Given an array arr[] consisting of positive, negative, and zero values, find the maximum product that can be obtained from any contiguous subarray of arr[].
Examples:
Input: arr[] = [-2, 6, -3, -10, 0, 2]
Output: 180
Explanation: The subarray with maximum product is [6, -3, -10] with product = 6 * (-3) * (-10) = 180.
Input: arr[] = [-1, -3, -10, 0, 6]
Output: 30
Explanation: The subarray with maximum product is [-3, -10] with product = (-3) * (-10) = 30.
Input: arr[] = [2, 3, 4]
Output: 24
Explanation: For an array with all positive elements, the result is product of all elements.
[Naive Approach] Using Two Nested Loops – O(n^2) Time and O(1) Space
The idea is to traverse over every contiguous subarray, find the product of each of these subarrays and return the maximum product among all the subarrays.
C++
#include <iostream>
#include<vector>
using namespace std;
int maxProduct(vector<int> &arr) {
int n = arr.size();
// Initializing result
int maxProd = arr[0];
for (int i = 0; i < n; i++) {
int mul = 1;
// traversing in current subarray
for (int j = i; j < n; j++) {
mul *= arr[j];
// updating result every time
// to keep track of the maximum product
maxProd = max(maxProd, mul);
}
}
return maxProd;
}
int main() {
vector<int> arr = { -2, 6, -3, -10, 0, 2 };
cout << maxProduct(arr);
return 0;
}
#include <stdio.h>
int maxProduct(int arr[], int n) {
// Initializing result
int maxProd = arr[0];
for (int i = 0; i < n; i++) {
int mul = 1;
// traversing in current subarray
for (int j = i; j < n; j++) {
mul *= arr[j];
// updating result every time
// to keep track of the maximum product
maxProd = (maxProd > mul) ? maxProd : mul;
}
}
return maxProd;
}
int main() {
int arr[] = { -2, 6, -3, -10, 0, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%lld\n", maxProduct(arr, n));
return 0;
}
class GfG {
static int maxProduct(int arr[]) {
int n = arr.length;
// Initializing result
int maxProd = arr[0];
for (int i = 0; i < n; i++) {
int mul = 1;
// traversing in current subarray
for (int j = i; j < n; j++) {
mul *= arr[j];
// updating result every time
// to keep track of the maximum product
maxProd = Math.max(maxProd, mul);
}
}
return maxProd;
}
public static void main(String[] args) {
int arr[] = { -2, 6, -3, -10, 0, 2 };
System.out.println(maxProduct(arr));
}
}
def maxProduct(arr):
n = len(arr)
# Initializing result
maxProd = arr[0]
for i in range(n):
mul = 1
# traversing in current subarray
for j in range(i, n):
mul *= arr[j]
# updating result every time
# to keep track of the maximum product
maxProd = max(maxProd, mul)
return maxProd
if __name__ == "__main__":
arr = [-2, 6, -3, -10, 0, 2]
print(maxProduct(arr))
using System;
class GfG {
static int maxProduct(int[] arr) {
int n = arr.Length;
// Initializing result
int maxProd = arr[0];
for (int i = 0; i < n; i++) {
int mul = 1;
// traversing in current subarray
for (int j = i; j < n; j++) {
mul *= arr[j];
// updating result every time
// to keep track of the maximum product
maxProd = Math.Max(maxProd, mul);
}
}
return maxProd;
}
public static void Main(String[] args) {
int[] arr = { -2, 6, -3, -10, 0, 2 };
Console.Write(maxProduct(arr));
}
}
function maxProduct(arr) {
const n = arr.length;
// Initializing result
let maxProd = arr[0];
for (let i = 0; i < n; i++) {
let mul = 1;
// Traversing in current subarray
for (let j = i; j < n; j++) {
mul *= arr[j];
// Update result to keep track of the maximum product
if(mul > maxProd)
maxProd = mul;
}
}
return maxProd;
}
// Driver Code
const arr = [-2, 6, -3, -10, 0, 2];
console.log(maxProduct(arr).toString());
[Expected Approach - 1] Greedy Min-Max Product - O(n) Time and O(1) Space
Let's assume that the input array has only positive elements. Then, we can simply iterate from left to right keeping track of the maximum running product ending at any index. The maximum product would be the product ending at the last index. The problem arises when we encounter zero or a negative element.
If we encounter zero, then all the subarrays containing this zero will have product = 0, so zero simply resets the product of the subarray.
If we encounter a negative number, we need to keep track of the minimum product as well as the maximum product ending at the previous index. This is because when we multiply the minimum product with a negative number, it can give us the maximum product. So, keeping track of minimum product ending at any index is important as it can lead to the maximum product on encountering a negative number.
Step By Step Approach:
- Initialize three variables: currMax, currMin, and maxProd with the first element of the array.
- Loop through the array from index 1 to end to evaluate every position’s contribution.
- For each index, calculate the temporary maximum using max of current value, current × currMax, and current × currMin.
- Update currMin using min of current value, current × currMax, and current × currMin.
- Assign currMax to the previously computed temporary maximum value.
- Update maxProd by comparing it with the updated currMax value.
- Finally, return maxProd which stores the maximum product of any subarray.
Working:
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maxProduct(vector<int> &arr) {
int n = arr.size();
// max product ending at the current index
int currMax = arr[0];
// min product ending at the current index
int currMin = arr[0];
// Initialize overall max product
int maxProd = arr[0];
// Iterate through the array
for (int i = 1; i < n; i++) {
// Temporary variable to store the maximum product ending
// at the current index
int temp = max({ arr[i], arr[i] * currMax,
arr[i] * currMin });
// Update the minimum product ending at the current index
currMin = min({ arr[i], arr[i] * currMax,
arr[i] * currMin });
// Update the maximum product ending at the current index
currMax = temp;
// Update the overall maximum product
maxProd = max(maxProd, currMax);
}
return maxProd;
}
int main() {
vector<int> arr = { -2, 6, -3, -10, 0, 2 };
cout << maxProduct(arr);
return 0;
}
#include <stdio.h>
#include <limits.h>
int max(int a, int b, int c) {
if (a >= b && a >= c) return a;
if (b >= a && b >= c) return b;
return c;
}
int min(int a, int b, int c) {
if (a <= b && a <= c) return a;
if (b <= a && b <= c) return b;
return c;
}
int maxProduct(int arr[], int n) {
// max product ending at the current index
int currMax = arr[0];
// min product ending at the current index
int currMin = arr[0];
// Initialize overall max product
int maxProd = arr[0];
// Iterate through the array
for (int i = 1; i < n; i++) {
// Temporary variable to store the maximum product ending at the
// current index
int temp = max(arr[i], arr[i] * currMax, arr[i] * currMin);
// Update the minimum product ending at the current index
currMin = min(arr[i], arr[i] * currMax, arr[i] * currMin);
// Update the maximum product ending at the current index
currMax = temp;
// Update the overall maximum product
maxProd = max(maxProd, currMax, maxProd);
}
return maxProd;
}
int main() {
int arr[] = { -2, 6, -3, -10, 0, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%lld\n", maxProduct(arr, n));
return 0;
}
class GfG {
static int max(int a, int b, int c) {
return Math.max(a, Math.max(b, c));
}
static int min(int a, int b, int c) {
return Math.min(a, Math.min(b, c));
}
static int maxProduct(int[] arr) {
int n = arr.length;
// max product ending at the current index
int currMax = arr[0];
// min product ending at the current index
int currMin = arr[0];
// Initialize overall max product
int maxProd = arr[0];
// Iterate through the array
for (int i = 1; i < n; i++) {
// Temporary variable to store the maximum product ending
// at the current index
int temp = max(arr[i], arr[i] * currMax, arr[i] * currMin);
// Update the minimum product ending at the current index
currMin = min(arr[i], arr[i] * currMax, arr[i] * currMin);
// Update the maximum product ending at the current index
currMax = temp;
// Update the overall maximum product
maxProd = Math.max(maxProd, currMax);
}
return maxProd;
}
public static void main(String[] args) {
int[] arr = { -2, 6, -3, -10, 0, 2 };
System.out.println(maxProduct(arr));
}
}
def maxProduct(arr):
n = len(arr)
# max product ending at the current index
currMax = arr[0]
# min product ending at the current index
currMin = arr[0]
# Initialize overall max product
maxProd = arr[0]
# Iterate through the array
for i in range(1, n):
# Temporary variable to store the maximum product ending
# at the current index
temp = max(arr[i], arr[i] * currMax, arr[i] * currMin)
# Update the minimum product ending at the current index
currMin = min(arr[i], arr[i] * currMax, arr[i] * currMin)
# Update the maximum product ending at the current index
currMax = temp
# Update the overall maximum product
maxProd = max(maxProd, currMax)
return maxProd
if __name__ == "__main__":
arr = [-2, 6, -3, -10, 0, 2]
print(maxProduct(arr))
using System;
class GfG {
static int maxProduct(int[] arr) {
int n = arr.Length;
// max product ending at the current index
int currMax = arr[0];
// min product ending at the current index
int currMin = arr[0];
// Initialize overall max product
int maxProd = arr[0];
// Iterate through the array
for (int i = 1; i < n; i++) {
// Temporary variable to store the maximum product ending
// at the current index
int temp = Math.Max(arr[i], Math.Max(arr[i] * currMax,
arr[i] * currMin));
// Update the minimum product ending at the current index
currMin = Math.Min(arr[i], Math.Min(arr[i] * currMax,
arr[i] * currMin));
// Update the maximum product ending at the current index
currMax = temp;
// Update the overall maximum product
maxProd = Math.Max(maxProd, currMax);
}
return maxProd;
}
static void Main() {
int[] arr = { -2, 6, -3, -10, 0, 2 };
Console.WriteLine(maxProduct(arr));
}
}
function max(a, b, c) {
return a > b ? (a > c ? a : c) : (b > c ? b : c);
}
function min(a, b, c) {
return a < b ? (a < c ? a : c) : (b < c ? b : c);
}
function maxProduct(arr) {
// Initialize the maximum and minimum products ending at
// the current index
let currMax = arr[0];
let currMin = arr[0];
// Initialize the overall maximum product
let maxProd = arr[0];
// Iterate through the array starting from the second element
for (let i = 1; i < arr.length; i++) {
// Calculate potential maximum product at this index
const temp = max(arr[i] * currMax, arr[i] * currMin, arr[i]);
// Update the minimum product ending at the current index
currMin = min(arr[i] * currMax, arr[i] * currMin, arr[i]);
// Update the maximum product ending at the current index
currMax = temp;
// Update the overall maximum product
maxProd = max(maxProd, maxProd, currMax);
}
return maxProd;
}
// Driver Code
const arr = [ -2, 6, -3, -10, 0, 2 ];
console.log(maxProduct(arr).toString());
[Expected Approach - 2] By Traversing in Both Directions - O(n) Time and O(1) Space
We will follow a simple approach that is to traverse from the start and keep track of the running product and if the running product is greater than the max product, then we update the max product. Also, if we encounter '0' then make product of all elements till now equal to 1 because from the next element, we will start a new subarray.
But what is the problem with this approach?
Problem will occur when our array will contain odd no. of negative elements. In that case, we have to reject one negative element so that we can even no. of negative elements and their product can be positive. Now, since subarray should be contiguous so we can't simply reject any one negative element. We have to either reject the first negative element or the last negative element.
Now, if we traverse from start 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 ends and find the maximum product subarray.
Illustration:
C++
#include <iostream>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
int maxProduct(vector<int> &arr) {
int n = arr.size();
int maxProd = INT_MIN;
// leftToRight to store product from left to Right
int leftToRight = 1;
// rightToLeft to store product from right to left
int rightToLeft = 1;
for (int i = 0; i < n; i++) {
if (leftToRight == 0)
leftToRight = 1;
if (rightToLeft == 0)
rightToLeft = 1;
// calculate product from index left to right
leftToRight *= arr[i];
// calculate product from index right to left
int j = n - i - 1;
rightToLeft *= arr[j];
maxProd = max({leftToRight, rightToLeft, maxProd});
}
return maxProd;
}
int main() {
vector<int> arr = { -2, 6, -3, -10, 0, 2 };
cout << maxProduct(arr);
return 0;
}
#include <stdio.h>
#include <limits.h>
int maxProduct(int arr[], int n) {
int maxProd = LLONG_MIN;
// leftToRight to store product from left to Right
int leftToRight = 1;
// rightToLeft to store product from right to left
int rightToLeft = 1;
for (int i = 0; i < n; i++) {
if (leftToRight == 0)
leftToRight = 1;
if (rightToLeft == 0)
rightToLeft = 1;
// calculate product from index left to right
leftToRight *= arr[i];
// calculate product from index right to left
int j = n - i - 1;
rightToLeft *= arr[j];
maxProd = (leftToRight > maxProd ? leftToRight : maxProd);
maxProd = (rightToLeft > maxProd ? rightToLeft : maxProd);
}
return maxProd;
}
int main() {
int arr[] = { -2, 6, -3, -10, 0, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%lld\n", maxProduct(arr, n));
return 0;
}
class GfG {
static int maxProduct(int[] arr) {
int n = arr.length;
int maxProd = Integer.MIN_VALUE;
// leftToRight to store product from left to Right
int leftToRight = 1;
// rightToLeft to store product from right to left
int rightToLeft = 1;
for (int i = 0; i < n; i++) {
if (leftToRight == 0)
leftToRight = 1;
if (rightToLeft == 0)
rightToLeft = 1;
// calculate product from index left to right
leftToRight *= arr[i];
// calculate product from index right to left
int j = n - i - 1;
rightToLeft *= arr[j];
maxProd = Math.max(leftToRight,
Math.max(rightToLeft, maxProd));
}
return maxProd;
}
public static void main(String[] args) {
int[] arr = { -2, 6, -3, -10, 0, 2 };
System.out.println(maxProduct(arr));
}
}
def maxProduct(arr):
n = len(arr)
maxProd = float('-inf')
# leftToRight to store product from left to Right
leftToRight = 1
# rightToLeft to store product from right to left
rightToLeft = 1
for i in range(n):
if leftToRight == 0:
leftToRight = 1
if rightToLeft == 0:
rightToLeft = 1
# calculate product from index left to right
leftToRight *= arr[i]
# calculate product from index right to left
j = n - i - 1
rightToLeft *= arr[j]
maxProd = max(leftToRight, rightToLeft, maxProd)
return maxProd
if __name__=="__main__":
arr = [-2, 6, -3, -10, 0, 2]
print(maxProduct(arr))
using System;
class GfG {
static int maxProduct(int[] arr) {
int n = arr.Length;
int maxProd = int.MinValue;
// leftToRight to store product from left to Right
int leftToRight = 1;
// rightToLeft to store product from right to left
int rightToLeft = 1;
for (int i = 0; i < n; i++) {
if (leftToRight == 0)
leftToRight = 1;
if (rightToLeft == 0)
rightToLeft = 1;
// calculate product from index left to right
leftToRight *= arr[i];
// calculate product from index right to left
int j = n - i - 1;
rightToLeft *= arr[j];
maxProd = Math.Max(leftToRight, Math.Max(rightToLeft, maxProd));
}
return maxProd;
}
static void Main() {
int[] arr = { -2, 6, -3, -10, 0, 2 };
Console.WriteLine(maxProduct(arr));
}
}
function maxProduct(arr) {
let n = arr.length;
let maxProd = Number.MIN_SAFE_INTEGER;
// leftToRight to store product from left to Right
let leftToRight = 1;
// rightToLeft to store product from right to left
let rightToLeft = 1;
for (let i = 0; i < n; i++) {
if (leftToRight === 0)
leftToRight = 1;
if (rightToLeft === 0)
rightToLeft = 1;
// calculate product from index left to right
leftToRight *= arr[i];
// calculate product from index right to left
let j = n - i - 1;
rightToLeft *= arr[j];
maxProd = Math.max(leftToRight, rightToLeft, maxProd);
}
return maxProd;
}
// Driver Code
let arr = [ -2, 6, -3, -10, 0, 2 ];
console.log(maxProduct(arr));
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