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

Approach:

  1. 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.
  2. Now, all the positive numbers can be reduced to 1 because they are more closer to 1 than they are closer to -1.
  3. Similarly, all the negative numbers can be updated to -1.
  4. 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).
  5. If the count of -ve numbers are even then they are always going to yield -1.
  6. 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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.

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Output:

7

Time Complexity: O(N)



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