# 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:** Follow the steps below to solve the problem:

- 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 closer to**1**than they are closer to**-1**. - Similarly, all the negative numbers can be updated to
**-1**. - If there are
**0**s 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 an 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**.

- First try to find the count

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 required 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 available 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 required 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 available 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 required 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 available 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 required 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 available 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.

## Javascript

<script> // Javascript implementation of the approach // Function to return the minimum // steps required function MinStep(a, n) { // To store the count of 0s, positive // and negative numbers let positive = 0, negative = 0, zero = 0; // To store the ans let step = 0; for (let 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 available 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 let a = [ 0, -2, -1, -3, 4 ]; let n = a.length; document.write(MinStep(a, n)); </script>

**Output:**

7

**Time Complexity:** O(N)

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