标签:1551 Operations arr elements target res Make array operation
题目如下:
You have an array
arr
of lengthn
wherearr[i] = (2 * i) + 1
for all valid values ofi
(i.e.0 <= i < n
).In one operation, you can select two indices
x
andy
where0 <= x, y < n
and subtract1
fromarr[x]
and add1
toarr[y]
(i.e. performarr[x] -=1
andarr[y] += 1
). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made equal using some operations.Given an integer
n
, the length of the array. Return the minimum number of operations needed to make all the elements of arr equal.Example 1:
Input: n = 3 Output: 2 Explanation: arr = [1, 3, 5] First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4] In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].Example 2:
Input: n = 6 Output: 9Constraints:
1 <= n <= 10^4
解题思路:每次操作都是加一减一,即数组的和是不变的,由此可知最后必定是所有元素的值等于数组的平均值。因此结果是所有元素与平均值的差的绝对值的和除以二。
代码如下:
class Solution(object): def minOperations(self, n): """ :type n: int :rtype: int """ res = 0 if n % 2 == 1: target = 2*(n/2) + 1 else: target = ((2*(n/2) + 1) + 2*(n/2-1) + 1)/2 for i in range(n/2): res += (target - 2*i -1) return res
标签:1551,Operations,arr,elements,target,res,Make,array,operation 来源: https://www.cnblogs.com/seyjs/p/14793562.html
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