Maximize the subarray sum after multiplying all elements of any subarray with X

Given an array arr[] of N integers and an integer X. We can choose any sub-array and multiply all its element by X. After multiplication, find the sub-array with the maximum sum. The task is to multiply the sub-array in such a way that the final sub-array sum is maximized. 
Examples:

Input: arr[] = { -3, 8, -2, 1, -6 }, X = -1 
Output: 15 
Choose the sub-array with {-2, 1, -6} and multiply X. 
Now the new array is {-3, 8, 2, -1, 6} whose maximum sub-array sum is 15. 
Input: arr[] = {1, 2, 4, -10}, X = 2 
Output: 14 

Approach: The problem cannot be solved using a greedy approach. Greedy approach fails in many cases one of them being a[] = { -3, 8, -2, 1, 6 } and x = -1. We will use Dynamic Programming to solve the above problem. Let dp[ind][state], where ind is the current index in the array and there are 3 states. 

• The first state defines that no sub-array has been chosen till index ind to be multiplied with X.
• The second state defines that the current element at index ind is in the sub-array which is multiplied with X.
• The third state defines that the sub-array has already been chosen.

Kadane’s algorithm has been used along with DP to get the maximum subarray sum simultaneously. 
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 5

// Function to return the maximum sum
int func(int idx, int cur, int a[], int dp[N][3], int n, int x)
{

    // Base case
    if (idx == n)
        return 0;

    // If already calculated
    if (dp[idx][cur] != -1)
        return dp[idx][cur];

    int ans = 0;

    // If no elements have been chosen
    if (cur == 0) {

        // Do not choose any element and use
        // Kadane's algorithm by taking max
        ans = max(ans, a[idx] + func(idx + 1, 0, a, dp, n, x));

        // Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x));
    }
    else if (cur == 1) {

        // Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x));

        // End the sub-array multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x));
    }
    else

        // No more multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x));

    // Memoize and return the answer
    return dp[idx][cur] = ans;
}

// Function to get the maximum sum
int getMaximumSum(int a[], int n, int x)
{

    // Initialize dp with -1
    int dp[n][3];
    memset(dp, -1, sizeof dp);

    // Iterate from every position and find the
    // maximum sum which is possible
    int maxi = 0;
    for (int i = 0; i < n; i++)
        maxi = max(maxi, func(i, 0, a, dp, n, x));

    return maxi;
}

// Driver code
int main()
{
    int a[] = { -3, 8, -2, 1, -6 };
    int n = sizeof(a) / sizeof(a[0]);
    int x = -1;

    cout << getMaximumSum(a, n, x);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    static int N = 5;

// Function to return the maximum sum
static int func(int idx, int cur, int a[],
                int dp[][], int n, int x)
{

    // Base case
    if (idx == n)
    {
        return 0;
    }

    // If already calculated
    if (dp[idx][cur] != -1)
    {
        return dp[idx][cur];
    }

    int ans = 0;

    // If no elements have been chosen
    if (cur == 0)
    {

        // Do not choose any element and use
        // Kadane's algorithm by taking max
        ans = Math.max(ans, a[idx] +
                func(idx + 1, 0, a, dp, n, x));

        // Choose the sub-array and multiply x
        ans = Math.max(ans, x * a[idx] +
                func(idx + 1, 1, a, dp, n, x));
    }
    else if (cur == 1)
    {

        // Choose the sub-array and multiply x
        ans = Math.max(ans, x * a[idx] +
                func(idx + 1, 1, a, dp, n, x));

        // End the sub-array multiplication
        ans = Math.max(ans, a[idx] +
                func(idx + 1, 2, a, dp, n, x));
    }
    else // No more multiplication
    {
        ans = Math.max(ans, a[idx] +
                func(idx + 1, 2, a, dp, n, x));
    }

    // Memoize and return the answer
    return dp[idx][cur] = ans;
}

// Function to get the maximum sum
static int getMaximumSum(int a[], int n, int x)
{

    // Initialize dp with -1
    int dp[][] = new int[n][3];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            dp[i][j] = -1;
        }
    }

    // Iterate from every position and find the
    // maximum sum which is possible
    int maxi = 0;
    for (int i = 0; i < n; i++)
    {
        maxi = Math.max(maxi, func(i, 0, a, dp, n, x));
    }

    return maxi;
}

// Driver code
public static void main(String[] args)
{
    int a[] = {-3, 8, -2, 1, -6};
    int n = a.length;
    int x = -1;
    System.out.println(getMaximumSum(a, n, x));

}
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

N = 5

# Function to return the maximum sum
def func(idx, cur, a, dp, n, x) :


    # Base case
    if (idx == n) :
        return 0

    # If already calculated
    if (dp[idx][cur] != -1):
        return dp[idx][cur]

    ans = 0

    # If no elements have been chosen
    if (cur == 0) :

        # Do not choose any element and use
        # Kadane's algorithm by taking max
        ans = max(ans, a[idx] + func(idx + 1, 0, a, dp, n, x))

        # Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x))

    elif (cur == 1) :

        # Choose the sub-array and multiply x
        ans = max(ans, x * a[idx] + func(idx + 1, 1, a, dp, n, x))

        # End the sub-array multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x))

    else :

        # No more multiplication
        ans = max(ans, a[idx] + func(idx + 1, 2, a, dp, n, x))

    # Memoize and return the answer
    dp[idx][cur] = ans

    return dp[idx][cur]


# Function to get the maximum sum
def getMaximumSum(a, n, x) :


    # Initialize dp with -1
    dp = [[-1 for i in range(3)] for j in range(n)]


    # Iterate from every position and find the
    # maximum sum which is possible
    maxi = 0
    for i in range (0, n) :
        maxi = max(maxi, func(i, 0, a, dp, n, x))

    return maxi


# Driver code

a =  [ -3, 8, -2, 1, -6 ]
n = len(a)
x = -1

print(getMaximumSum(a, n, x))

# This code is contributed by ihritik

C#

// C# implementation of the approach
using System;

class GFG
{

    static int N = 5;

// Function to return the maximum sum
static int func(int idx, int cur, int []a,
                int [,]dp, int n, int x)
{

    // Base case
    if (idx == n)
    {
        return 0;
    }

    // If already calculated
    if (dp[idx,cur] != -1)
    {
        return dp[idx,cur];
    }

    int ans = 0;

    // If no elements have been chosen
    if (cur == 0)
    {

        // Do not choose any element and use
        // Kadane's algorithm by taking max
        ans = Math.Max(ans, a[idx] +
                func(idx + 1, 0, a, dp, n, x));

        // Choose the sub-array and multiply x
        ans = Math.Max(ans, x * a[idx] +
                func(idx + 1, 1, a, dp, n, x));
    }
    else if (cur == 1)
    {

        // Choose the sub-array and multiply x
        ans = Math.Max(ans, x * a[idx] +
                func(idx + 1, 1, a, dp, n, x));

        // End the sub-array multiplication
        ans = Math.Max(ans, a[idx] +
                func(idx + 1, 2, a, dp, n, x));
    }
    else // No more multiplication
    {
        ans = Math.Max(ans, a[idx] +
                func(idx + 1, 2, a, dp, n, x));
    }

    // Memoize and return the answer
    return dp[idx,cur] = ans;
}

// Function to get the maximum sum
static int getMaximumSum(int []a, int n, int x)
{

    // Initialize dp with -1
    int [,]dp = new int[n,3];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            dp[i,j] = -1;
        }
    }

    // Iterate from every position and find the
    // maximum sum which is possible
    int maxi = 0;
    for (int i = 0; i < n; i++)
    {
        maxi = Math.Max(maxi, func(i, 0, a, dp, n, x));
    }

    return maxi;
}

// Driver code
public static void Main(String[] args)
{
    int []a = {-3, 8, -2, 1, -6};
    int n = a.Length;
    int x = -1;
    Console.WriteLine(getMaximumSum(a, n, x));

}
}

/* This code contributed by PrinciRaj1992 */

PHP

<?php
// PHP implementation of the approach

$N = 5;

// Function to return the maximum sum
function func($idx, $cur, $a, $dp, $n, $x)
{

    // Base case
    if ($idx == $n)
        return 0;

    // If already calculated
    if ($dp[$idx][$cur] != -1)
        return $dp[$idx][$cur];

    $ans = 0;

    // If no elements have been chosen
    if ($cur == 0)
    {

        // Do not choose any element and use
        // Kadane's algorithm by taking max
        $ans = max($ans, $a[$idx] +
                func($idx + 1, 0, $a, $dp, $n, $x));

        // Choose the sub-array and multiply x
        $ans = max($ans, $x * $a[$idx] +
                func($idx + 1, 1, $a, $dp, $n, $x));
    }
    else if ($cur == 1)
    {

        // Choose the sub-array and multiply x
        $ans = max($ans, $x * $a[$idx] +
                func($idx + 1, 1, $a, $dp, $n, $x));

        // End the sub-array multiplication
        $ans = max($ans, $a[$idx] +
                func($idx + 1, 2, $a, $dp, $n, $x));
    }
    else

        // No more multiplication
        $ans = max($ans, $a[$idx] +
                func($idx + 1, 2, $a, $dp,$n, $x));

    // Memoize and return the answer
    return $dp[$idx][$cur] = $ans;
}

// Function to get the maximum sum
function getMaximumSum($a, $n, $x)
{

    // Initialize dp with -1
    $dp = array(array()) ;

    for($i = 0; $i < $n; $i++)
    {
        for($j = 0; $j < 3; $j++)
        {
            $dp[$i][$j] = -1;
        }
    }

    // Iterate from every position and find the
    // maximum sum which is possible
    $maxi = 0;
    for ($i = 0; $i < $n; $i++)
        $maxi = max($maxi, func($i, 0, $a, $dp, $n, $x));

    return $maxi;
}

    // Driver code
    $a = array( -3, 8, -2, 1, -6 );
    $n = count($a) ;
    $x = -1;

    echo getMaximumSum($a, $n, $x);

    // This code is contributed by Ryuga
?>

Javascript

<script>

// JavaScript implementation of the approach
var N = 5;

    // Function to return the maximum sum
    function func(idx , cur , a , dp , n , x) {

        // Base case
        if (idx == n) {
            return 0;
        }

        // If already calculated
        if (dp[idx][cur] != -1) {
            return dp[idx][cur];
        }

        var ans = 0;

        // If no elements have been chosen
        if (cur == 0) {

            // Do not choose any element and use
            // Kadane's algorithm by taking max
            ans = Math.max(ans, a[idx] +
            func(idx + 1, 0, a, dp, n, x));

            // Choose the sub-array and multiply x
            ans = Math.max(ans, x * a[idx] +
            func(idx + 1, 1, a, dp, n, x));
        } else if (cur == 1) {

            // Choose the sub-array and multiply x
            ans = Math.max(ans, x * a[idx] +
            func(idx + 1, 1, a, dp, n, x));

            // End the sub-array multiplication
            ans = Math.max(ans, a[idx] +
            func(idx + 1, 2, a, dp, n, x));
        } else // No more multiplication
        {
            ans = Math.max(ans, a[idx] +
            func(idx + 1, 2, a, dp, n, x));
        }

        // Memoize and return the answer
        return dp[idx][cur] = ans;
    }

    // Function to get the maximum sum
    function getMaximumSum(a , n , x) {

        // Initialize dp with -1
        var dp = Array(n).fill().map(()=>Array(3).fill(0));
        for (i = 0; i < n; i++) {
            for (j = 0; j < 3; j++) {
                dp[i][j] = -1;
            }
        }

        // Iterate from every position and find the
        // maximum sum which is possible
        var maxi = 0;
        for (i = 0; i < n; i++) {
            maxi = Math.max(maxi, func(i, 0, a, dp, n, x));
        }

        return maxi;
    }

    // Driver code

        var a = [ -3, 8, -2, 1, -6 ];
        var n = a.length;
        var x = -1;
        document.write(getMaximumSum(a, n, x));

// This code contributed by Rajput-Ji

</script>

Output:

15

Time Complexity: O(N * 3) 
Auxiliary Space: O(N * 3)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a DP to store the solution of the subproblems.
• Initialize the DP with base cases
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
• Create a variable ans to keep track of values of DP in each condition.
• After iteration from loop create a variable maxi to store the maximum sum and update it by iterate over Dp.
• Return the final solution stored in maxi.

Implementation :

C++

#include <bits/stdc++.h>
using namespace std;
#define N 5

// Function to return the maximum sum
int getMaximumSum(int a[], int n, int x)
{
    int dp[n][3];

    // Initializing the dp array
    for(int i=0; i<n; i++){
        dp[i][0] = a[i];
        dp[i][1] = a[i]*x;
        dp[i][2] = a[i];
    }

    // Applying the recurrence relation
    for(int i=n-2; i>=0; i--){
        for(int j=0; j<3; j++){

              // to track the answer in each condition
            int ans = 0;
            if(j==0){
                ans = max(dp[i][j], dp[i+1][j]+a[i]);
                ans = max(ans, dp[i+1][j+1]+x*a[i]);
            }
            else if(j==1){
                ans = max(dp[i][j], dp[i+1][j]+x*a[i]);
                ans = max(ans, dp[i+1][j+1]+a[i]);
            }
            else{
                ans = max(dp[i][j], dp[i+1][j]+a[i]);
            }

              // update DP
            dp[i][j] = ans;
        }
    }

    // Finding the maximum sum from dp array
    int maxi = 0;
    for (int i = 0; i < n; i++)
        maxi = max(maxi, dp[i][0]);

      // return final answer
    return maxi;
}

// Driver code
int main()
{
    int a[] = { -3, 8, -2, 1, -6 };
    int n = sizeof(a) / sizeof(a[0]);
    int x = -1;
    // function call
    cout << getMaximumSum(a, n, x);

    return 0;
}

Java

import java.util.*;

public class Main {

  public static int getMaximumSum(int[] a, int n, int x)
  {

    // Initializing the dp array
    int[][] dp = new int[n][3];
    for (int i = 0; i < n; i++) {
      dp[i][0] = a[i];
      dp[i][1] = a[i] * x;
      dp[i][2] = a[i];
    }

    // Applying the recurrence relation
    for (int i = n - 2; i >= 0; i--) {
      for (int j = 0; j < 3; j++) {

        // to track the answer in each condition
        int ans = 0;
        if (j == 0) {
          ans = Math.max(dp[i][j],
                         dp[i + 1][j] + a[i]);
          ans = Math.max(ans, dp[i + 1][j + 1]
                         + x * a[i]);
        }
        else if (j == 1) {
          ans = Math.max(dp[i][j],
                         dp[i + 1][j] + x * a[i]);
          ans = Math.max(ans,
                         dp[i + 1][j + 1] + a[i]);
        }
        else {
          ans = Math.max(dp[i][j],
                         dp[i + 1][j] + a[i]);
        }

        // update DP
        dp[i][j] = ans;
      }
    }

    // Finding the maximum sum from dp array
    int maxi = 0;
    for (int i = 0; i < n; i++) {
      maxi = Math.max(maxi, dp[i][0]);
    }

    // return final answer
    return maxi;
  }

  public static void main(String[] args)
  {
    int[] a = { -3, 8, -2, 1, -6 };
    int n = a.length;
    int x = -1;

    // function call
    System.out.println(getMaximumSum(a, n, x));
  }
}

// This code is contributed by user_dtewbxkn77n

Python3

# Function to return the maximum sum
def getMaximumSum(a, n, x):

    # Initializing the dp array
    dp = [[0 for j in range(3)] for i in range(n)]
    for i in range(n):
        dp[i][0] = a[i]
        dp[i][1] = a[i] * x
        dp[i][2] = a[i]

    # Applying the recurrence relation
    for i in range(n-2, -1, -1):
        for j in range(3):

            # to track the answer in each condition
            ans = 0
            if j == 0:
                ans = max(dp[i][j], dp[i+1][j]+a[i])
                ans = max(ans, dp[i+1][j+1]+x*a[i])
            elif j == 1:
                ans = max(dp[i][j], dp[i+1][j]+x*a[i])
                ans = max(ans, dp[i+1][j+1]+a[i])
            else:
                ans = max(dp[i][j], dp[i+1][j]+a[i])

            # update DP
            dp[i][j] = ans

    # Finding the maximum sum from dp array
    maxi = 0
    for i in range(n):
        maxi = max(maxi, dp[i][0])

    # return final answer
    return maxi

# Driver code
if __name__ == '__main__':
    a = [-3, 8, -2, 1, -6]
    n = len(a)
    x = -1

    # function call
    print(getMaximumSum(a, n, x))

Output: 

15

Time Complexity: O(N * 3) 
Auxiliary Space: O(N * 3)