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 valuevoid 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 Codeint main(){ int arr[] = { 5, -1, -5, -3, 2, 9, -4 }; findMax(arr, 7);} |
Java
// Java implementation of the above approach import java.util.*;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 |
Javascript
<script>// javascript implementation of the above approach // A function to calculate the maximum value function findMax(a , n) { var dp = Array(n).fill().map(()=>Array(2).fill(0)); var 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]; } document.write(Math.max(dp[n - 2][0], dp[n - 2][1])); } // Driver Code var arr = [ 5, -1, -5, -3, 2, 9, -4 ]; findMax(arr, 7);// This code contributed by Rajput-Ji </script> |
Output
33
Time complexity: O(n) where n is size of given array
Auxiliary space: O(n) because it is using extra space for array “dp”
Efficient approach : Space optimization O(1)
In previous approach the current value ( dp[i] ) is depend upon the ( dp[i-1] ) so which is just the previous computation of DP array. So to optimize the space complexity we only require the computation of previous index.
Implementation Steps:
- Create variables prev_sum and prev_prod to get the computation of just previous index of DP.
- Initialize them with a[0] + a[1] and a[0] * a[1] resp.
- Now create variables curr_sum and curr_prod to get the current value from previous computation.
- Create variable max_val to store the final result.
- Now iterate over subproblems and get the current value from prev_sum and prev_prod.
- After every iteration assign values of curr_sum to prev_sum and curr_prod to prev_prod to iterate further.
- At last print answer stored in max_val.
Implementation:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// A function to calculate the maximum valuevoid findMax(int a[], int n){ // initialize variables for previous sum and product int prev_sum = a[0] + a[1]; int prev_prod = a[0] * a[1]; // initialize variables for current sum and product int curr_sum, curr_prod; // to store the final result int max_val; // Loop to iterate and update the max value for (int i = 1; i <= n - 2; i++) { // get current value from previous computation // store in previous sum and previous product curr_sum = max(prev_sum, prev_prod) + a[i + 1]; curr_prod = prev_sum - a[i] + a[i] * a[i + 1]; // update answer max_val = max(curr_sum, curr_prod); // assigning values of currsum and currproduct // to prev sum and product to iterate further prev_sum = curr_sum; prev_prod = curr_prod; } // print the final result cout << max_val;}// Driver Codeint main(){ int arr[] = { 5, -1, -5, -3, 2, 9, -4 }; // function call findMax(arr, 7);} |
Java
import java.util.*;class Main { // A function to calculate the maximum value static void findMax(int a[], int n) { // initialize variables for previous sum and product int prev_sum = a[0] + a[1]; int prev_prod = a[0] * a[1]; // initialize variables for current sum and product int curr_sum, curr_prod; // to store the final result int max_val = 0; // Loop to iterate and update the max value for (int i = 1; i <= n - 2; i++) { // get current value from previous computation // store in previous sum and previous product curr_sum = Math.max(prev_sum, prev_prod) + a[i + 1]; curr_prod = prev_sum - a[i] + a[i] * a[i + 1]; // update answer max_val = Math.max(curr_sum, curr_prod); // assigning values of currsum and currproduct // to prev sum and product to iterate further prev_sum = curr_sum; prev_prod = curr_prod; } // print the final result System.out.println(max_val); } // Driver Code public static void main(String[] args) { int arr[] = { 5, -1, -5, -3, 2, 9, -4 }; // function call findMax(arr, 7); }} |
Python3
# Python program for above approach# A function to calculate the maximum valuedef findMax(a, n): # initialize variables for previous sum and product prev_sum = a[0] + a[1] prev_prod = a[0] * a[1] # initialize variables for current sum and product curr_sum, curr_prod = 0, 0 # to store the final result max_val = 0 # Loop to iterate and update the max value for i in range(1, n - 1): # get current value from previous computation # store in previous sum and previous product curr_sum = max(prev_sum, prev_prod) + a[i + 1] curr_prod = prev_sum - a[i] + a[i] * a[i + 1] # update answer max_val = max(curr_sum, curr_prod) # assigning values of currsum and currproduct # to prev sum and product to iterate further prev_sum = curr_sum prev_prod = curr_prod # print the final result print(max_val)# Driver Codearr = [5, -1, -5, -3, 2, 9, -4]# function callfindMax(arr, 7) |
C#
using System;public class Program{ public static void Main() { int[] arr = { 5, -1, -5, -3, 2, 9, -4 }; findMax(arr, 7); } // A function to calculate the maximum value static void findMax(int[] a, int n) { // initialize variables for previous sum and product int prev_sum = a[0] + a[1]; int prev_prod = a[0] * a[1]; // initialize variables for current sum and product int curr_sum, curr_prod; // to store the final result int max_val = 0; // Loop to iterate and update the max value for (int i = 1; i <= n - 2; i++) { // get current value from previous computation // store in previous sum and previous product curr_sum = Math.Max(prev_sum, prev_prod) + a[i + 1]; curr_prod = prev_sum - a[i] + a[i] * a[i + 1]; // update answer max_val = Math.Max(curr_sum, curr_prod); // assigning values of currsum and currproduct // to prev sum and product to iterate further prev_sum = curr_sum; prev_prod = curr_prod; } // print the final result Console.WriteLine(max_val); }} |
Javascript
function findMax(arr) { // Initialize variables for previous sum and product let prev_sum = arr[0] + arr[1]; let prev_prod = arr[0] * arr[1]; // Initialize variables for current sum and product let curr_sum, curr_prod; // Initialize a variable to store the final result let max_val; // Loop to iterate and update the max value for (let i = 1; i < arr.length - 1; i++) { // Get the current value from previous computation // Store it in current sum and current product curr_sum = Math.max(prev_sum, prev_prod) + arr[i + 1]; curr_prod = prev_sum - arr[i] + arr[i] * arr[i + 1]; // Update the answer max_val = Math.max(curr_sum, curr_prod); // Assign values of curr_sum and curr_prod // to prev_sum and prev_prod to iterate further prev_sum = curr_sum; prev_prod = curr_prod; } // Print the final result console.log(max_val);}// Driver Codeconst arr = [5, -1, -5, -3, 2, 9, -4];// Function callfindMax(arr); |
Output
33
Time complexity: O(n) where n is size of given array
Auxiliary space: O(1) because no extra space is used.
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