Given an array of n integers, and three integers x, y and z. maximize 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
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 precompute values and stores 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 <bits/stdc++.h> 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 function 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 function 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 = 1y = 2z = -3print(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 function 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
<?php // PHP program to find the // maximum value of // x*arr[i]+ y*arr[j] + z*arr[k] // function to maximize // the condition function maximizeExpr($a, $n, $x, $y, $z) { // Traverse the whole array // and compute left maximum // for every index. $L = array(); $L[0] = $x * $a[0]; for ($i = 1; $i < $n; $i++) $L[$i] = max($L[$i - 1], $x * $a[$i]); // Compute right maximum // for every index. $R = array(); $R[$n - 1] = $z * $a[$n - 1]; for ($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. $ans = PHP_INT_MIN; for ($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. ?> |
Javascript
<script> // javascript program to find the maximum value // of x*arr[i] + y*arr[j] + z*arr[k] // function to maximize the condition function maximizeExpr(a , n , x , y , z) { // Traverse the whole array and compute // left maximum for every index. var L = Array(n).fill(0); L[0] = x * a[0]; for (i = 1; i < n; i++) L[i] = Math.max(L[i - 1], x * a[i]); // Compute right maximum for every index. var R = Array(n).fill(0); R[n - 1] = z * a[n - 1]; for (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. var ans = Number.MIN_VALUE; for (i = 0; i < n; i++) ans = Math.max(ans, L[i] + y * a[i] + R[i]); return ans; } // Driver program to test the above function var a = [ -1, -2, -3, -4, -5 ]; var n = a.length; var x = 1, y = 2, z = -3; document.write(maximizeExpr(a, n, x, y, z)); // This code is contributed by Rajput-Ji </script> |
Output
12
Time complexity: O(n)
Auxiliary Space: O(n)
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