# 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:

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++

 `// C++ implementation of the approach ` `#include ` `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:

```7
```

Time Complexity: O(N)

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