Given an array that contains both positive and negative integers, the task is to find the product of the maximum product subarray.
Examples:
Input: arr[] = {6, -3, -10, 0, 2}
Output: 180
Explanation: The subarray is {6, -3, -10}Input: arr[] = {-1, -3, -10, 0, 60}
Output: 60
Explanation: The subarray is {60}
Maximum Product Subarray by Traverse Over Every Contiguous Subarray:
The idea is to traverse over every contiguous subarray, find the product of each of these subarrays and return the maximum product from these results.
Follow the below steps to solve the problem:
-
Run a nested for loop to generate every subarray
- Calculate the product of elements in the current subarray
- Return the maximum of these products calculated from the subarrays
Below is the implementation of the above approach:
// C++ program to find Maximum Product Subarray #include <bits/stdc++.h> using namespace std;
/* Returns the product of max product subarray.*/ int maxSubarrayProduct( int arr[], int n)
{ // Initializing result
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
result = max(result, mul);
mul *= arr[j];
}
// updating the result for (n-1)th index.
result = max(result, mul);
}
return result;
} // Driver code int main()
{ int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Maximum Sub array product is "
<< maxSubarrayProduct(arr, n);
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// C program to find Maximum Product Subarray #include <stdio.h> // Find maximum between two numbers. int max( int num1, int num2)
{ return (num1 > num2) ? num1 : num2;
} /* Returns the product of max product subarray.*/ int maxSubarrayProduct( int arr[], int n)
{ // Initializing result
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
result = max(result, mul);
mul *= arr[j];
}
// updating the result for (n-1)th index.
result = max(result, mul);
}
return result;
} // Driver code int main()
{ int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
int n = sizeof (arr) / sizeof (arr[0]);
printf ( "Maximum Sub array product is %d " ,
maxSubarrayProduct(arr, n));
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// Java program to find maximum product subarray import java.io.*;
class GFG {
/* Returns the product of max product subarray.*/
static int maxSubarrayProduct( int arr[])
{
// Initializing result
int result = arr[ 0 ];
int n = arr.length;
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
result = Math.max(result, mul);
mul *= arr[j];
}
// updating the result for (n-1)th index.
result = Math.max(result, mul);
}
return result;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1 , - 2 , - 3 , 0 , 7 , - 8 , - 2 };
System.out.println( "Maximum Sub array product is "
+ maxSubarrayProduct(arr));
}
} // This code is contributed by Aditya Kumar (adityakumar129) |
# Python3 program to find Maximum Product Subarray # Returns the product of max product subarray. def maxSubarrayProduct(arr, n):
# Initializing result
result = arr[ 0 ]
for i in range (n):
mul = arr[i]
# traversing in current subarray
for j in range (i + 1 , n):
# updating result every time
# to keep an eye over the maximum product
result = max (result, mul)
mul * = arr[j]
# updating the result for (n-1)th index.
result = max (result, mul)
return result
# Driver code arr = [ 1 , - 2 , - 3 , 0 , 7 , - 8 , - 2 ]
n = len (arr)
print ( "Maximum Sub array product is" , maxSubarrayProduct(arr, n))
# This code is contributed by divyeshrabadiya07 |
// C# program to find maximum product subarray using System;
class GFG {
// Returns the product of max product subarray
static int maxSubarrayProduct( int [] arr)
{
// Initializing result
int result = arr[0];
int n = arr.Length;
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
result = Math.Max(result, mul);
mul *= arr[j];
}
// Updating the result for (n-1)th index
result = Math.Max(result, mul);
}
return result;
}
// Driver Code
public static void Main(String[] args)
{
int [] arr = { 1, -2, -3, 0, 7, -8, -2 };
Console.Write( "Maximum Sub array product is "
+ maxSubarrayProduct(arr));
}
} // This code is contributed by shivanisinghss2110 |
<script> // Javascript program to find Maximum Product Subarray /* Returns the product of max product subarray.*/ function maxSubarrayProduct(arr, n)
{ // Initializing result
let result = arr[0];
for (let i = 0; i < n; i++)
{
let mul = arr[i];
// traversing in current subarray
for (let j = i + 1; j < n; j++)
{
// updating result every time
// to keep an eye over the maximum product
result = Math.max(result, mul);
mul *= arr[j];
}
// updating the result for (n-1)th index.
result = Math.max(result, mul);
}
return result;
} // Driver code let arr = [ 1, -2, -3, 0, 7, -8, -2 ];
let n = arr.length;
document.write( "Maximum Sub array product is "
+ maxSubarrayProduct(arr, n));
// This code is contributed by Mayank Tyagi </script> |
Maximum Sub array product is 112
Time Complexity: O(N2)
Auxiliary Space: O(1)
Maximum Product Subarray using Kadane’s Algorithm
The idea is to use Kadane’s algorithm and maintain 3 variables max_so_far, max_ending_here & min_ending_here. Iterate the indices 0 to N-1 and update the variables such that:
- max_ending_here = maximum(arr[i], max_ending_here * arr[i], min_ending_here[i]*arr[i])
- min_ending_here = minimum(arr[i], max_ending_here * arr[i], min_ending_here[i]*arr[i])
- update the max_so_far with the maximum value for each index.
return max_so_far as the result.
Follow the below steps to solve the problem:
- Use 3 variables, 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 the maximum product subarray
Below is the implementation of the above approach:
// C++ program to find Maximum Product Subarray #include <bits/stdc++.h> using namespace std;
/* Returns the product of max product subarray. */ int maxSubarrayProduct( int arr[], int n)
{ // max positive product
// ending at the current position
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];
/* Traverse through the array.
the maximum product subarray ending at an index
will be the maximum of the element itself,
the product of element and max product ending previously
and the min product ending previously. */
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);
}
return max_so_far;
} // Driver code int main()
{ int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Maximum Sub array product is "
<< maxSubarrayProduct(arr, n);
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// C program to find Maximum Product Subarray #include <stdio.h> // Find maximum between two numbers. int max( int num1, int num2)
{ return (num1 > num2) ? num1 : num2;
} // Find minimum between two numbers. int min( int num1, int num2)
{ return (num1 > num2) ? num2 : num1;
} /* Returns the product of max product subarray. */ int maxSubarrayProduct( int arr[], int n)
{ // max positive product
// ending at the current position
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];
/* Traverse through the array.
the maximum product subarray ending at an index
will be the maximum of the element itself,
the product of element and max product ending previously
and the min product ending previously. */
for ( int i = 1; i < n; i++) {
int temp
= max(max(arr[i], arr[i] * max_ending_here),
arr[i] * min_ending_here);
min_ending_here
= min(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);
}
return max_so_far;
} // Driver code int main()
{ int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
int n = sizeof (arr) / sizeof (arr[0]);
printf ( "Maximum Sub array product is %d" ,
maxSubarrayProduct(arr, n));
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
/*package whatever //do not write package name here */ import java.io.*;
class GFG {
// Java program to find Maximum Product Subarray
// Returns the product
// of max product subarray.
static int maxSubarrayProduct( int arr[], int n)
{
// max positive product
// ending at the current position
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 ];
// /* Traverse through the array.
// the maximum product subarray ending at an index
// will be the maximum of the element itself,
// the product of element and max product ending
// previously and the min product ending previously.
// */
for ( int i = 1 ; i < n; i++) {
int temp = Math.max(
Math.max(arr[i], arr[i] * max_ending_here),
arr[i] * min_ending_here);
min_ending_here = Math.min(
Math.min(arr[i], arr[i] * max_ending_here),
arr[i] * min_ending_here);
max_ending_here = temp;
max_so_far
= Math.max(max_so_far, max_ending_here);
}
return max_so_far;
}
// Driver code
public static void main(String args[])
{
int [] arr = { 1 , - 2 , - 3 , 0 , 7 , - 8 , - 2 };
int n = arr.length;
System.out.printf( "Maximum Sub array product is %d" ,
maxSubarrayProduct(arr, n));
}
} // This code is contributed by shinjanpatra |
# Python3 program to find Maximum Product Subarray # Returns the product # of max product subarray. def maxSubarrayProduct(arr, n):
# max positive product
# ending at the current position
max_ending_here = arr[ 0 ]
# min negative product ending
# at the current position
min_ending_here = arr[ 0 ]
# Initialize overall max product
max_so_far = arr[ 0 ]
# /* Traverse through the array.
# the maximum product subarray ending at an index
# will be the maximum of the element itself,
# the product of element and max product ending previously
# and the min product ending previously. */
for i in range ( 1 , n):
temp = max ( max (arr[i], arr[i] * max_ending_here),
arr[i] * min_ending_here)
min_ending_here = min (
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)
return max_so_far
# Driver code arr = [ 1 , - 2 , - 3 , 0 , 7 , - 8 , - 2 ]
n = len (arr)
print (f "Maximum Sub array product is {maxSubarrayProduct(arr, n)}" )
# This code is contributed by shinjanpatra |
// C# program to find maximum product subarray using System;
class GFG {
/* Returns the product of max product subarray.
Assumes that the given array always has a subarray
with product more than 1 */
static int maxSubarrayProduct( int [] arr)
{
// max positive product
// ending at the current position
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];
/* Traverse through the array.
the maximum product subarray ending at an index
will be the maximum of the element itself,
the product of element and max product ending
previously and the min product ending previously. */
for ( int i = 1; i < arr.Length; i++) {
int temp = Math.Max(
Math.Max(arr[i], arr[i] * max_ending_here),
arr[i] * min_ending_here);
min_ending_here = Math.Min(
Math.Min(arr[i], arr[i] * max_ending_here),
arr[i] * min_ending_here);
max_ending_here = temp;
max_so_far
= Math.Max(max_so_far, max_ending_here);
}
return max_so_far;
}
// Driver Code
public static void Main()
{
int [] arr = { 1, -2, -3, 0, 7, -8, -2 };
Console.WriteLine( "Maximum Sub array product is "
+ maxSubarrayProduct(arr));
}
} // This code is contributed by CodeWithMini |
<script> // JavaScript program to find Maximum Product Subarray /* Returns the product of max product subarray. */ function maxSubarrayProduct(arr, n)
{ // max positive product
// ending at the current position
let max_ending_here = arr[0];
// min negative product ending
// at the current position
let min_ending_here = arr[0];
// Initialize overall max product
let max_so_far = arr[0];
/* Traverse through the array.
the maximum product subarray ending at an index
will be the maximum of the element itself,
the product of element and max product ending previously
and the min product ending previously. */
for (let i = 1; i < n; i++)
{
let temp = Math.max(Math.max(arr[i], arr[i] * max_ending_here), arr[i] * min_ending_here);
min_ending_here = Math.min(Math.min(arr[i], arr[i] * max_ending_here), arr[i] * min_ending_here);
max_ending_here = temp;
max_so_far = Math.max(max_so_far, max_ending_here);
}
return max_so_far;
} // Driver code let arr = [ 1, -2, -3, 0, 7, -8, -2 ] let n = arr.length document.write( "Maximum Sub array product is " +maxSubarrayProduct(arr, n));
// This code is contributed by shinjanpatra </script> |
Maximum Sub array product is 112
Time Complexity: O(N)
Auxiliary Space: O(1)
Maximum Product Subarray using Traversal From Starting and End of an Array:
We will follow a simple approach that is to traverse and multiply elements and if our value is greater than the previously stored value then store this value in place of the previously stored value. If we encounter “0” then make products of all elements till now equal to 1 because from the next element, we will start a new subarray.
But what can be the problem with that?
Problem will occur when our array will contain odd no. of negative elements. In that case, we have to reject anyone negative element so that we can even no. of negative elements and their product can be positive. Now since we are considering subarray so we can’t simply reject any one negative element. We have to either reject the first negative element or the last negative element.
But if we will traverse from starting then only the last negative element can be rejected and if we traverse from the last then the first negative element can be rejected. So we will traverse from both the end and from both the traversal we will take answer from that traversal only which will give maximum product subarray.
So actually we will reject that negative element whose rejection will give us the maximum product’s subarray.
Below is the implementation of the above approach:
// C++ program to find Maximum Product Subarray #include <bits/stdc++.h> using namespace std;
/* Returns the product of max product subarray. */ long long int maxSubarrayProduct( int arr[], int n)
{ long long ans=INT_MIN;
long long product=1;
for ( int i=0;i<n;i++){
product*=arr[i];
ans=max(ans,product);
if (arr[i]==0){product=1;}
}
product=1;
for ( int i=n-1;i>=0;i--){
product*=arr[i];
ans=max(ans,product);
if (arr[i]==0){product=1;}
}
return ans;
} // Driver code int main()
{ int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Maximum Sub array product is "
<< maxSubarrayProduct(arr, n);
return 0;
} |
// Java program to find Maximum Product Subarray import java.util.*;
public class Main {
/* Returns the product of max product subarray. */
public static long maxSubarrayProduct( int [] arr, int n)
{
long ans = Integer.MIN_VALUE;
long product = 1 ;
// Traverse the array from left to right
for ( int i = 0 ; i < n; i++) {
product *= arr[i];
ans = Math.max(ans, product);
if (arr[i] == 0 ) {
product = 1 ;
}
}
product = 1 ;
// Traverse the array from right to left
for ( int i = n - 1 ; i >= 0 ; i--) {
product *= arr[i];
ans = Math.max(ans, product);
if (arr[i] == 0 ) {
product = 1 ;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int [] arr = { 1 , - 2 , - 3 , 0 , 7 , - 8 , - 2 };
int n = arr.length;
System.out.println( "Maximum Subarray product is "
+ maxSubarrayProduct(arr, n));
}
} |
# Python program to find Maximum Product Subarray import sys
# Returns the product of max product subarray. def maxSubarrayProduct(arr, n):
ans = - sys.maxsize - 1 # Initialize the answer to the minimum possible value
product = 1
for i in range (n):
product * = arr[i]
ans = max (ans, product) # Update the answer with the maximum of the current answer and product
if arr[i] = = 0 :
product = 1 # Reset the product to 1 if the current element is 0
product = 1
for i in range (n - 1 , - 1 , - 1 ):
product * = arr[i]
ans = max (ans, product)
if arr[i] = = 0 :
product = 1
return ans
# Driver code arr = [ 1 , - 2 , - 3 , 0 , 7 , - 8 , - 2 ]
n = len (arr)
print ( "Maximum Subarray product is" , maxSubarrayProduct(arr, n))
|
using System;
public class MainClass {
// Returns the product of max product subarray.
public static int MaxSubarrayProduct( int [] arr, int n)
{
int ans
= int .MinValue; // Initialize the answer to the
// minimum possible value
int product = 1;
for ( int i = 0; i < n; i++) {
product *= arr[i];
ans = Math.Max(
ans, product); // Update the answer with the
// maximum of the current
// answer and product
if (arr[i] == 0) {
product = 1; // Reset the product to 1 if
// the current element is 0
}
}
product = 1;
for ( int i = n - 1; i >= 0; i--) {
product *= arr[i];
ans = Math.Max(ans, product);
if (arr[i] == 0) {
product = 1;
}
}
return ans;
}
// Driver code
public static void Main()
{
int [] arr = { 1, -2, -3, 0, 7, -8, -2 };
int n = arr.Length;
Console.WriteLine( "Maximum Subarray product is "
+ MaxSubarrayProduct(arr, n));
}
} |
// JavaScript program to find Maximum Product Subarray // Function to find the maximum product subarray function maxSubarrayProduct(arr, n) {
let ans = -Infinity;
let product = 1;
for (let i = 0; i < n; i++) {
product *= arr[i];
ans = Math.max(ans, product);
if (arr[i] === 0) {
product = 1;
}
}
product = 1;
for (let i = n - 1; i >= 0; i--) {
product *= arr[i];
ans = Math.max(ans, product);
if (arr[i] === 0) {
product = 1;
}
}
return ans;
} // Driver code const arr = [1, -2, -3, 0, 7, -8, -2]; const n = arr.length; console.log(`Maximum Subarray product is ${maxSubarrayProduct(arr, n)}`); //This code is written by Sundaram |
Maximum Sub array product is 112
Time Complexity: O(N)
Auxiliary Space: O(1)