimize the maximum, etc. can be solved by binary search.

class Solution:
    def maxDistance(self, position: List[int], m: int) -> int:
        """
        二分
        """
        if m==2: return max(position) - min(position)  # 特判
        def check(mid: int) -> bool:
            cnt = 1
            pre = arr[0]
            for i in range(1, len(arr)):
                if arr[i] - pre >= mid:
                    cnt += 1
                    pre = arr[i]
            return cnt >= m  # cnt太多 可以试着增大mid的值

        arr = list(sorted(position))
        left = min([abs(arr[i]-arr[i-1]) for i in range(1, len(arr))])
        right= arr[-1] - arr[0]
        while left < right:
            print(left, right)
            mid = left + (right - left) // 2
            if check(mid):
                left = mid + 1
            else:
                right = mid
        # 由于每次满足条件后左边界右移，所以左边界的左边一个数值是一定满足条件的
        # 但是 最后一次的left = mid+1 不会满足条件，证明如下：
        # 反证法：若left = mid+1也满足条件，即选择left为最大化的最小磁力，则下次二分时，
        # 还是会执行 left = mid+1，与最后一次矛盾。
        # 并且 一定会执行一次 left = mid + 1 操作
        # 因为存在解，left为最小值时一定满足check要求
        # 所以最后返回值为 l - 1
        return left-1