Maximize product of subarray sum with its minimum element
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) = 60Input: 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> |
60
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array