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 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 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.
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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