给定两个整数X和Y以及一个长度为N的二进制数组arr[]其第一个和最后一个元素为1 ，任务是最小化将所有数组元素转换为0的成本，其中X和Y表示转换子数组的成本的所有1秒至0秒和分别任何元素转换为0的成本。
例子：
Input: arr[] = {1, 1, 1, 0, 1, 1}, X = 10, Y = 4
Output: 14
Explanation: To minimize the cost to convert all elements to 0, perform the following operations:
- Change element at index 3 to 1. Now the array modifies to {1, 1, 1, 1, 1, 1}. The cost of this operation is 4.
- Change all element of the array to 0. The cost of this operation is 10.
Therefore, the total cost is 4 + 10 + 14.
Input: arr[] = {1, 0, 0, 1, 1, 0, 1}, X = 2, Y = 3
Output: 6
Explanation: To minimize the cost of changing all array elements to 0, perform the following operations:
- Change all element of the subarray over the range [3, 4] to 0. Now the array modifies to {1, 0, 0, 0, 0, 0, 1}. The cost of this operation is 2.
- Change all element of the subarray over the range [0, 0] to 0. Now the array modifies to {0, 0, 0, 0, 0, 0, 1}. The cost of this operation is 2.
- Change all element of the subarray over the range [6, 6] to 0. Now the array modifies to {0, 0, 0, 0, 0, 0, 0}. The cost of this operation is 2.
Therefore, the total cost is 2 + 2 + 2 = 3.
方法：按照以下步骤操作：
- 初始化一个变量，比如ans ，以存储将所有数组元素转换为0的最小成本。
- 计算并存储所有仅由0组成的子数组的长度，并将其存储在一个向量中，并按升序对向量进行排序。
- 现在，计算仅由1组成的子数组的数量。
- 使用变量i遍历给定数组，其中i表示Y成本操作的数量，并执行以下操作：
- 对于成本为Y 的每个可能的操作次数，通过执行X操作来找到成本。
- 由于在两组1之间设置位时，组的总数会减少，因此首先合并两组连续的1以减少最小操作次数。
- 为每个索引找到完成上述步骤的最小成本作为currCost并更新ans以存储ans和currCost的最小值。
- 完成上述步骤后，打印ans的值作为最小成本。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate the minimum cost
// of converting all array elements to 0s
void minimumCost(int* binary, int n,
int a, int b)
{
// Stores subarrays of 0s only
vector groupOfZeros;
int len = 0, i = 0;
bool increment_need = true;
// Traverse the array
while (i < n) {
increment_need = true;
// If consecutive 0s occur
while (i < n && binary[i] == 0) {
len++;
i++;
increment_need = false;
}
// Increment if needed
if (increment_need == true) {
i++;
}
// Push the current length of
// consecutive 0s in a vector
if (len != 0) {
groupOfZeros.push_back(len);
}
// Update lengths as 0
len = 0;
}
// Sorting vector
sort(groupOfZeros.begin(),
groupOfZeros.end());
i = 0;
bool found_ones = false;
// Stores the number of
// subarrays consisting of 1s
int NumOfOnes = 0;
// Traverse the array
while (i < n) {
found_ones = false;
// If current element is 1
while (i < n && binary[i] == 1) {
i++;
found_ones = true;
}
if (found_ones == false)
i++;
// Otherwise
else
// Increment count of
// consecutive ones
NumOfOnes++;
}
// Stores the minimum cost
int ans = INT_MAX;
// Traverse the array
for (int i = 0; i < n; i++) {
int curr = 0, totalOnes = NumOfOnes;
// First element
if (i == 0) {
curr = totalOnes * a;
}
else {
int mark = i, num_of_changes = 0;
// Traverse the subarray sizes
for (int x : groupOfZeros) {
if (mark >= x) {
totalOnes--;
mark -= x;
// Update cost
num_of_changes += x;
}
else {
break;
}
}
// Cost of performing X
// and Y operations
curr = (num_of_changes * b)
+ (totalOnes * a);
}
// Find the minimum cost
ans = min(ans, curr);
}
// Print the minimum cost
cout << ans;
}
// Driver Code
int main()
{
int arr[] = { 1, 1, 1, 0, 1, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
int X = 10, Y = 4;
// Function Call
minimumCost(arr, N, X, Y);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to calculate the minimum cost
// of converting all array elements to 0s
public static void minimumCost(int[] binary, int n,
int a, int b)
{
// Stores subarrays of 0s only
List groupOfZeros = new ArrayList();
int len = 0, i = 0;
boolean increment_need = true;
// Traverse the array
while (i < n)
{
increment_need = true;
// If consecutive 0s occur
while (i < n && binary[i] == 0)
{
len++;
i++;
increment_need = false;
}
// Increment if needed
if (increment_need == true)
{
i++;
}
// Push the current length of
// consecutive 0s in a vector
if (len != 0)
{
groupOfZeros.add(len);
}
// Update lengths as 0
len = 0;
}
// Sorting List
Collections.sort(groupOfZeros);
i = 0;
boolean found_ones = false;
// Stores the number of
// subarrays consisting of 1s
int NumOfOnes = 0;
// Traverse the array
while (i < n)
{
found_ones = false;
// If current element is 1
while (i < n && binary[i] == 1)
{
i++;
found_ones = true;
}
if (found_ones == false)
i++;
// Otherwise
else
// Increment count of
// consecutive ones
NumOfOnes++;
}
// Stores the minimum cost
int ans = Integer.MAX_VALUE;
// Traverse the array
for(int i1 = 0; i1 < n; i1++)
{
int curr = 0, totalOnes = NumOfOnes;
// First element
if (i1 == 0)
{
curr = totalOnes * a;
}
else
{
int mark = i1, num_of_changes = 0;
// Traverse the subarray sizes
for(int x : groupOfZeros)
{
if (mark >= x)
{
totalOnes--;
mark -= x;
// Update cost
num_of_changes += x;
}
else
{
break;
}
}
// Cost of performing X
// and Y operations
curr = (num_of_changes * b) +
(totalOnes * a);
}
// Find the minimum cost
ans = Math.min(ans, curr);
}
// Print the minimum cost
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 1, 1, 0, 1, 1 };
int N = 6;
int X = 10, Y = 4;
// Function Call
minimumCost(arr, N, X, Y);
}
}
// This code is contributed by RohitOberoi Python3
# Python3 program for the above approach
import sys
# Function to calculate the minimum cost
# of converting all array elements to 0s
def minimumCost(binary, n,
a, b):
# Stores subarrays of 0s only
groupOfZeros = []
length = 0
i = 0
increment_need = True
# Traverse the array
while (i < n):
increment_need = True
# If consecutive 0s occur
while (i < n and binary[i] == 0):
length += 1
i += 1
increment_need = False
# Increment if needed
if (increment_need == True):
i += 1
# Push the current length of
# consecutive 0s in a vector
if (length != 0):
groupOfZeros.append(length)
# Update lengths as 0
length = 0
# Sorting vector
groupOfZeros.sort()
i = 0
found_ones = False
# Stores the number of
# subarrays consisting of 1s
NumOfOnes = 0
# Traverse the array
while (i < n):
found_ones = False
# If current element is 1
while (i < n and binary[i] == 1):
i += 1
found_ones = True
if (found_ones == False):
i += 1
# Otherwise
else:
# Increment count of
# consecutive ones
NumOfOnes += 1
# Stores the minimum cost
ans = sys.maxsize
# Traverse the array
for i in range(n):
curr = 0
totalOnes = NumOfOnes
# First element
if (i == 0):
curr = totalOnes * a
else:
mark = i
num_of_changes = 0
# Traverse the subarray sizes
for x in groupOfZeros:
if (mark >= x):
totalOnes -= 1
mark -= x
# Update cost
num_of_changes += x
else:
break
# Cost of performing X
# and Y operations
curr = ((num_of_changes * b)
+ (totalOnes * a))
# Find the minimum cost
ans = min(ans, curr)
# Print the minimum cost
print(ans)
# Driver Code
if __name__ == "__main__":
arr = [1, 1, 1, 0, 1, 1]
N = len(arr)
X = 10
Y = 4
# Function Call
minimumCost(arr, N, X, Y)
# This code is contributed by chitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate the minimum cost
// of converting all array elements to 0s
public static void minimumCost(int[] binary, int n,
int a, int b)
{
// Stores subarrays of 0s only
List groupOfZeros = new List();
int len = 0, i = 0;
bool increment_need = true;
// Traverse the array
while (i < n)
{
increment_need = true;
// If consecutive 0s occur
while (i < n && binary[i] == 0)
{
len++;
i++;
increment_need = false;
}
// Increment if needed
if (increment_need == true)
{
i++;
}
// Push the current length of
// consecutive 0s in a vector
if (len != 0)
{
groupOfZeros.Add(len);
}
// Update lengths as 0
len = 0;
}
// Sorting List
groupOfZeros.Sort();
i = 0;
bool found_ones = false;
// Stores the number of
// subarrays consisting of 1s
int NumOfOnes = 0;
// Traverse the array
while (i < n)
{
found_ones = false;
// If current element is 1
while (i < n && binary[i] == 1)
{
i++;
found_ones = true;
}
if (found_ones == false)
i++;
// Otherwise
else
// Increment count of
// consecutive ones
NumOfOnes++;
}
// Stores the minimum cost
int ans = int.MaxValue;
// Traverse the array
for(int i1 = 0; i1 < n; i1++)
{
int curr = 0, totalOnes = NumOfOnes;
// First element
if (i1 == 0)
{
curr = totalOnes * a;
}
else
{
int mark = i1, num_of_changes = 0;
// Traverse the subarray sizes
foreach(int x in groupOfZeros)
{
if (mark >= x)
{
totalOnes--;
mark -= x;
// Update cost
num_of_changes += x;
}
else
{
break;
}
}
// Cost of performing X
// and Y operations
curr = (num_of_changes * b) +
(totalOnes * a);
}
// Find the minimum cost
ans = Math.Min(ans, curr);
}
// Print the minimum cost
Console.WriteLine(ans);
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 1, 1, 0, 1, 1 };
int N = 6;
int X = 10, Y = 4;
// Function Call
minimumCost(arr, N, X, Y);
}
}
// This code is contributed by Amit Katiyar Javascript
输出：
14时间复杂度： O(N*log N)
辅助空间： O(1)
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