Given an array that contains both positive and negative integers, find the product of the maximum product subarray. Expected Time complexity is O(n) and only O(1) extra space can be used.
Examples :
Input: arr[] = {6, -3, -10, 0, 2} Output: 180 // The subarray is {6, -3, -10} Input: arr[] = {-1, -3, -10, 0, 60} Output: 60 // The subarray is {60} Input: arr[] = {-1, -2, -3, 4} Output: 24 // The subarray is {-2, -3, 4} Input: arr[] = {-10} Output: 0 // An empty array is also subarray // and product of empty subarray is // considered as 0.
We have discussed a solution of this problem here.
In this post an interesting solution is discussed. The idea is based on the fact that overall maximum product is maximum of following two:
- Maximum product in left to right traversal.
- Maximum product in right to left traversal
For example, consider the above third sample input {-1, -2, -3, 4}. If we traverse the array only in forward direction (considering -1 as part of output), maximum product will be 2. If we traverse the array in backward direction (considering 4 as part of output), maximum product will be 24 i.e; { -2, -3, 4}.
One important thing is to handle 0’s. We need to compute fresh forward (or backward) sum whenever we see 0.
Below is the implementation of above idea :
// C++ program to find maximum product subarray #include<bits/stdc++.h> using namespace std;
// Function for maximum product int max_product( int arr[], int n)
{ // Initialize maximum products in forward and
// backward directions
int max_fwd = INT_MIN, max_bkd = INT_MIN;
// Initialize current product
int max_till_now = 1;
// max_fwd for maximum contiguous product in
// forward direction
// max_bkd for maximum contiguous product in
// backward direction
// iterating within forward direction in array
for ( int i=0; i<n; i++)
{
// if arr[i]==0, it is breaking condition
// for contiguous subarray
max_till_now = max_till_now*arr[i];
if (max_till_now == 0)
{
max_till_now = 1;
continue ;
}
if (max_fwd < max_till_now) // update max_fwd
max_fwd = max_till_now;
}
max_till_now = 1;
// iterating within backward direction in array
for ( int i=n-1; i>=0; i--)
{
max_till_now = max_till_now * arr[i];
if (max_till_now == 0)
{
max_till_now = 1;
continue ;
}
// update max_bkd
if (max_bkd < max_till_now)
max_bkd = max_till_now;
}
// return max of max_fwd and max_bkd
int res = max(max_fwd, max_bkd);
// Product should not be nagative.
// (Product of an empty subarray is
// considered as 0)
return max(res, 0);
} // Driver Program to test above function int main()
{ int arr[] = {-1, -2, -3, 4};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << max_product(arr, n) << endl;
return 0;
} |
// Java program to find // maximum product subarray import java.io.*;
class GFG
{ // Function for maximum product static int max_product( int arr[], int n)
{ // Initialize maximum products in
// forward and backward directions
int max_fwd = Integer.MIN_VALUE,
max_bkd = Integer.MIN_VALUE;
// Initialize current product
int max_till_now = 1 ;
// max_fwd for maximum contiguous
// product in forward direction
// max_bkd for maximum contiguous
// product in backward direction
// iterating within forward
// direction in array
for ( int i = 0 ; i < n; i++)
{
// if arr[i]==0, it is breaking
// condition for contiguous subarray
max_till_now = max_till_now * arr[i];
if (max_till_now == 0 )
{
max_till_now = 1 ;
continue ;
}
// update max_fwd
if (max_fwd < max_till_now)
max_fwd = max_till_now;
}
max_till_now = 1 ;
// iterating within backward
// direction in array
for ( int i = n - 1 ; i >= 0 ; i--)
{
max_till_now = max_till_now * arr[i];
if (max_till_now == 0 )
{
max_till_now = 1 ;
continue ;
}
// update max_bkd
if (max_bkd < max_till_now)
max_bkd = max_till_now;
}
// return max of max_fwd and max_bkd
int res = Math. max(max_fwd, max_bkd);
// Product should not be nagative.
// (Product of an empty subarray is
// considered as 0)
return Math.max(res, 0 );
} // Driver Code public static void main (String[] args)
{ int arr[] = {- 1 , - 2 , - 3 , 4 };
int n = arr.length;
System.out.println( max_product(arr, n) );
} } // This code is contributed by anuj_67. |
# Python3 program to find # maximum product subarray import sys
# Function for maximum product def max_product(arr, n):
# Initialize maximum products
# in forward and backward directions
max_fwd = - sys.maxsize - 1
max_bkd = - sys.maxsize - 1
# Initialize current product
max_till_now = 1
# max_fwd for maximum contiguous
# product in forward direction
# max_bkd for maximum contiguous
# product in backward direction
# iterating within forward
# direction in array
for i in range (n):
# if arr[i]==0, it is breaking
# condition for contiguous subarray
max_till_now = max_till_now * arr[i]
if (max_till_now = = 0 ):
max_till_now = 1 ;
continue
if (max_fwd < max_till_now): #update max_fwd
max_fwd = max_till_now
max_till_now = 1
# iterating within backward
# direction in array
for i in range (n - 1 , - 1 , - 1 ):
max_till_now = max_till_now * arr[i]
if (max_till_now = = 0 ):
max_till_now = 1
continue
# update max_bkd
if (max_bkd < max_till_now) :
max_bkd = max_till_now
# return max of max_fwd and max_bkd
res = max (max_fwd, max_bkd)
# Product should not be nagative.
# (Product of an empty subarray is
# considered as 0)
return max (res, 0 )
# Driver Code arr = [ - 1 , - 2 , - 3 , 4 ]
n = len (arr)
print (max_product(arr, n))
# This code is contributed # by Yatin Gupta |
// C# program to find maximum product // subarray using System;
class GFG {
// Function for maximum product
static int max_product( int []arr, int n)
{
// Initialize maximum products in
// forward and backward directions
int max_fwd = int .MinValue,
max_bkd = int .MinValue;
// Initialize current product
int max_till_now = 1;
// max_fwd for maximum contiguous
// product in forward direction
// max_bkd for maximum contiguous
// product in backward direction
// iterating within forward
// direction in array
for ( int i = 0; i < n; i++)
{
// if arr[i]==0, it is breaking
// condition for contiguous subarray
max_till_now = max_till_now * arr[i];
if (max_till_now == 0)
{
max_till_now = 1;
continue ;
}
// update max_fwd
if (max_fwd < max_till_now)
max_fwd = max_till_now;
}
max_till_now = 1;
// iterating within backward
// direction in array
for ( int i = n - 1; i >= 0; i--)
{
max_till_now = max_till_now * arr[i];
if (max_till_now == 0)
{
max_till_now = 1;
continue ;
}
// update max_bkd
if (max_bkd < max_till_now)
max_bkd = max_till_now;
}
// return max of max_fwd and max_bkd
int res = Math. Max(max_fwd, max_bkd);
// Product should not be nagative.
// (Product of an empty subarray is
// considered as 0)
return Math.Max(res, 0);
}
// Driver Code
public static void Main ()
{
int []arr = {-1, -2, -3, 4};
int n = arr.Length;
Console.Write( max_product(arr, n) );
}
} // This code is contributed by nitin mittal. |
<?php // PHP program to find maximum // product subarray // Function for maximum product function max_product( $arr , $n )
{ // Initialize maximum products
// in forward and backward
// directions
$max_fwd = PHP_INT_MIN;
$max_bkd = PHP_INT_MIN;
// Initialize current product
$max_till_now = 1;
// max_fwd for maximum contiguous
// product in forward direction
// max_bkd for maximum contiguous
// product in backward direction
// iterating within forward direction
// in array
for ( $i = 0; $i < $n ; $i ++)
{
// if arr[i]==0, it is
// breaking condition
// for contiguous subarray
$max_till_now = $max_till_now * $arr [ $i ];
if ( $max_till_now == 0)
{
$max_till_now = 1;
continue ;
}
// update max_fwd
if ( $max_fwd < $max_till_now )
$max_fwd = $max_till_now ;
}
$max_till_now = 1;
// iterating within backward
// direction in array
for ( $i = $n - 1; $i >= 0; $i --)
{
$max_till_now = $max_till_now * $arr [ $i ];
if ( $max_till_now == 0)
{
$max_till_now = 1;
continue ;
}
// update max_bkd
if ( $max_bkd < $max_till_now )
$max_bkd = $max_till_now ;
}
// return max of max_fwd
// and max_bkd
$res = max( $max_fwd , $max_bkd );
// Product should not be nagative.
// (Product of an empty subarray is
// considered as 0)
return max( $res , 0);
} // Driver Code
$arr = array (-1, -2, -3, 4);
$n = count ( $arr );
echo max_product( $arr , $n );
// This code is contributed by anuj_67. ?> |
Output :
24
Time Complexity : O(n)
Auxiliary Space : O(1)
Note that the above solution requires two traversals of array while the previous solution requires only one traversal.
This article is contributed by Shashank Mishra (Gullu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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