Maximize product of subarray sum with its minimum element
Last Updated :
11 Jul, 2022
Given an array arr[] consisting of N positive integers, the task is to find the maximum product of subarray sum with the minimum element of that subarray.
Examples:
Input: arr[] = {3, 1, 6, 4, 5, 2}
Output: 60
Explanation:
The required maximum product can be obtained using subarray {6, 4, 5}
Therefore, maximum product = (6 + 4 + 5) * (4) = 60
Input: arr[] = {4, 1, 2, 9, 3}
Output: 81
Explanation:
The required maximum product can be obtained using subarray {9}
Maximum product = (9)* (9) = 81
Naive Approach: The simplest approach to solve the problem is to generate all subarrays of the given array and for each subarray, calculate the sum of the subarray, and multiply it with the minimum element in the subarray. Update the maximum product by comparing it with the product calculated. Finally, print the maximum product obtained after processing all the subarray.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized using a Stack and Prefix Sum Array. The idea is to use the stack to get the index of nearest smaller elements on the left and right of each element. Now, using these, the required product can be obtained. Follow the steps below to solve the problem:
- Initialize an array presum[] to store all the resultant prefix sum array of the given array.
- Initialize two arrays l[] and r[] to store the index of the nearest left and right smaller elements respectively.
- For every element arr[i], calculate l[i] and r[i] using a stack.
- Traverse the given array and for each index i, the product can be calculated by:
arr[i] * (presum[r[i]] - presum[l[i]-1])
- Print the maximum product after completing all the above steps
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the
// maximum product possible
void maxValue(int a[], int n)
{
// Stores prefix sum
int presum[n];
presum[0] = a[0];
// Find the prefix sum array
for(int i = 1; i < n; i++)
{
presum[i] = presum[i - 1] + a[i];
}
// l[] and r[] stores index of
// nearest smaller elements on
// left and right respectively
int l[n], r[n];
stack<int> st;
// Find all left index
for(int i = 1; i < n; i++)
{
// Until stack is non-empty
// & top element is greater
// than the current element
while (!st.empty() &&
a[st.top()] >= a[i])
st.pop();
// If stack is empty
if (!st.empty())
l[i] = st.top() + 1;
else
l[i] = 0;
// Push the current index i
st.push(i);
}
// Reset stack
while(!st.empty())
st.pop();
// Find all right index
for(int i = n - 1; i >= 0; i--)
{
// Until stack is non-empty
// & top element is greater
// than the current element
while (!st.empty() &&
a[st.top()] >= a[i])
st.pop();
if (!st.empty())
r[i] = st.top() - 1;
else
r[i] = n - 1;
// Push the current index i
st.push(i);
}
// Stores the maximum product
int maxProduct = 0;
int tempProduct;
// Iterate over the range [0, n)
for(int i = 0; i < n; i++)
{
// Calculate the product
tempProduct = a[i] * (presum[r[i]] -
(l[i] == 0 ? 0 :
presum[l[i] - 1]));
// Update the maximum product
maxProduct = max(maxProduct,
tempProduct);
}
// Return the maximum product
cout << maxProduct;
}
// Driver Code
int main()
{
// Given array
int n = 6;
int arr[] = { 3, 1, 6, 4, 5, 2 };
// Function call
maxValue(arr, n);
}
// This code is contributed by grand_master
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Function to find the
// maximum product possible
public static void
maxValue(int[] a, int n)
{
// Stores prefix sum
int[] presum = new int[n];
presum[0] = a[0];
// Find the prefix sum array
for (int i = 1; i < n; i++) {
presum[i] = presum[i - 1] + a[i];
}
// l[] and r[] stores index of
// nearest smaller elements on
// left and right respectively
int[] l = new int[n], r = new int[n];
Stack<Integer> st = new Stack<>();
// Find all left index
for (int i = 1; i < n; i++) {
// Until stack is non-empty
// & top element is greater
// than the current element
while (!st.isEmpty()
&& a[st.peek()] >= a[i])
st.pop();
// If stack is empty
if (!st.isEmpty())
l[i] = st.peek() + 1;
else
l[i] = 0;
// Push the current index i
st.push(i);
}
// Reset stack
st.clear();
// Find all right index
for (int i = n - 1; i >= 0; i--) {
// Until stack is non-empty
// & top element is greater
// than the current element
while (!st.isEmpty()
&& a[st.peek()] >= a[i])
st.pop();
if (!st.isEmpty())
r[i] = st.peek() - 1;
else
r[i] = n - 1;
// Push the current index i
st.push(i);
}
// Stores the maximum product
int maxProduct = 0;
int tempProduct;
// Iterate over the range [0, n)
for (int i = 0; i < n; i++) {
// Calculate the product
tempProduct
= a[i]
* (presum[r[i]]
- (l[i] == 0 ? 0
: presum[l[i] - 1]));
// Update the maximum product
maxProduct
= Math.max(maxProduct,
tempProduct);
}
// Return the maximum product
System.out.println(maxProduct);
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 3, 1, 6, 4, 5, 2 };
// Function Call
maxValue(arr, arr.length);
}
}
# Python3 program to implement
# the above approach
# Function to find the
# maximum product possible
def maxValue(a, n):
# Stores prefix sum
presum = [0 for i in range(n)]
presum[0] = a[0]
# Find the prefix sum array
for i in range(1, n, 1):
presum[i] = presum[i - 1] + a[i]
# l[] and r[] stores index of
# nearest smaller elements on
# left and right respectively
l = [0 for i in range(n)]
r = [0 for i in range(n)]
st = []
# Find all left index
for i in range(1, n):
# Until stack is non-empty
# & top element is greater
# than the current element
while (len(st) and
a[st[len(st) - 1]] >= a[i]):
st.remove(st[len(st) - 1])
# If stack is empty
if (len(st)):
l[i] = st[len(st) - 1] + 1;
else:
l[i] = 0
# Push the current index i
st.append(i)
# Reset stack
while(len(st)):
st.remove(st[len(st) - 1])
# Find all right index
i = n - 1
while(i >= 0):
# Until stack is non-empty
# & top element is greater
# than the current element
while (len(st) and
a[st[len(st) - 1]] >= a[i]):
st.remove(st[len(st) - 1])
if (len(st)):
r[i] = st[len(st) - 1] - 1
else:
r[i] = n - 1
# Push the current index i
st.append(i)
i -= 1
# Stores the maximum product
maxProduct = 0
# Iterate over the range [0, n)
for i in range(n):
# Calculate the product
if l[i] == 0:
tempProduct = (a[i] *
presum[r[i]])
else:
tempProduct = (a[i] *
(presum[r[i]] -
presum[l[i] - 1]))
# Update the maximum product
maxProduct = max(maxProduct,
tempProduct)
# Return the maximum product
print(maxProduct)
# Driver Code
if __name__ == '__main__':
# Given array
n = 6
arr = [ 3, 1, 6, 4, 5, 2 ]
# Function call
maxValue(arr, n)
# This code is contributed by SURENDRA_GANGWAR
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the
// maximum product possible
public static void maxValue(int[] a,
int n)
{
// Stores prefix sum
int[] presum = new int[n];
presum[0] = a[0];
// Find the prefix sum array
for(int i = 1; i < n; i++)
{
presum[i] = presum[i - 1] + a[i];
}
// l[] and r[] stores index of
// nearest smaller elements on
// left and right respectively
int[] l = new int[n], r = new int[n];
Stack<int> st = new Stack<int>();
// Find all left index
for(int i = 1; i < n; i++)
{
// Until stack is non-empty
// & top element is greater
// than the current element
while (st.Count > 0 &&
a[st.Peek()] >= a[i])
st.Pop();
// If stack is empty
if (st.Count > 0)
l[i] = st.Peek() + 1;
else
l[i] = 0;
// Push the current index i
st.Push(i);
}
// Reset stack
st.Clear();
// Find all right index
for(int i = n - 1; i >= 0; i--)
{
// Until stack is non-empty
// & top element is greater
// than the current element
while (st.Count > 0 &&
a[st.Peek()] >= a[i])
st.Pop();
if (st.Count > 0)
r[i] = st.Peek() - 1;
else
r[i] = n - 1;
// Push the current index i
st.Push(i);
}
// Stores the maximum product
int maxProduct = 0;
int tempProduct;
// Iterate over the range [0, n)
for(int i = 0; i < n; i++)
{
// Calculate the product
tempProduct = a[i] * (presum[r[i]] -
(l[i] == 0 ? 0 :
presum[l[i] - 1]));
// Update the maximum product
maxProduct = Math.Max(maxProduct,
tempProduct);
}
// Return the maximum product
Console.WriteLine(maxProduct);
}
// Driver code
static void Main()
{
// Given array
int[] arr = { 3, 1, 6, 4, 5, 2 };
// Function call
maxValue(arr, arr.Length);
}
}
// This code is contributed by divyeshrabadiya07
<script>
// Javascript program to implement
// the above approach
// Function to find the
// maximum product possible
function maxValue(a, n)
{
// Stores prefix sum
var presum = Array(n);
presum[0] = a[0];
// Find the prefix sum array
for(var i = 1; i < n; i++)
{
presum[i] = presum[i - 1] + a[i];
}
// l[] and r[] stores index of
// nearest smaller elements on
// left and right respectively
var l = Array(n).fill(0), r = Array(n).fill(0);
var st = [];
// Find all left index
for(var i = 1; i < n; i++)
{
// Until stack is non-empty
// & top element is greater
// than the current element
while (st.length!=0 &&
a[st[st.length-1]] >= a[i])
st.pop();
// If stack is empty
if (st.length!=0)
l[i] = st[st.length-1] + 1;
else
l[i] = 0;
// Push the current index i
st.push(i);
}
// Reset stack
while(st.length!=0)
st.pop();
// Find all right index
for(var i = n - 1; i >= 0; i--)
{
// Until stack is non-empty
// & top element is greater
// than the current element
while (st.length!=0 &&
a[st[st.length-1]] >= a[i])
st.pop();
if (st.length!=0)
r[i] = st[st.length-1] - 1;
else
r[i] = n - 1;
// Push the current index i
st.push(i);
}
// Stores the maximum product
var maxProduct = 0;
var tempProduct;
// Iterate over the range [0, n)
for(var i = 0; i < n; i++)
{
// Calculate the product
tempProduct = a[i] * (presum[r[i]] -
(l[i] == 0 ? 0 :
presum[l[i] - 1]));
// Update the maximum product
maxProduct = Math.max(maxProduct,
tempProduct);
}
// Return the maximum product
document.write( maxProduct);
}
// Driver Code
// Given array
var n = 6;
var arr = [3, 1, 6, 4, 5, 2];
// Function call
maxValue(arr, n);
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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