给定大小为N的数组arr [] ，任务是通过任意次数向左或向右移动数组元素，最大程度地增加偶数索引元素之和与奇数索引元素之和之间的绝对差。
例子：
Input: arr[] = {332, 421, 215, 584, 232}
Output: 658
Explanation:
Convert the array to {233, 421, 152, 845, 223},
Sum of odd indexed elements = 608.
Sum of even indexed elements = 1266.
Therefore, the difference equals 658.
Input: arr[] = {11, 22, 33, 44, 55}
Output: 33
方法：想法是最小化偶数或奇数索引数组元素中任何一个的值，并最大化另一个元素的值，以最大化它们的绝对差。请按照以下步骤解决问题：
- 存在两种可能的情况。或者最小化偶数索引的数组元素并最大化奇数索引的数组元素，或者最小化奇数索引的数组元素并最大化偶数索引的数组元素。
- 为了使元素最小化，请应用所有移位操作并取尽可能小的值。同样，要使元素最大化，请应用所有移位操作并采用最大可能值。
- 两种情况的差异最大。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to minimize array
// elements by shift operations
int minimize(int n)
{
int optEle = n;
string strEle = to_string(n);
// For checking all the
// left shift operations
for (int idx = 0; idx < strEle.length();idx++)
{
// Left shift
int temp = stoi(strEle.substr(idx) +
strEle.substr(0, idx));
// Consider the minimum possible value
optEle = min(optEle, temp);
}
return optEle;
}
// Function to maximize array
// elements by shift operations
int maximize(int n)
{
int optEle = n;
string strEle = to_string(n);
// For checking all the
// left shift operations
for (int idx = 0; idx < strEle.length();idx++)
{
// Left shift
int temp = stoi(strEle.substr(idx) +
strEle.substr(0, idx));
// Consider the maximum possible value
optEle = max(optEle, temp);
}
return optEle;
}
// Function to maximize the absolute
// difference between even and odd
// indexed array elements
void minimumDifference(int arr[], int N)
{
int caseOne = 0;
int minVal = 0;
int maxVal = 0;
// To calculate the difference of
// odd indexed elements
// and even indexed elements
for (int i = 0; i < N; i++)
{
if (i % 2 == 0)
minVal += minimize(arr[i]);
else
maxVal += maximize(arr[i]);
}
caseOne = abs(maxVal - minVal);
int caseTwo = 0;
minVal = 0;
maxVal = 0;
// To calculate the difference
// between odd and even indexed
// array elements
for (int i = 0; i < N; i++)
{
if (i % 2 == 0)
maxVal += maximize(arr[i]);
else
minVal += minimize(arr[i]);
caseTwo = abs(maxVal - minVal);
}
// Print the maximum value
cout << max(caseOne, caseTwo) << endl;
}
// Driver code
int main()
{
int arr[] = { 332, 421, 215, 584, 232 };
int N = sizeof(arr) / sizeof(arr[0]);
minimumDifference(arr, N);
return 0;
}
// This code is contributed by divyesh072019. Java
// Java program for the above approach
import java.io.*;
class GFG
{
// Function to minimize array
// elements by shift operations
static int minimize(int n)
{
int optEle = n;
String strEle = Integer.toString(n);
// For checking all the
// left shift operations
for (int idx = 0; idx < strEle.length(); idx++)
{
// Left shift
int temp
= Integer.parseInt(strEle.substring(idx)
+ strEle.substring(0, idx));
// Consider the minimum possible value
optEle = Math.min(optEle, temp);
}
return optEle;
}
// Function to maximize array
// elements by shift operations
static int maximize(int n)
{
int optEle = n;
String strEle = Integer.toString(n);
// For checking all the
// left shift operations
for (int idx = 0; idx < strEle.length(); idx++)
{
// Left shift
int temp
= Integer.parseInt(strEle.substring(idx)
+ strEle.substring(0, idx));
// Consider the maximum possible value
optEle = Math.max(optEle, temp);
}
return optEle;
}
// Function to maximize the absolute
// difference between even and odd
// indexed array elements
static void minimumDifference(int[] arr)
{
int caseOne = 0;
int minVal = 0;
int maxVal = 0;
// To calculate the difference of
// odd indexed elements
// and even indexed elements
for (int i = 0; i < arr.length; i++)
{
if (i % 2 == 0)
minVal += minimize(arr[i]);
else
maxVal += maximize(arr[i]);
}
caseOne = Math.abs(maxVal - minVal);
int caseTwo = 0;
minVal = 0;
maxVal = 0;
// To calculate the difference
// between odd and even indexed
// array elements
for (int i = 0; i < arr.length; i++)
{
if (i % 2 == 0)
maxVal += maximize(arr[i]);
else
minVal += minimize(arr[i]);
caseTwo = Math.abs(maxVal - minVal);
}
// Print the maximum value
System.out.println(Math.max(caseOne, caseTwo));
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 332, 421, 215, 584, 232 };
minimumDifference(arr);
}
}
// This code is contributed by Dharanendra L V.Python3
# Python3 program for the above approach
# Function to minimize array
# elements by shift operations
def minimize(n):
optEle = n
strEle = str(n)
# For checking all the
# left shift operations
for idx in range(len(strEle)):
# Left shift
temp = int(strEle[idx:] + strEle[:idx])
# Consider the minimum possible value
optEle = min(optEle, temp)
return optEle
# Function to maximize array
# elements by shift operations
def maximize(n):
optEle = n
strEle = str(n)
# For checking all the
# left shift operations
for idx in range(len(strEle)):
# Left shift
temp = int(strEle[idx:] + strEle[:idx])
# Consider the maximum possible value
optEle = max(optEle, temp)
return optEle
# Function to maximize the absolute
# difference between even and odd
# indexed array elements
def minimumDifference(arr):
caseOne = 0
minVal = 0
maxVal = 0
# To calculate the difference of
# odd indexed elements
# and even indexed elements
for i in range(len(arr)):
if i % 2:
minVal += minimize(arr[i])
else:
maxVal += maximize(arr[i])
caseOne = abs(maxVal - minVal)
caseTwo = 0
minVal = 0
maxVal = 0
# To calculate the difference
# between odd and even indexed
# array elements
for i in range(len(arr)):
if i % 2:
maxVal += maximize(arr[i])
else:
minVal += minimize(arr[i])
caseTwo = abs(maxVal - minVal)
# Print the maximum value
print (max(caseOne, caseTwo))
# Given array
arr = [ 332, 421, 215, 584, 232 ]
minimumDifference(arr)C#
// C# program for the above approach
using System;
class GFG
{
// Function to minimize array
// elements by shift operations
static int minimize(int n)
{
int optEle = n;
string strEle = n.ToString();
// For checking all the
// left shift operations
for (int idx = 0; idx < strEle.Length;idx++)
{
// Left shift
int temp = Int32.Parse(strEle.Substring(idx) +
strEle.Substring(0, idx));
// Consider the minimum possible value
optEle = Math.Min(optEle, temp);
}
return optEle;
}
// Function to maximize array
// elements by shift operations
static int maximize(int n)
{
int optEle = n;
string strEle = n.ToString();
// For checking all the
// left shift operations
for (int idx = 0; idx < strEle.Length;idx++)
{
// Left shift
int temp = Int32.Parse(strEle.Substring(idx) +
strEle.Substring(0, idx));
// Consider the maximum possible value
optEle = Math.Max(optEle, temp);
}
return optEle;
}
// Function to maximize the absolute
// difference between even and odd
// indexed array elements
static void minimumDifference(int[] arr)
{
int caseOne = 0;
int minVal = 0;
int maxVal = 0;
// To calculate the difference of
// odd indexed elements
// and even indexed elements
for (int i = 0; i < arr.Length; i++)
{
if (i % 2 == 0)
minVal += minimize(arr[i]);
else
maxVal += maximize(arr[i]);
}
caseOne = Math.Abs(maxVal - minVal);
int caseTwo = 0;
minVal = 0;
maxVal = 0;
// To calculate the difference
// between odd and even indexed
// array elements
for (int i = 0; i < arr.Length; i++)
{
if (i % 2 == 0)
maxVal += maximize(arr[i]);
else
minVal += minimize(arr[i]);
caseTwo = Math.Abs(maxVal - minVal);
}
// Print the maximum value
Console.WriteLine(Math.Max(caseOne, caseTwo));
}
// Given array
public static void Main()
{
int[] arr = { 332, 421, 215, 584, 232 };
minimumDifference(arr);
}
}
// This code is contributed by chitranayal.输出：
658时间复杂度： O(N)
辅助空间： O(1)