给定一个由正整数组成的数组A [] ，任务是在多次执行以下运算后，计算给定数组A []的最小可能偏差：
- 操作1：如果数组元素为偶数，则将其除以2 。
- 操作2：如果数组元素为奇数，则将其乘以2 。
数组A []的偏差是在该阵列A中的最大和最小元件本之间的差[]。
例子：
Input: A[] = {4, 1, 5, 20, 3}
Output: 3
Explanation: Array modifies to {4, 2, 5, 5, 3} after performing given operations. Therefore, deviation = 5 – 2 = 3.
Input: A[] = {1, 2, 3, 4}
Output: 1
Explanation: Array modifies to after two operations to {2, 2, 3, 2}. Therefore, deviation = 3 – 2 = 1.
方法：可以根据以下观察结果解决问题：
- 偶数可以被除以多次，直到转换为奇数为止。
- 奇数转换为偶数时只能加倍一次。
- 因此，偶数永远不会增加。
请按照以下步骤解决问题：
- 遍历数组并将所有奇数数组元素加倍。这使第二次操作的要求无效。
- 现在，减少最大的数组元素，使其均匀。
- 要以排序方式存储数组元素，请将所有数组元素插入Set中。
- 贪婪地减少集合中存在的最大元素
- 如果集合中存在的最大元素为奇数，请中断循环。
- 打印获得的最小偏差。
下面是上述方法的实现：
C++
// C++ implementation of the
// above approach
#include
using namespace std;
// Function to find the minimum
// deviation of the array A[]
void minimumDeviation(int A[], int N)
{
// Store all array elements
// in sorted order
set s;
for (int i = 0; i < N; i++) {
if (A[i] % 2 == 0)
s.insert(A[i]);
// Odd number are transformed
// using 2nd operation
else
s.insert(2 * A[i]);
}
// (Maximum - Minimum)
int diff = *s.rbegin() - *s.begin();
// Check if the size of set is > 0 and
// the maximum element is divisible by 2
while ((int)s.size()
&& *s.rbegin() % 2 == 0) {
// Maximum element of the set
int maxEl = *s.rbegin();
// Erase the maximum element
s.erase(maxEl);
// Using operation 1
s.insert(maxEl / 2);
// (Maximum - Minimum)
diff = min(diff, *s.rbegin() - *s.begin());
}
// Print the Minimum
// Deviation Obtained
cout << diff;
}
// Driver Code
int main()
{
int A[] = { 4, 1, 5, 20, 3 };
int N = sizeof(A) / sizeof(A[0]);
// Function Call to find
// Minimum Deviation of A[]
minimumDeviation(A, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to find the minimum
// deviation of the array A[]
static void minimumDeviation(int A[], int N)
{
// Store all array elements
// in sorted order
TreeSet s = new TreeSet();
for (int i = 0; i < N; i++)
{
if (A[i] % 2 == 0)
s.add(A[i]);
// Odd number are transformed
// using 2nd operation
else
s.add(2 * A[i]);
}
// (Maximum - Minimum)
int diff = s.last() - s.first() ;
// Check if the size of set is > 0 and
// the maximum element is divisible by 2
while ((s.last() % 2 == 0))
{
// Maximum element of the set
int maxEl = s.last();
// Erase the maximum element
s.remove(maxEl);
// Using operation 1
s.add(maxEl / 2);
// (Maximum - Minimum)
diff = Math.min(diff, s.last() - s.first());
}
// Print the Minimum
// Deviation Obtained
System.out.print(diff);
}
// Driver code
public static void main(String[] args)
{
int A[] = { 4, 1, 5, 20, 3 };
int N = A.length;
// Function Call to find
// Minimum Deviation of A[]
minimumDeviation(A, N);
}
}
// This code is contributed by susmitakundugoaldanga. Python3
# Python 3 implementation of the
# above approach
# Function to find the minimum
# deviation of the array A[]
def minimumDeviation(A, N):
# Store all array elements
# in sorted order
s = set([])
for i in range(N):
if (A[i] % 2 == 0):
s.add(A[i])
# Odd number are transformed
# using 2nd operation
else:
s.add(2 * A[i])
# (Maximum - Minimum)
s = list(s)
diff = s[-1] - s[0]
# Check if the size of set is > 0 and
# the maximum element is divisible by 2
while (len(s) and s[-1] % 2 == 0):
# Maximum element of the set
maxEl = s[-1]
# Erase the maximum element
s.remove(maxEl)
# Using operation 1
s.append(maxEl // 2)
# (Maximum - Minimum)
diff = min(diff, s[-1] - s[0])
# Print the Minimum
# Deviation Obtained
print(diff)
# Driver Code
if __name__ == "__main__":
A = [4, 1, 5, 20, 3]
N = len(A)
# Function Call to find
# Minimum Deviation of A[]
minimumDeviation(A, N)
# This code is contributed by chitranayal.C#
// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{
// Function to find the minimum
// deviation of the array A[]
static void minimumDeviation(int[] A, int N)
{
// Store all array elements
// in sorted order
HashSet s = new HashSet();
for (int i = 0; i < N; i++)
{
if (A[i] % 2 == 0)
s.Add(A[i]);
// Odd number are transformed
// using 2nd operation
else
s.Add(2 * A[i]);
}
List S = s.ToList();
S.Sort();
// (Maximum - Minimum)
int diff = S[S.Count - 1] - S[0];
// Check if the size of set is > 0 and
// the maximum element is divisible by 2
while ((int)S.Count != 0 && S[S.Count - 1] % 2 == 0) {
// Maximum element of the set
int maxEl = S[S.Count - 1];
// Erase the maximum element
S.RemoveAt(S.Count - 1);
// Using operation 1
S.Add(maxEl / 2);
S.Sort();
// (Maximum - Minimum)
diff = Math.Min(diff, S[S.Count - 1] - S[0]);
}
// Print the Minimum
// Deviation Obtained
Console.Write(diff);
}
// Driver code
static void Main()
{
int[] A = { 4, 1, 5, 20, 3 };
int N = A.Length;
// Function Call to find
// Minimum Deviation of A[]
minimumDeviation(A, N);
}
}
// This code is contributed by divyeshrabadiya07. 输出：
3时间复杂度：O(N * log(N))
辅助空间：O(N)