Maximize product of subarray sum with its minimum element

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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

// 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

# 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#

// 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

Javascript

<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>

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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