leetcode1300. Sum of Mutated Array Closest to Target

Posted 2020-11-26 seyjs

题目如下：

Given an integer array arr and a target value target, return the integer value such that when we change all the integers larger than value in the given array to be equal to value, the sum of the array gets as close as possible (in absolute difference) to target.

In case of a tie, return the minimum such integer.

Notice that the answer is not neccesarilly a number from arr.

Example 1:

Input: arr = [4,9,3], target = 10
Output: 3
Explanation: When using 3 arr converts to [3, 3, 3] which sums 9 and that‘s the optimal answer.

Example 2:

Input: arr = [2,3,5], target = 10
Output: 5

Example 3:

Input: arr = [60864,25176,27249,21296,20204], target = 56803
Output: 11361

Constraints:

• 1 <= arr.length <= 10^4
• 1 <= arr[i], target <= 10^5

解题思路：首先对arr进行排序。如果要把数组中所有大于value的数替换成value，那么可以通过二分查找的方法找出value在arr中出现的位置，左半部分的元素不需要改变，直接求和，右半部分的元素的和为 length * value。

代码如下：

class Solution(object):
    def findBestValue(self, arr, target):
        """
        :type arr: List[int]
        :type target: int
        :rtype: int
        """
        import bisect
        diff = float(‘inf‘)
        res = 0
        arr.sort()
        val = []
        count = 0
        for i in arr:
            count += i
            val.append(count)
        for v in range(0,arr[-1] + 1):
            inx = bisect.bisect_right(arr,v)
            amount = v * (len(arr) - inx)
            if inx > 0:amount += val[inx-1]
            if diff > abs(amount - target):
                diff = abs(amount - target)
                res = v
        return res