给定一个由N 个整数组成的数组arr[]和一个整数K ，任务是在删除K 个数组元素后找到给定数组可能的最大反转次数。
例子：
Input: arr[] = {2, 3, 4, 1}, K = 2
Output: 1
Explanation: Removing the 1st and 2nd array elements modifies the array to {4, 1}. Therefore, the maximum number of inversions is 1.
Input: arr[] = {4, 3, 2, 1}, K = 3
Output: 0
Explanation: Since the array after K removals will have only 1 element remaining, the maximum number of inversions is 0.
处理方法：按照以下步骤解决问题：
- 从数组arr[] 中删除K 个元素后，初始化变量res以存储最大反转次数。
- 初始化长度为N的二进制字符串S以存储要从数组arr[] 中考虑的(N – K) 个元素。
- 在[0, K – 1]位置存储0表示已删除的K 个元素，在[K, N – 1]位置存储1表示字符串S 中的(N – K) 个选定元素。
- 生成字符串S 的所有排列并执行以下操作：
- 初始化数组v以根据字符串S存储从arr[]中选择的(N – K) 个元素。
- 计算数组v 中的反转次数并存储在cnt 中。
- 将res更新为res和cnt 的最大值。
- 完成上述步骤后，打印res的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum number
// of inversions after K removals
void maximizeInversions(int a[], int n,
int k)
{
// Initialize a binary string
string s;
// Interate over range [0, K]
for (int i = 0; i < k; i++)
{
// Append 0 to the string
s += "0";
}
for (int i = k; i < n; i++)
{
// Append 1 to the string
s += "1";
}
// Stores the maximum number of
// inversions after K removals
int res = 0;
// Generate all permutations
// of string s
do
{
// Stores the current array
vector v;
// Stores number of inversions
// of the current array
int cnt = 0;
for (int i = 0; i < n; i++)
{
if (s[i] == '1')
v.push_back(a[i]);
}
// Find the number of inversions
// in the array v
for (int i = 0;
i < v.size() - 1; i++)
{
for (int j = i + 1;
j < v.size(); j++)
{
// Condition to check if the
// number of pairs satisfy
// required condition
if (v[i] >= v[j])
// Increment the count
cnt++;
}
}
// Update res
res = max(res, cnt);
}
while (next_permutation(s.begin(),
s.end()));
// Print the result
cout << res;
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 4, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 2;
// Function Call
maximizeInversions(arr, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the maximum number
// of inversions after K removals
static void maximizeInversions(int a[], int n,
int k)
{
// Initialize a binary String
String s="";
// Interate over range [0, K]
for (int i = 0; i < k; i++)
{
// Append 0 to the String
s += "0";
}
for (int i = k; i < n; i++)
{
// Append 1 to the String
s += "1";
}
// Stores the maximum number of
// inversions after K removals
int res = 0;
// Generate all permutations
// of String s
do
{
// Stores the current array
Vector v = new Vector<>();
// Stores number of inversions
// of the current array
int cnt = 0;
for (int i = 0; i < n; i++)
{
if (s.charAt(i) == '1')
v.add(a[i]);
}
// Find the number of inversions
// in the array v
for (int i = 0;
i < v.size() - 1; i++)
{
for (int j = i + 1;
j < v.size(); j++)
{
// Condition to check if the
// number of pairs satisfy
// required condition
if (v.get(i) >= v.get(j))
// Increment the count
cnt++;
}
}
// Update res
res = Math.max(res, cnt);
}
while (next_permutation(s));
// Print the result
System.out.print(res);
}
static boolean next_permutation(String str)
{
char []p = str.toCharArray();
for (int a = p.length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (int b = p.length - 1;; --b)
if (p[b] > p[a])
{
char t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1;
a < b; ++a, --b)
{
t = p[a];
p[a] = p[b];
p[b] = t;
}
return false;
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 1 };
int N = arr.length;
int K = 2;
// Function Call
maximizeInversions(arr, N, K);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the above approach
def next_permutation(Str):
p = list(Str)
for a in range(len(p) - 2, -1, -1):
if p[a] < p[a + 1]:
b = len(p) - 1
while True:
if p[b] > p[a]:
t = p[a]
p[a] = p[b]
p[b] = t
a += 1
b = len(p) - 1
while a < b:
t = p[a]
p[a] = p[b]
p[b] = t
a += 1
b -= 1
return False
b -= 1
return True
# Function to find the maximum number
# of inversions after K removals
def maximizeInversions(a, n, k):
# Initialize a binary string
s = ""
# Interate over range [0, K]
for i in range(k):
# Append 0 to the string
s += "0"
for i in range(k, n):
# Append 1 to the string
s += "1"
# Stores the maximum number of
# inversions after K removals
res = 0
# Generate all permutations
# of string s
while (True):
# Stores the current array
v = []
# Stores number of inversions
# of the current array
cnt = 0
for i in range(n):
if (s[i] == '1'):
v.append(a[i])
# Find the number of inversions
# in the array v
for i in range(len(v) - 1):
for j in range(i + 1, len(v)):
# Condition to check if the
# number of pairs satisfy
# required condition
if (v[i] >= v[j]):
# Increment the count
cnt += 1
# Update res
res = max(res, cnt)
if (not next_permutation(s)):
break
# Print the result
print(res)
# Driver Code
arr = [ 2, 3, 4, 1 ]
N = len(arr)
K = 2
# Function Call
maximizeInversions(arr, N, K)
# This code is contributed by divyeshrabadiya07C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the maximum number
// of inversions after K removals
static void maximizeInversions(int []a, int n,
int k)
{
// Initialize a binary String
String s="";
// Interate over range [0, K]
for (int i = 0; i < k; i++)
{
// Append 0 to the String
s += "0";
}
for (int i = k; i < n; i++)
{
// Append 1 to the String
s += "1";
}
// Stores the maximum number of
// inversions after K removals
int res = 0;
// Generate all permutations
// of String s
do
{
// Stores the current array
List v = new List();
// Stores number of inversions
// of the current array
int cnt = 0;
for (int i = 0; i < n; i++)
{
if (s[i] == '1')
v.Add(a[i]);
}
// Find the number of inversions
// in the array v
for (int i = 0;
i < v.Count - 1; i++)
{
for (int j = i + 1;
j < v.Count; j++)
{
// Condition to check if the
// number of pairs satisfy
// required condition
if (v[i] >= v[j])
// Increment the count
cnt++;
}
}
// Update res
res = Math.Max(res, cnt);
}
while (next_permutation(s));
// Print the result
Console.Write(res);
}
static bool next_permutation(String str)
{
char []p = str.ToCharArray();
for (int a = p.Length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (int b = p.Length - 1;; --b)
if (p[b] > p[a])
{
char t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.Length - 1;
a < b; ++a, --b)
{
t = p[a];
p[a] = p[b];
p[b] = t;
}
return false;
}
return true;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 3, 4, 1 };
int N = arr.Length;
int K = 2;
// Function Call
maximizeInversions(arr, N, K);
}
}
// This code contributed by Princi Singh 输出：
1时间复杂度： O((N!)*(N – K) 2 )
辅助空间： O(N)
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