Maximize Sum possible by subtracting same value from all elements of a Subarray of the given Array

Last Updated : 28 Nov, 2022

Given an array a[] consisting of N integers, the task is to find the maximum possible sum that can be achieved by deducting any value, say X, from all elements of a subarray.

Examples:

Input: N = 3, a[] = {80, 48, 82}
Output: 144
Explanation:
48 can be deducted from each array element. Therefore, sum obtained = 48 * 3 = 144
Input: N = a[] = {8, 40, 77}
Output: 80
Explanation:
Subtracting 8 from all array elements generates sum 24.
Subtracting 40 from a[1] and a[2] generates sum 80.
Subtracting 77 from a[2] generates sum 77.
Therefore, maximum possible sum is 80.

Approach:
Follow the steps below to solve the problem:

Traverse the array
For every element, find the element which is nearest smaller on its left and nearest smaller on its right.
Calculate the sum possible by that element calculating current_element * ( j - i - 1 ) where j and i are the indices of the nearest smaller numbers on the left and right respectively.
Find the maximum possible sum among all of them.

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 generate previous smaller 
// element for each array element
vector<int> findPrevious(vector<int> a, int n)
{
    vector<int> ps(n);

    // The first element has no
    // previous smaller
    ps[0] = -1;

    // Stack to keep track of elements
    // that have occurred previously
    stack<int> Stack;

    // Push the first index
    Stack.push(0);
    
    for(int i = 1; i < n; i++)
    {
        
        // Pop all the elements until the previous
        // element is smaller than current element
        while (Stack.size() > 0 && 
             a[Stack.top()] >= a[i])
            Stack.pop();

        // Store the previous smaller element
        ps[i] = Stack.size() > 0 ? 
                Stack.top() : -1;

        // Push the index of the current element
        Stack.push(i);
    }

    // Return the array
    return ps;
}

// Function to generate next smaller element
// for each array element
vector<int> findNext(vector<int> a, int n)
{
    vector<int> ns(n);

    ns[n - 1] = n;

    // Stack to keep track of elements
    // that have occurring next
    stack<int> Stack;
    Stack.push(n - 1);

    // Iterate in reverse order
    // for calculating next smaller
    for(int i = n - 2; i >= 0; i--)
    {
        
        // Pop all the elements until the
        // next element is smaller
        // than current element
        while (Stack.size() > 0 && 
             a[Stack.top()] >= a[i])
            Stack.pop();

        // Store the next smaller element
        ns[i] = Stack.size() > 0 ? 
                Stack.top() : n;

        // Push the index of the current element
        Stack.push(i);
    }

    // Return the array
    return ns;
}

// Function to find the maximum sum by
// subtracting same value from all
// elements of a Subarray
int findMaximumSum(vector<int> a, int n)
{
    
    // Stores previous smaller element
    vector<int> prev_smaller = findPrevious(a, n);

    // Stores next smaller element
    vector<int> next_smaller = findNext(a, n);

    int max_value = 0;
    for(int i = 0; i < n; i++)
    {
        
        // Calculate contribution
        // of each element
        max_value = max(max_value, a[i] * 
                       (next_smaller[i] - 
                        prev_smaller[i] - 1));
    }

    // Return answer
    return max_value;
}

// Driver Code    
int main()
{
    int n = 3;
    vector<int> a{ 80, 48, 82 };
    
    cout << findMaximumSum(a, n);
    
    return 0;
}

// This code is contributed by divyeshrabadiya07

Java

// Java Program to implement
// the above approach
import java.util.*;

public class GFG {

    // Function to find the maximum sum by
    // subtracting same value from all
    // elements of a Subarray
    public static int findMaximumSum(int[] a, int n)
    {
        // Stores previous smaller element
        int prev_smaller[] = findPrevious(a, n);

        // Stores next smaller element
        int next_smaller[] = findNext(a, n);

        int max_value = 0;
        for (int i = 0; i < n; i++) {

            // Calculate contribution
            // of each element
            max_value
                = Math.max(max_value,
                        a[i] * (next_smaller[i]
                                - prev_smaller[i] - 1));
        }

        // Return answer
        return max_value;
    }

    // Function to generate previous smaller element
    // for each array element
    public static int[] findPrevious(int[] a, int n)
    {
        int ps[] = new int[n];

        // The first element has no
        // previous smaller
        ps[0] = -1;

        // Stack to keep track of elements
        // that have occurred previously
        Stack<Integer> stack = new Stack<>();

        // Push the first index
        stack.push(0);
        for (int i = 1; i < a.length; i++) {

            // Pop all the elements until the previous
            // element is smaller than current element
            while (stack.size() > 0
                && a[stack.peek()] >= a[i])
                stack.pop();

            // Store the previous smaller element
            ps[i] = stack.size() > 0 ? stack.peek() : -1;

            // Push the index of the current element
            stack.push(i);
        }

        // Return the array
        return ps;
    }

    // Function to generate next smaller element
    // for each array element
    public static int[] findNext(int[] a, int n)
    {
        int ns[] = new int[n];

        ns[n - 1] = n;

        // Stack to keep track of elements
        // that have occurring next
        Stack<Integer> stack = new Stack<>();
        stack.push(n - 1);

        // Iterate in reverse order
        // for calculating next smaller
        for (int i = n - 2; i >= 0; i--) {

            // Pop all the elements until the
            // next element is smaller
            // than current element
            while (stack.size() > 0
                && a[stack.peek()] >= a[i])
                stack.pop();

            // Store the next smaller element
            ns[i] = stack.size() > 0 ? stack.peek()
                                    : a.length;

            // Push the index of the current element
            stack.push(i);
        }

        // Return the array
        return ns;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 3;
        int a[] = { 80, 48, 82 };
        System.out.println(findMaximumSum(a, n));
    }
}

Python3

# Python3 program to implement 
# the above approach 

# Function to find the maximum sum by 
# subtracting same value from all 
# elements of a Subarray 
def findMaximumSum(a, n):
    
    # Stores previous smaller element 
    prev_smaller = findPrevious(a, n)
    
    # Stores next smaller element
    next_smaller = findNext(a, n)
    
    max_value = 0
    for i in range(n):
        
        # Calculate contribution 
        # of each element 
        max_value = max(max_value, a[i] *
                    (next_smaller[i] -
                        prev_smaller[i] - 1))
        
    # Return answer
    return max_value

# Function to generate previous smaller 
# element for each array element 
def findPrevious(a, n):
    
    ps = [0] * n
    
    # The first element has no 
    # previous smaller 
    ps[0] = -1
    
    # Stack to keep track of elements 
    # that have occurred previously 
    stack = []
    
    # Push the first index
    stack.append(0)
    
    for i in range(1, n):
        
        # Pop all the elements until the previous 
        # element is smaller than current element 
        while len(stack) > 0 and a[stack[-1]] >= a[i]:
            stack.pop()
            
        # Store the previous smaller element 
        ps[i] = stack[-1] if len(stack) > 0 else -1
        
        # Push the index of the current element
        stack.append(i)
        
    # Return the array 
    return ps

# Function to generate next smaller 
# element for each array element 
def findNext(a, n):
    
    ns = [0] * n
    ns[n - 1] = n
    
    # Stack to keep track of elements 
    # that have occurring next 
    stack = []
    stack.append(n - 1)
    
    # Iterate in reverse order 
    # for calculating next smaller
    for i in range(n - 2, -1, -1):
        
        # Pop all the elements until the 
        # next element is smaller 
        # than current element 
        while (len(stack) > 0 and
                a[stack[-1]] >= a[i]):
            stack.pop()
        
        # Store the next smaller element 
        ns[i] = stack[-1] if len(stack) > 0 else n
        
        # Push the index of the current element
        stack.append(i)
        
    # Return the array
    return ns

# Driver code
n = 3
a = [ 80, 48, 82 ]

print(findMaximumSum(a, n))

# This code is contributed by Stuti Pathak

// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{

// Function to find the maximum sum by
// subtracting same value from all
// elements of a Subarray
public static int findMaximumSum(int[] a, int n)
{
    // Stores previous smaller element
    int []prev_smaller = findPrevious(a, n);

    // Stores next smaller element
    int []next_smaller = findNext(a, n);

    int max_value = 0;
    for (int i = 0; i < n; i++)
    {

    // Calculate contribution
    // of each element
    max_value = Math.Max(max_value,
                a[i] * (next_smaller[i] - 
                        prev_smaller[i] - 1));
    }

    // Return answer
    return max_value;
}

// Function to generate previous smaller element
// for each array element
public static int[] findPrevious(int[] a, int n)
{
    int []ps = new int[n];

    // The first element has no
    // previous smaller
    ps[0] = -1;

    // Stack to keep track of elements
    // that have occurred previously
    Stack<int> stack = new Stack<int>();

    // Push the first index
    stack.Push(0);
    for (int i = 1; i < a.Length; i++)
    {

    // Pop all the elements until the previous
    // element is smaller than current element
    while (stack.Count > 0 && 
            a[stack.Peek()] >= a[i])
        stack.Pop();

    // Store the previous smaller element
    ps[i] = stack.Count > 0 ? stack.Peek() : -1;

    // Push the index of the current element
    stack.Push(i);
    }

    // Return the array
    return ps;
}

// Function to generate next smaller element
// for each array element
public static int[] findNext(int[] a, int n)
{
    int []ns = new int[n];

    ns[n - 1] = n;

    // Stack to keep track of elements
    // that have occurring next
    Stack<int> stack = new Stack<int>();
    stack.Push(n - 1);

    // Iterate in reverse order
    // for calculating next smaller
    for (int i = n - 2; i >= 0; i--) 
    {

    // Pop all the elements until the
    // next element is smaller
    // than current element
    while (stack.Count > 0 && 
            a[stack.Peek()] >= a[i])
        stack.Pop();

    // Store the next smaller element
    ns[i] = stack.Count > 0 ? stack.Peek()
        : a.Length;

    // Push the index of the current element
    stack.Push(i);
    }

    // Return the array
    return ns;
}

// Driver Code
public static void Main(String []args)
{
    int n = 3;
    int []a = { 80, 48, 82 };
    Console.WriteLine(findMaximumSum(a, n));
}
}

// This code is contributed by Amit Katiyar

JavaScript

<script>
// javascript Program to implement
// the above approach 

    // Function to find the maximum sum by
    // subtracting same value from all
    // elements of a Subarray
    function findMaximumSum(a ,n) 
    {
    
        // Stores previous smaller element
        var prev_smaller = findPrevious(a, n);

        // Stores next smaller element
        var next_smaller = findNext(a, n);

        var max_value = 0;
        for (var i = 0; i < n; i++)
        {

            // Calculate contribution
            // of each element
            max_value = Math.max(max_value, a[i] * (next_smaller[i] - prev_smaller[i] - 1));
        }

        // Return answer
        return max_value;
    }

    // Function to generate previous smaller element
    // for each array element
      function findPrevious(a , n) {
        var ps = Array(n).fill(0);

        // The first element has no
        // previous smaller
        ps[0] = -1;

        // Stack to keep track of elements
        // that have occurred previously
        let stack = Array();

        // Push the first index
        stack.push(0);
        for (var i = 1; i < a.length; i++) {

            // Pop all the elements until the previous
            // element is smaller than current element
            while (stack.length > 0 && a[stack[stack.length-1]] >= a[i])
                stack.pop();

            // Store the previous smaller element
            ps[i] = stack.length > 0 ?stack[stack.length-1] : -1;

            // Push the index of the current element
            stack.push(i);
        }

        // Return the array
        return ps;
    }

    // Function to generate next smaller element
    // for each array element
      function findNext(a , n) {
        var ns = Array(n).fill(0);

        ns[n - 1] = n;

        // Stack to keep track of elements
        // that have occurring next
        var stack = Array();
        stack.push(n - 1);

        // Iterate in reverse order
        // for calculating next smaller
        for (var i = n - 2; i >= 0; i--) {

            // Pop all the elements until the
            // next element is smaller
            // than current element
            while (stack.length > 0 && a[stack[stack.length-1]] >= a[i])
                stack.pop();

            // Store the next smaller element
            ns[i] = stack.length > 0 ? stack[stack.length-1] : a.length;

            // Push the index of the current element
            stack.push(i);
        }

        // Return the array
        return ns;
    }

    // Driver Code
        var n = 3;
        var a = [ 80, 48, 82 ];
        document.write(findMaximumSum(a, n));

// This code is contributed by gauravrajput1 
</script>

Output:

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

Maximize subarray sum by inverting sign of elements of any subarray at most twice

hemanthgadarla007

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