Given an array arr of N integers, the task is to find the minimum steps in which the sum and product of all elements of the array can be made non-zero. In one step any element of the array can be incremented by 1.
Examples:
Input: N = 4, arr[] = {0, 1, 2, 3}
Output: 1
Explanation:
As product of all elements of the array is zero
Increment the array element 0 by 1, such that array sum and product is not equal to zero.
Input: N = 4, arr[] = {-1, -1, 0, 0}
Output: 3
Explanation:
As product of all elements of the array is zero
Increment the array element 2 and 3 by 1, such that array sum and product is not equal to zero
Approach: The idea is to break problem into two parts that is –
- Minimum steps required to make the array product not equal to zero.
- Minimum steps required to make the array sum not equal to zero.
For the product of all elements of the array not equal to zero, then every element of the array should be non-zero and to get the array sum not equal to zero increment any element by 1 if the array sum is zero.
For Example:
N = 4, arr[] = {0, 1, 2, 3}
Iterate over the array to find,
If there is an element that is zero.
If yes, then increment it by 1 and also
increment the number of steps by 1.
Again, Iterate over the updated array,
To check if the array sum is zero.
If the array sum of the updated array
is zero then increment any element by 1.
Algorithm:
- Iterate over the array to check if there is an element that is zero, then increment the element by 1 and also increment the number of steps by 1
- Again, Iterate over the array and find the sum of the array if the sum of the array is zero then increment any element by 1 and also increment the number of steps by 1.
Below is the implementation of the above approach:
C++
// C++ implementation to find the // minimum steps to make the array sum // and the product not equal to zero #include <bits/stdc++.h> using namespace std; int sum(int arr[], int n) { int sum = 0; for(int i= 0; i < n; i++) sum += arr[i]; return sum; } // Function to find the // minimum steps to make the array sum // and the product not equal to zero int steps(int n, int a[]) { // Variable to store the minimum // number of steps required int count_steps = 0; // Loop to iterate over the array to // find if there is any element in // array which is zero for(int i = 0; i < n; i++) { if(a[i] == 0) { a[i] += 1; count_steps += 1; } } // Condition to check if the sum // of array elements is zero if( sum(a, n) != 0) return count_steps; else return count_steps + 1; } // Driver code int main() { int n = 4; int a[] = {-1, -1, 0, 0}; int count = steps(n, a); cout<<(count); return 0; } // This code is contributed by Rajput-Ji |
Java
// Java implementation to find the // minimum steps to make the array sum // and the product not equal to zero class GFG { // Function to find the // minimum steps to make the array sum // and the product not equal to zero static int steps(int n, int []a) { // Variable to store the minimum // number of steps required int count_steps = 0; // Loop to iterate over the array to // find if there is any element in // array which is zero for(int i = 0; i < n; i++) { if(a[i] == 0) { a[i] += 1; count_steps += 1; } } // Condition to check if the sum // of array elements is zero if( sum(a) != 0) return count_steps; else return count_steps + 1; } static int sum(int[] arr) { int sum = 0; for(int i= 0; i < arr.length; i++) sum += arr[i]; return sum; } // Driver code public static void main(String []args) { int n = 4; int []a = {-1, -1, 0, 0}; int count = steps(n, a); System.out.println(count); } } // This code is contributed by Rajput-Ji |
Python
# Python implementation to find the # minimum steps to make the array sum # and the product not equal to zero # Function to find the # minimum steps to make the array sum # and the product not equal to zero def steps(n, a): # Variable to store the minimum # number of steps required count_steps = 0 # Loop to iterate over the array to # find if there is any element in # array which is zero for i in range(n): if a[i]== 0: a[i] += 1 count_steps += 1 # Condition to check if the sum # of array elements is zero if sum(a)!= 0: return count_steps else: return count_steps + 1 # Driver code if __name__ == "__main__": n = 4 a = [-1, -1, 0, 0] count = steps(n, a) print(count) |
C#
// C# implementation to find the // minimum steps to make the array sum // and the product not equal to zero using System; class GFG { // Function to find the // minimum steps to make the array sum // and the product not equal to zero static int steps(int n, int []a) { // Variable to store the minimum // number of steps required int count_steps = 0; // Loop to iterate over the array to // find if there is any element in // array which is zero for(int i = 0; i < n; i++) { if(a[i] == 0) { a[i] += 1; count_steps += 1; } } // Condition to check if the sum // of array elements is zero if( sum(a) != 0) return count_steps; else return count_steps + 1; } static int sum(int[] arr) { int sum = 0; for(int i= 0; i < arr.Length; i++) sum += arr[i]; return sum; } // Driver code public static void Main(String []args) { int n = 4; int []a = {-1, -1, 0, 0}; int count = steps(n, a); Console.WriteLine(count); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation to find the // minimum steps to make the array sum // and the product not equal to zero // Function to find the // minimum steps to make the array sum // and the product not equal to zero function steps(n, a) { // Variable to store the minimum // number of steps required let count_steps = 0; // Loop to iterate over the array to // find if there is any element in // array which is zero for(let i = 0; i < n; i++) { if(a[i] == 0) { a[i] += 1; count_steps += 1; } } // Condition to check if the sum // of array elements is zero if( sum(a) != 0) return count_steps; else return count_steps + 1; } function sum(arr) { let sum = 0; for(let i= 0; i < arr.length; i++) sum += arr[i]; return sum; } // Driver Code let n = 4; let a = [-1, -1, 0, 0]; let count = steps(n, a); document.write(count); </script> |
Output:
3
Performance Analysis:
- Time Complexity: In the given approach, there are two iterations to compute the minimum steps required to make the product to non-zero and another iteration to compute the sum of the array. O(N)
- Space Complexity: In the given approach, there is no extra space used. O(1)
