# C program to Find maximum product sub array in a given array

## Maximum product of sub-array in C

Here, in this page we will discuss the program to find the maximum product of sub-array in C programming language. We will discuss the two different ways to find the maximum product.

Here, we will discuss the following two methods, and compare the complexity of them,

• Method 1 : Naive solution
• Method 2 : Efficient solution

Let’s discuss above two methods in brief.

## Method 1 :

• Create a variable say result, set result = arr[0], this variable hold the required maximum product.
• Run a loop for range(n)
• Create a variable mul = arr[i], this variable hold the product of sub-array.
• Run a inner loop, set result = max(result, mul)
• And, mul *= arr[j]
• Update, result = max(result, mul)
• Return result.

### Time and Space Complexity :

• Time Complexity : O(n2)
• Space Complexity : O(1)

### Method 1 : Code in C

Run
```#include<stdio.h>

int main(){
int arr[] = { 10, -20, -30, 0, 70, -80, -20 };
int n=sizeof(arr)/sizeof(arr[0]);
int result = arr[0];

for (int i = 0; i < n; i++)
{
int mul = arr[i];
// traversing in current subarray
for (int j = i + 1; j < n; j++) { // updating result every time // to keep an eye over the // maximum product if(mul>result)
result = mul;
mul *= arr[j];
}
if(mul>result)
result = mul;
}

printf("Maximum Product of sub-array is %d", result);
}```

### Output

`Maximum Product of sub-array is 1600`

## Method 2 :

This is the efficient solution and is also similar to Largest Sum Contiguous Subarray problem which uses Kadane’s algorithm.

• Declare three variables say max_so_far, max_ending_here & min_ending_here.
• For every index the maximum number ending at that index will be the maximum(arr[i], max_ending_here * arr[i], min_ending_here[i]*arr[i]).
• Similarly the minimum number ending here will be the minimum of these 3.
• Thus we get the final value for maximum product subarray.

### Time and Space Complexity :

• Time Complexity : O(n)
• Space Complexity : O(1)

### Method 2 : Code in C

Run
```#include<stdio.h>

int max(int a, int b, int c){

if(a>=b && a>=c)
return a;
if(b>=a && b>=a)
return b;
return c;

}

int min(int a, int b, int c){

if(a<=b && a<=c)
return a;
if(b<=a && b<=a)
return b;
return c;

}

int main(){
int arr[] = { 1, -2, -3, 0, 7, -8, 2};
int n=sizeof(arr)/sizeof(arr[0]);
int max_ending_here = arr[0];

// min negative product ending
// at the current position
int min_ending_here = arr[0];

// Initialize overall max product
int max_so_far = arr[0];

for (int i = 1; i < n; i++) { int temp = max(arr[i], arr[i] * max_ending_here, arr[i] * min_ending_here); min_ending_here = min(arr[i], arr[i] * max_ending_here, arr[i] * min_ending_here); max_ending_here = temp; //max_so_far = max(max_so_far, max_ending_here); if(max_ending_here>max_so_far)
max_so_far = max_ending_here;
}

printf("Maximum Product of sub-array is %d", max_so_far);
}```

### Output

`Maximum Product of sub-array is 2`