【leetcode】1300. Sum of Mutated Array Closest to Target

Topics are as follows:

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

Problem-solving ideas: First arr sort. If the array should be greater than the number of all the replaced value value, you can find the position of the value appearing in arr by the binary search method, the left half of the elements need not be changed, the direct sum of the right half of the element and for the length * value.

code show as below:

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