Minimum steps to make the product of the array equal to 1

Given an array arr[] containing N integers. In one step, any element of the array can either be increased or decreased by one. The task is to find minimum steps required such that the product of the array elements becomes 1.

Examples:

Input: arr[] = { -2, 4, 0 }
Output: 5
We can change -2 to -1, 0 to -1 and 4 to 1.
So, a total of 5 steps are required to update the elements
such that the product of the final array is 1.

Input: arr[] = { -1, 1, -1 }
Output: 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

The product of the array elements can only be equal to 1 when there are only 1s and -1s in the array and the count of -1s is even.
Now, all the positive numbers can be reduced to 1 because they are more closer to 1 than they are closer to -1.
Similarly, all the negative numbers can be updated to -1.
If there are 0s present in the array then they can be reduced to either 1 or -1 according to the situation (the count of -1s must be even).
If the count of -ve numbers are even then they are always going to yield -1.
But if there are odd number of -ve numbers then they are going to yield odd number of -1s. To fix that, there are two possibilities:
- First try to find the count 0s in the array because it will take 1 operation to be -1.
- If there are no zeros in the array then just add 2 in the answer because it will take two steps to make 1 to -1.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the minimum 
// steps requried 
int MinStep(int a[], int n) 
{ 
  
    // To store the count of 0s, positive 
    // and negative numbers 
    int positive = 0, 
        negative = 0, 
        zero = 0; 
  
    // To store the ans 
    int step = 0; 
  
    for (int i = 0; i < n; i++) { 
  
        // If array element is 
        // equal to 0 
        if (a[i] == 0) { 
            zero++; 
        } 
  
        // If array element is 
        // a negative number 
        else if (a[i] < 0) { 
            negative++; 
  
            // Extra cost needed 
            // to make it -1 
            step = step + (-1 - a[i]); 
        } 
  
        // If array element is 
        // a positive number 
        else { 
            positive++; 
  
            // Extra cost needed 
            // to make it 1 
            step = step + (a[i] - 1); 
        } 
    } 
  
    // Now the array will 
    // have -1, 0 and 1 only 
    if (negative % 2 == 0) { 
  
        // As count of negative is even 
        // so we will change all 0 to 1 
        // total cost here will be 
        // count of 0s 
        step = step + zero; 
    } 
    else { 
  
        // If there are zeroes present 
        // in the array 
        if (zero > 0) { 
  
            // Change one zero to -1 
            // and rest of them to 1 
            // Total cost here will 
            // be count of '0' 
            step = step + zero; 
        } 
  
        // If there are no zeros in the array 
        else { 
  
            // As no 0s are availabe so we 
            // have to change one -1 to 1 
            // which will cost 2 to 
            // change -1 to 1 
            step = step + 2; 
        } 
    } 
  
    return step; 
} 
  
// Driver code 
int main() 
{ 
    int a[] = { 0, -2, -1, -3, 4 }; 
    int n = sizeof(a) / sizeof(a[0]); 
  
    cout << MinStep(a, n); 
  
    return 0; 
} 

Java

// Java implementation of the approach  
class GFG 
{ 
      
    // Function to return the minimum  
    // steps requried  
    static int MinStep(int a[], int n)  
    {  
      
        // To store the count of 0s, positive  
        // and negative numbers  
        int positive = 0,  
            negative = 0,  
            zero = 0;  
      
        // To store the ans  
        int step = 0;  
      
        for (int i = 0; i < n; i++)  
        {  
      
            // If array element is  
            // equal to 0  
            if (a[i] == 0)  
            {  
                zero++;  
            }  
      
            // If array element is  
            // a negative number  
            else if (a[i] < 0)  
            {  
                negative++;  
      
                // Extra cost needed  
                // to make it -1  
                step = step + (-1 - a[i]);  
            }  
      
            // If array element is  
            // a positive number  
            else 
            {  
                positive++;  
      
                // Extra cost needed  
                // to make it 1  
                step = step + (a[i] - 1);  
            }  
        }  
      
        // Now the array will  
        // have -1, 0 and 1 only  
        if (negative % 2 == 0)  
        {  
      
            // As count of negative is even  
            // so we will change all 0 to 1  
            // total cost here will be  
            // count of 0s  
            step = step + zero;  
        }  
        else 
        {  
      
            // If there are zeroes present  
            // in the array  
            if (zero > 0) 
            {  
      
                // Change one zero to -1  
                // and rest of them to 1  
                // Total cost here will  
                // be count of '0'  
                step = step + zero;  
            }  
      
            // If there are no zeros in the array  
            else 
            {  
      
                // As no 0s are availabe so we  
                // have to change one -1 to 1  
                // which will cost 2 to  
                // change -1 to 1  
                step = step + 2;  
            }  
        }  
      
        return step;  
    }  
      
    // Driver code  
    public static void main (String[] args)  
    {  
        int a[] = { 0, -2, -1, -3, 4 };  
        int n = a.length;  
      
        System.out.println(MinStep(a, n));  
    }  
} 
  
// This code is contributed by AnkitRai01 

Python3

# Python3 implementation of the approach 
  
# Function to return the minimum 
# steps requried 
def MinStep(a, n): 
      
    # To store the count of 0s, positive 
    # and negative numbers 
    positive = 0; 
    negative = 0; 
    zero = 0; 
  
    # To store the ans 
    step = 0; 
  
    for i in range(n): 
          
        # If array element is 
        # equal to 0 
        if (a[i] == 0): 
            zero += 1; 
              
        # If array element is 
        # a negative number 
        elif (a[i] < 0): 
  
            negative += 1; 
  
            # Extra cost needed 
            # to make it -1 
            step = step + (-1 - a[i]); 
  
        # If array element is 
        # a positive number 
        else: 
            positive += 1; 
  
            # Extra cost needed 
            # to make it 1 
            step = step + (a[i] - 1); 
  
    # Now the array will 
    # have -1, 0 and 1 only 
    if (negative % 2 == 0): 
  
        # As count of negative is even 
        # so we will change all 0 to 1 
        # total cost here will be 
        # count of 0s 
        step = step + zero; 
  
    else: 
  
        # If there are zeroes present 
        # in the array 
        if (zero > 0): 
  
            # Change one zero to -1 
            # and rest of them to 1 
            # Total cost here will 
            # be count of '0' 
            step = step + zero; 
  
        # If there are no zeros in the array 
        else: 
  
            # As no 0s are availabe so we 
            # have to change one -1 to 1 
            # which will cost 2 to 
            # change -1 to 1 
            step = step + 2; 
    return step; 
  
# Driver code 
if __name__ == '__main__': 
    a = [0, -2, -1, -3, 4]; 
    n = len(a); 
  
    print(MinStep(a, n)); 
  
# This code is contributed by PrinciRaj1992 

C#

// C# implementation of the approach  
using System; 
  
class GFG 
{ 
          
    // Function to return the minimum  
    // steps requried  
    static int MinStep(int []a, int n)  
    {  
      
        // To store the count of 0s,  
        // positive and negative numbers  
        int positive = 0,  
            negative = 0,  
            zero = 0;  
      
        // To store the ans  
        int step = 0;  
      
        for (int i = 0; i < n; i++)  
        {  
      
            // If array element is  
            // equal to 0  
            if (a[i] == 0)  
            {  
                zero++;  
            }  
      
            // If array element is  
            // a negative number  
            else if (a[i] < 0)  
            {  
                negative++;  
      
                // Extra cost needed  
                // to make it -1  
                step = step + (-1 - a[i]);  
            }  
      
            // If array element is  
            // a positive number  
            else
            {  
                positive++;  
      
                // Extra cost needed  
                // to make it 1  
                step = step + (a[i] - 1);  
            }  
        }  
      
        // Now the array will  
        // have -1, 0 and 1 only  
        if (negative % 2 == 0)  
        {  
      
            // As count of negative is even  
            // so we will change all 0 to 1  
            // total cost here will be  
            // count of 0s  
            step = step + zero;  
        }  
        else
        {  
      
            // If there are zeroes present  
            // in the array  
            if (zero > 0) 
            {  
      
                // Change one zero to -1  
                // and rest of them to 1  
                // Total cost here will  
                // be count of '0'  
                step = step + zero;  
            }  
      
            // If there are no zeros in the array  
            else
            {  
      
                // As no 0s are availabe so we  
                // have to change one -1 to 1  
                // which will cost 2 to  
                // change -1 to 1  
                step = step + 2;  
            }  
        }  
      
        return step;  
    }  
      
    // Driver code  
    static public void Main () 
    { 
        int []a = { 0, -2, -1, -3, 4 };  
        int n = a.Length;  
      
        Console.Write(MinStep(a, n));  
    }  
} 
  
// This code is contributed by ajit. 

Output:

Time Complexity: O(N)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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