Given an array arr[], the task is to find out the maximum obtainable value. The user is allowed to add or multiply the two consecutive elements. However, there has to be at least one addition operation between two multiplication operations (i.e), two consecutive multiplication operations are not allowed.
Let the array elements be 1, 2, 3, 4 then 1 * 2 + 3 + 4 is a valid operation whereas 1 + 2 * 3 * 4 is not a a valid operation as there are consecutive multiplication operations.
Examples:
Input : 5 -1 -5 -3 2 9 -4 Output : 33 Explanation: The maximum value obtained by following the above conditions is 33. The sequence of operations are given as: 5 + (-1) + (-5) * (-3) + 2 * 9 + (-4) = 33 Input : 5 -3 -5 2 3 9 4 Output : 62
Approach:
This problem can be solved by using dynamic programming.
- Assuming 2D array dp[][] of dimensions n * 2.
- dp[i][0] represents the maximum value of the array up to ith position if the last operation is addition.
- dp[i][1] represents the maximum value of the array up to ith position if the last operation is multiplication.
Now, since consecutive multiplication operation is not allowed, the recurrence relation can be considered as :
dp[i][0] = max(dp[ i - 1][0], dp[ i - 1][1]) + a[ i + 1]; dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1];
The base cases are:
dp[0][0] = a[0] + a[1]; dp[0][1] = a[0] * a[1];
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // A function to calculate the maximum value void findMax(int a[], int n) { int dp[n][2]; memset(dp, 0, sizeof(dp)); // basecases dp[0][0] = a[0] + a[1]; dp[0][1] = a[0] * a[1]; //Loop to iterate and add the max value in the dp array for (int i = 1; i <= n - 2; i++) { dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1]; dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1]; } cout << max(dp[n - 2][0], dp[n - 2][1]); } // Driver Code int main() { int arr[] = { 5, -1, -5, -3, 2, 9, -4 }; findMax(arr, 7); } |
Java
// Java implementation of the above approach class GFG { // A function to calculate the maximum value static void findMax(int []a, int n) { int dp[][] = new int[n][2]; int i, j; for (i = 0; i < n ; i++) for(j = 0; j < 2; j++) dp[i][j] = 0; // basecases dp[0][0] = a[0] + a[1]; dp[0][1] = a[0] * a[1]; // Loop to iterate and add the // max value in the dp array for (i = 1; i <= n - 2; i++) { dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1]; dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1]; } System.out.println(Math.max(dp[n - 2][0], dp[n - 2][1])); } // Driver Code public static void main (String[] args) { int arr[] = { 5, -1, -5, -3, 2, 9, -4 }; findMax(arr, 7); } } // This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the above approach import numpy as np # A function to calculate the maximum value def findMax(a, n) : dp = np.zeros((n, 2)); # basecases dp[0][0] = a[0] + a[1]; dp[0][1] = a[0] * a[1]; # Loop to iterate and add the max value in the dp array for i in range(1, n - 1) : dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1]; dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1]; print(max(dp[n - 2][0], dp[n - 2][1]), end =""); # Driver Code if __name__ == "__main__" : arr = [ 5, -1, -5, -3, 2, 9, -4 ]; findMax(arr, 7); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the above approach using System; class GFG { // A function to calculate the maximum value static void findMax(int []a, int n) { int [,]dp = new int[n, 2]; int i, j; for (i = 0; i < n ; i++) for(j = 0; j < 2; j++) dp[i, j] = 0; // basecases dp[0, 0] = a[0] + a[1]; dp[0, 1] = a[0] * a[1]; // Loop to iterate and add the // max value in the dp array for (i = 1; i <= n - 2; i++) { dp[i, 0] = Math.Max(dp[i - 1, 0], dp[i - 1, 1]) + a[i + 1]; dp[i, 1] = dp[i - 1, 0] - a[i] + a[i] * a[i + 1]; } Console.WriteLine(Math.Max(dp[n - 2, 0], dp[n - 2, 1])); } // Driver Code public static void Main() { int []arr = { 5, -1, -5, -3, 2, 9, -4 }; findMax(arr, 7); } } // This code is contributed by AnkitRai01 |
33
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Improved By : AnkitRai01
