# Given an array and three numbers, maximize (x * a[i]) + (y * a[j]) + (z * a[k])

Given an array of n integers, and three integers x, y and z. maximise the value of (x * a[i]) + (y * a[j]) + (z * a[k]) where i ≤ j ≤ k.

Examples :

```Input : arr[] = {-1, -2, -3, -4, -5}
x = 1
y = 2
z = -3
Output: 12
Explanation: The maximized values is
(1 * -1) + (2 * -1) + ( -3 * -5) = 12

Input: arr[] = {1, 2, 3, 4, 5}
x = 1
y = 2
z = 3
Output: 30
(1*5 + 2*5 + 3*5) = 30
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to run three nested loops to iterate through all triplets. For every triplet, compute the required value and keep track of maximum and finally return the same.

An efficient solution is to preocompute values and store them using extra space. The first key observation is i ≤ j ≤ k, so x*a[i] will always be the left maximum, and z*a[k] will always be the right maximum. Create a left array where we store the left maximums for every element. Create a right array where we store the right maximums for every element. Then for every element, calculate the maximum value of the function possible. For any index ind, the maximum at that position will always be (left[ind] + j * a[ind] + right[ind]), find the maximum of this value for every element in the array and that will be your answer.

Below is the implementation of the above approach

## C++

 `// CPP program to find the maximum value of ` `// x*arr[i] + y*arr[j] + z*arr[k] ` `#include ` `using` `namespace` `std; ` ` `  `// function to maximize the condition ` `int` `maximizeExpr(``int` `a[], ``int` `n, ``int` `x, ``int` `y,  ` `                                        ``int` `z) ` `{ ` `    ``// Traverse the whole array and compute ` `    ``// left maximum for every index.  ` `    ``int` `L[n]; ` `    ``L[0] = x * a[0]; ` `    ``for` `(``int` `i = 1; i < n; i++)  ` `        ``L[i] = max(L[i - 1], x * a[i]); ` ` `  `    ``// Compute right maximum for every index.  ` `    ``int` `R[n]; ` `    ``R[n-1] = z * a[n-1]; ` `    ``for` `(``int` `i = n - 2; i >= 0; i--) ` `        ``R[i] = max(R[i + 1], z * a[i]); ` ` `  `    ``// Traverse through the whole array to  ` `    ``// maximize the required expression. ` `    ``int` `ans = INT_MIN;  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `          ``ans = max(ans, L[i] + y * a[i] + R[i]);  ` ` `  `    ``return` `ans; ` `} ` `     `  `// driver program to test the above funcion  ` `int` `main()  ` `{ ` `   ``int` `a[] = {-1, -2, -3, -4, -5}; ` `   ``int` `n = ``sizeof``(a)/``sizeof``(a[0]); ` `   ``int` `x = 1, y = 2 , z = -3; ` `   ``cout << maximizeExpr(a, n, x, y, z) << endl; ` `   ``return` `0; ` `} `

## Java

 `// Java program to find the maximum value  ` `// of x*arr[i] + y*arr[j] + z*arr[k] ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// function to maximize the condition ` `    ``static` `int` `maximizeExpr(``int` `a[], ``int` `n, ``int` `x,  ` `                             ``int` `y, ``int` `z) ` `    ``{ ` `        ``// Traverse the whole array and compute ` `        ``// left maximum for every index.  ` `        ``int` `L[] = ``new` `int``[n]; ` `        ``L[``0``] = x * a[``0``]; ` `        ``for` `(``int` `i = ``1``; i < n; i++)  ` `            ``L[i] = Math.max(L[i - ``1``], x * a[i]); ` ` `  `        ``// Compute right maximum for every index.  ` `        ``int` `R[] = ``new` `int``[n]; ` `        ``R[n - ``1``] = z * a[n - ``1``]; ` `        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) ` `            ``R[i] = Math.max(R[i + ``1``], z * a[i]); ` ` `  `        ``// Traverse through the whole array to  ` `        ``// maximize the required expression. ` `        ``int` `ans = Integer.MIN_VALUE;  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``ans = Math.max(ans, L[i] + y * a[i] + ` `                                         ``R[i]);  ` ` `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// driver program to test the above funcion  ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `    ``int` `a[] = {-``1``, -``2``, -``3``, -``4``, -``5``}; ` `    ``int` `n = a.length; ` `    ``int` `x = ``1``, y = ``2` `, z = -``3``; ` `    ``System.out.println(maximizeExpr(a, n, x, y, z)); ` `    ``} ` `} ` `// This code is contributed by Prerna Saini `

## Python3

 `# Python3 program to find  ` `# the maximum value of ` `# x*arr[i] + y*arr[j] + z*arr[k] ` `import` `sys ` ` `  `# function to maximize ` `# the condition ` `def` `maximizeExpr(a, n, x, y, z): ` ` `  `    ``# Traverse the whole array  ` `    ``# and compute left maximum  ` `    ``# for every index.  ` `    ``L ``=` `[``0``] ``*` `n ` `    ``L[``0``] ``=` `x ``*` `a[``0``] ` `    ``for` `i ``in` `range``(``1``, n): ` `        ``L[i] ``=` `max``(L[i ``-` `1``], x ``*` `a[i]) ` ` `  `    ``# Compute right maximum ` `    ``# for every index.  ` `    ``R ``=` `[``0``] ``*` `n ` `    ``R[n ``-` `1``] ``=` `z ``*` `a[n ``-` `1``] ` `    ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``): ` `        ``R[i] ``=` `max``(R[i ``+` `1``], z ``*` `a[i]) ` ` `  `    ``# Traverse through the whole  ` `    ``# array to maximize the ` `    ``# required expression. ` `    ``ans ``=` `-``sys.maxsize  ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``ans ``=` `max``(ans, L[i] ``+` `y ``*`  `                       ``a[i] ``+` `R[i])  ` ` `  `    ``return` `ans ` ` `  `# Driver Code ` `a ``=` `[``-``1``, ``-``2``, ``-``3``, ``-``4``, ``-``5``] ` `n ``=` `len``(a) ` `x ``=` `1` `y ``=` `2` `z ``=` `-``3` `print``(maximizeExpr(a, n, x, y, z)) ` ` `  `# This code is contributed ` `# by Smitha `

## C#

 `// C# program to find the maximum value  ` `// of x*arr[i] + y*arr[j] + z*arr[k] ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// function to maximize the condition ` `    ``static` `int` `maximizeExpr(``int` `[]a, ``int` `n, ` `                       ``int` `x, ``int` `y, ``int` `z) ` `    ``{ ` `         `  `        ``// Traverse the whole array and ` `        ``// compute left maximum for every ` `        ``// index.  ` `        ``int` `[]L = ``new` `int``[n]; ` `        ``L[0] = x * a[0]; ` `        ``for` `(``int` `i = 1; i < n; i++)  ` `            ``L[i] = Math.Max(L[i - 1], x * a[i]); ` ` `  `        ``// Compute right maximum for ` `        ``// every index.  ` `        ``int` `[]R = ``new` `int``[n]; ` `        ``R[n - 1] = z * a[n - 1]; ` `        ``for` `(``int` `i = n - 2; i >= 0; i--) ` `            ``R[i] = Math.Max(R[i + 1], z * a[i]); ` ` `  `        ``// Traverse through the whole array to  ` `        ``// maximize the required expression. ` `        ``int` `ans = ``int``.MinValue;  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``ans = Math.Max(ans, L[i] +  ` `                             ``y * a[i] + R[i]);  ` ` `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// driver program to test the ` `    ``// above funcion  ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]a = {-1, -2, -3, -4, -5}; ` `        ``int` `n = a.Length; ` `        ``int` `x = 1, y = 2 , z = -3; ` `         `  `        ``Console.WriteLine( ` `              ``maximizeExpr(a, n, x, y, z)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `= 0; ``\$i``--) ` `        ``\$R``[``\$i``] = max(``\$R``[``\$i` `+ 1],  ` `                     ``\$z` `* ``\$a``[``\$i``]); ` ` `  `    ``// Traverse through the whole  ` `    ``// array to maximize the  ` `    ``// required expression. ` `    ``\$ans` `= PHP_INT_MIN;  ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)  ` `        ``\$ans` `= max(``\$ans``, ``\$L``[``\$i``] +  ` `                   ``\$y` `* ``\$a``[``\$i``] + ``\$R``[``\$i``]);  ` ` `  `    ``return` `\$ans``; ` `} ` `     `  `// Driver Code ` `\$a` `= ``array``(-1, -2, -3, -4, -5); ` `\$n` `= ``count``(``\$a``); ` `\$x` `= 1; ``\$y` `= 2 ; ``\$z` `= -3; ` `echo` `maximizeExpr(``\$a``, ``\$n``, ``\$x``, ``\$y``, ``\$z``); ` ` `  `// This code is contributed by anuj_67. ` `?> `

Output :

```12
```

Time complexity : O(n)
Auxiliary Space : O(n)

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Improved By : vt_m, Smitha Dinesh Semwal

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