给定一个大小为N的数组arr[ ] ，考虑一个数组prefix[ ] ，其中prefix[i]是arr的前i 个元素的总和。任务是找到修改给定数组所需的最少操作次数，使得前缀数组中任意两个相邻元素的乘积为负。在一个操作中，您可以将任何元素的值增加或减少 1。
Input: arr[] = {1, -3, 1, 0}
Output: 4
Explanation: The sequence can be transformed into 1, -2, 2, -2 by 4 operations, the sum of the prefixes are 1, -1, 1, -1 which satisfy all the conditions.
Input: arr[] = {-1 4 3 2 -5 4}
Output: 8
方法：这个想法是尝试两种独立的可能性，即偶数长度前缀和为正，奇数长度前缀和为负，反之亦然。请按照以下步骤解决问题：
- 初始化一个变量res = INT_MAX来存储最小操作次数。
- 使用变量r遍历范围 [0, 1] 。
- 初始化变量ans = 0和sum = 0分别存储总操作数和当前前缀和。
- 使用变量i遍历范围 [0, N-1] ，
- 将arr[i]添加到sum的值。
- 如果(i+r) 的值为奇数且sum不是正数，则将sum+1添加到ans并将sum的值更新为 1。
- 否则，如果第（i + r）为偶数，如果总和不为负然后添加总和+ 1到ANS更新的值的总和为-1。
- 将 res 的值更新为 min(res, ans)。
- 完成上述步骤后，打印res的值。
C++
// C++ code for the above approach
#include
using namespace std;
// Function to find minimum operations
// needed to make the product of any
// two adjacent elements in prefix
// sum array negative
void minOperations(vector a)
{
// Stores the minimum operations
int res = INT_MAX;
int N = a.size();
for (int r = 0; r < 2; r++) {
// Stores the prefix sum
// and number of operations
int sum = 0, ans = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update the value of sum
sum += a[i];
// Check if i+r is odd
if ((i + r) % 2) {
// Check if prefix sum
// is not positive
if (sum <= 0) {
// Update the value of
// ans and sum
ans += -sum + 1;
sum = 1;
}
}
else {
// Check if prefix sum is
// not negative
if (sum >= 0) {
// Update the value of
// ans and sum
ans += sum + 1;
sum = -1;
}
}
}
// Update the value of res
res = min(res, ans);
}
// Print the value of res
cout << res;
}
// Driver Code
int main()
{
vector a{ 1, -3, 1, 0 };
minOperations(a);
return 0;
} Java
// Java code for the above approach
import java.util.*;
class GFG{
// Function to find minimum operations
// needed to make the product of any
// two adjacent elements in prefix
// sum array negative
static void minOperations(ArrayList a)
{
// Stores the minimum operations
int res = Integer.MAX_VALUE;
int N = a.size();
for(int r = 0; r < 2; r++)
{
// Stores the prefix sum
// and number of operations
int sum = 0, ans = 0;
// Traverse the array
for(int i = 0; i < N; i++)
{
// Update the value of sum
sum += a.get(i);
// Check if i+r is odd
if ((i + r) % 2 == 1)
{
// Check if prefix sum
// is not positive
if (sum <= 0)
{
// Update the value of
// ans and sum
ans += -sum + 1;
sum = 1;
}
}
else
{
// Check if prefix sum is
// not negative
if (sum >= 0)
{
// Update the value of
// ans and sum
ans += sum + 1;
sum = -1;
}
}
}
// Update the value of res
res = Math.min(res, ans);
}
// Print the value of res
System.out.print(res);
}
// Driver Code
public static void main(String args[])
{
ArrayList a = new ArrayList();
a.add(1);
a.add(-3);
a.add(1);
a.add(0);
minOperations(a);
}
}
// This code is contributed by SURENDRA_GANGWAR Python3
# python code for the above approach
# // Function to find minimum operations
# // needed to make the product of any
# // two adjacent elements in prefix
# // sum array negative
def minOperations(a):
#Stores the minimum operations
res = 100000000000
N = len(a)
for r in range(0,2):
# Stores the prefix sum
# and number of operations
sum = 0
ans = 0
# Traverse the array
for i in range (0,N):
# Update the value of sum
sum += a[i]
# Check if i+r is odd
if ((i + r) % 2):
# Check if prefix sum
# is not positive
if (sum <= 0):
# Update the value of
# ans and sum
ans += -sum + 1
sum = 1
else:
# Check if prefix sum is
# not negative
if (sum >= 0):
# Update the value of
# ans and sum
ans += sum + 1;
sum = -1;
# Update the value of res
res = min(res, ans)
# // Print the value of res
print(res)
# Driver code
a = [1, -3, 1, 0]
minOperations(a);
# This code is contributed by Stream_CipherC#
// C# code for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find minimum operations
// needed to make the product of any
// two adjacent elements in prefix
// sum array negative
static void minOperations(List a)
{
// Stores the minimum operations
int res = Int32.MaxValue;
int N = a.Count;
for (int r = 0; r < 2; r++)
{
// Stores the prefix sum
// and number of operations
int sum = 0, ans = 0;
// Traverse the array
for (int i = 0; i < N; i++)
{
// Update the value of sum
sum += a[i];
// Check if i+r is odd
if ((i + r) % 2 == 1) {
// Check if prefix sum
// is not positive
if (sum <= 0) {
// Update the value of
// ans and sum
ans += -sum + 1;
sum = 1;
}
}
else {
// Check if prefix sum is
// not negative
if (sum >= 0) {
// Update the value of
// ans and sum
ans += sum + 1;
sum = -1;
}
}
}
// Update the value of res
res = Math.Min(res, ans);
}
// Print the value of res
Console.Write(res);
}
// Driver Code
public static void Main()
{
List a = new List(){ 1, -3, 1, 0 };
minOperations(a);
}
}
// This code is contributed by bgangwar59. Javascript
输出
4时间复杂度： O(N)
辅助空间： O(1)
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