# Minimum elements to change so that for an index i all elements on the left are -ve and all elements on the right are +ve

Given an array arr of size n, the task is to find the minimum number of elements that should be changed (element value can be changed to anything) so that there exists an index 0 ? i ? n-2 such that:

1. All the numbers in range 0 to i (inclusive) are < 0.
2. All numbers in range i+1 to n-1 (inclusive) are > 0.

Examples:

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

Input: arr[] = {3, -5}
Output:
Explanation: Change 3 to -3 and -5 to 5

Approach: Fix the value of i, what changes would we need to make index i the required index? Change all the positive elements on the left of i to negative and all negative elements to the right of i to positive. Hence, the number of operations required would be:

(Number of positive terms on the left of i) + (Number of negative terms on the right of i)

To find the required terms, we can pre-compute them using suffix sum.
Hence, we try each i as the required index and choose the one which needs minimum changes.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to return the minimum number // of required changesint 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 functionint 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 changesstatic 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 changesdef 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 Codeif __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



Output
1

Complexity Analysis:

• Time Complexity: O(N), where N is the size of the given array.
• Auxiliary Space: O(N), for creating an additional array of size N+1.

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