要更改的最小元素，以便对于索引 i，左侧的所有元素都是 -ve，右侧的所有元素都是 +ve

给定一个大小为n的数组arr ，任务是找到应该更改的最小元素数（元素值可以更改为任何值），以便存在索引0 ≤ i ≤ n-2 ，使得：

1. 0到i （含）范围内的所有数字均< 0 。
2. 在i+1到n-1 （包括）范围内的所有数字都是> 0 。

例子：

Input: arr[] = {-1, -2, -3, 3, -5, 3, 4} 
Output: 1 
Change -5 to 5 and the array becomes {-1, -2, -3, 3, 5, 3, 4}
Input: arr[] = {3, -5} 
Output: 2 
Change 3 to -3 and -5 to 5 
 

方法：固定i的值，我们需要进行哪些更改才能使索引i成为所需的索引？将i左侧的所有正元素更改为负，将i右侧的所有负元素更改为正。因此，所需的操作数量为：
(i 左边的正项数) + (i 右边的负数)
为了找到所需的术语，我们可以使用后缀 sum 预先计算它们。
因此，我们尝试将每个i作为所需的索引，并选择需要最小更改的索引。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum number
// of required changes
int minimumChanges(int n, int a[])
{
    int i, sf[n + 1];
    sf[n] = 0;
    for (i = n - 1; i >= 0; i--) {
        sf[i] = sf[i + 1];
        if (a[i] <= 0)
            sf[i]++;
    }
 
    // number of elements >=0 in prefix
    int pos = 0;
 
    // Minimum elements to change
    int mn = n;
    for (i = 0; i < n - 1; i++) {
        if (a[i] >= 0)
            pos++;
        mn = min(mn, pos + sf[i + 1]);
    }
    return mn;
}
 
// Driver Program to test above function
int main()
{
    int a[] = { -1, -2, -3, 3, -5, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimumChanges(n, a);
}

Java

// Java implementation of the approach
 
import java.io.*;
 
class GFG {
 
// Function to return the minimum number
// of required changes
static int minimumChanges(int n, int a[])
{
    int i;
    int []sf= new int[n+1];
    sf[n] = 0;
    for (i = n - 1; i >= 0; i--) {
        sf[i] = sf[i + 1];
        if (a[i] <= 0)
            sf[i]++;
    }
 
    // number of elements >=0 in prefix
    int pos = 0;
 
    // Minimum elements to change
    int mn = n;
    for (i = 0; i < n - 1; i++) {
        if (a[i] >= 0)
            pos++;
        mn = Math.min(mn, pos + sf[i + 1]);
    }
    return mn;
}
 
    // Driver Program to test above function
    public static void main (String[] args) {
    int []a = { -1, -2, -3, 3, -5, 3, 4 };
    int n = a.length;
    System.out.println( minimumChanges(n, a));
    }
}
// This code is contributed by inder_verma.

Python 3

# Python 3 implementation of the approach
 
# Function to return the minimum
# number of required changes
def minimumChanges(n, a):
 
    sf = [0] * (n + 1)
    sf[n] = 0
    for i in range(n - 1, -1, -1) :
        sf[i] = sf[i + 1]
        if (a[i] <= 0):
            sf[i] += 1
 
    # number of elements >=0 in prefix
    pos = 0
 
    # Minimum elements to change
    mn = n
    for i in range(n - 1) :
        if (a[i] >= 0):
            pos += 1
        mn = min(mn, pos + sf[i + 1])
     
    return mn
 
# Driver Code
if __name__ == "__main__":
     
    a = [ -1, -2, -3, 3, -5, 3, 4 ]
    n = len(a)
    print(minimumChanges(n, a))
 
# This code is contributed
# by ChitraNayal

C#

using System;
 
// C# implementation of the approach
 
public class GFG
{
 
// Function to return the minimum number 
// of required changes
public static int minimumChanges(int n, int[] a)
{
    int i;
    int[] sf = new int[n + 1];
    sf[n] = 0;
    for (i = n - 1; i >= 0; i--)
    {
        sf[i] = sf[i + 1];
        if (a[i] <= 0)
        {
            sf[i]++;
        }
    }
 
    // number of elements >=0 in prefix
    int pos = 0;
 
    // Minimum elements to change
    int mn = n;
    for (i = 0; i < n - 1; i++)
    {
        if (a[i] >= 0)
        {
            pos++;
        }
        mn = Math.Min(mn, pos + sf[i + 1]);
    }
    return mn;
}
 
    // Driver Program to test above function
    public static void Main(string[] args)
    {
    int[] a = new int[] {-1, -2, -3, 3, -5, 3, 4};
    int n = a.Length;
    Console.WriteLine(minimumChanges(n, a));
    }
}
 
// This code is contributed by Shrikant13

PHP

= 0; $i--)
    {
        $sf[$i] = $sf[$i + 1];
        if ($a[$i] <= 0)
            $sf[$i]++;
    }
 
    // number of elements >=0 in prefix
    $pos = 0;
 
    // Minimum elements to change
    $mn = $n;
    for ($i = 0; $i < $n - 1; $i++)
    {
        if ($a[$i] >= 0)
            $pos++;
        $mn = min($mn, $pos + $sf[$i + 1]);
    }
    return $mn;
}
 
// Driver Code
$a = array(-1, -2, -3, 3, -5, 3, 4 );
$n = sizeof($a);
echo minimumChanges($n, $a);
 
// This code is contributed by ajit
?>

Javascript

 
    // Javascript implementation of the approach
     
    // Function to return the minimum number 
    // of required changes
    function minimumChanges(n, a)
    {
        let i;
        let sf = new Array(n + 1);
        sf.fill(0);
        sf[n] = 0;
        for (i = n - 1; i >= 0; i--)
        {
            sf[i] = sf[i + 1];
            if (a[i] <= 0)
            {
                sf[i]++;
            }
        }
 
        // number of elements >=0 in prefix
        let pos = 0;
 
        // Minimum elements to change
        let mn = n;
        for (i = 0; i < n - 1; i++)
        {
            if (a[i] >= 0)
            {
                pos++;
            }
            mn = Math.min(mn, pos + sf[i + 1]);
        }
        return mn;
    }
     
    let a = [-1, -2, -3, 3, -5, 3, 4];
    let n = a.length;
    document.write(minimumChanges(n, a));
     

输出：

1

时间复杂度： O(n)