Maximize array product by changing any array element arr[i] to (-1)*arr[i] – 1 any number of times

• Last Updated : 13 Jan, 2022

Given an array arr[] consisting of N integers, the task is to modify the array elements by changing each array element arr[i] to (-1)*arr[i] – 1 any number of times such that the product of the array is maximized.

Examples:

Input: arr[] = {-3, 0, 1}
Output: 2 -1 -2
Explanation:
Performing the following operations on the array elements maximizes the product:
Applying the operation on arr once, arr = (- 1) * (- 3) – 1 = 2.
Applying the operation on arr once, arr = 0 – 1 = -1.
Applying the operation on arr once, arr = -1 – 1 = -2.
The modified array is {2, -1, -2} with product 4, which is maximum possible.

Input: arr[] = {-9, 2, 0, 1, -5, 3, 16}
Output: -9 -3 -1 -2 -5 -4 16

Approach: The given problem can be solved based on the following observations:

• For the product to be maximum, the absolute value of elements should be higher as the operation decreases the absolute value of negative elements. Therefore, the given operations should only be applied to positive values.
• If the number of elements is odd, then perform an operation on the highest absolute valued element as the product of an odd number of negative numbers is negative. Therefore, change one positive number for the product to be positive and maximum.
• Perform the operation on a negative number decreases its absolute value. Therefore, find the element with the greatest absolute value and apply the operation on it once as that would decrease the product by the least.

Follow the steps below to solve the problem:

• Traverse the given array arr[] and for each element arr[i], check if arr[i] ≥ 0 or not. If found to be true, then update arr[i] = (-1)*arr[i] – 1. Otherwise, continue.
• After the array traversal, check if the value of N is odd or not. If found to be true, then find the element with the greatest absolute value and apply the operation to it once.
• After completing the above steps, print the modified array arr[].

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to find array with maximum// product by changing array// elements to (-1)arr[i] - 1void findArrayWithMaxProduct(int arr[],                             int N){    // Traverse the array    for (int i = 0; i < N; i++)         // Applying the operation on        // all the positive elements        if (arr[i] >= 0) {            arr[i] = -arr[i] - 1;        }     if (N % 2 == 1) {         // Stores maximum element in array        int max_element = -1;         // Stores index of maximum element        int index = -1;         for (int i = 0; i < N; i++)             // Check if current element            // is greater than the            // maximum element            if (abs(arr[i]) > max_element) {                 max_element = abs(arr[i]);                 // Find index of the                // maximum element                index = i;            }         // Perform the operation on the        // maximum element        arr[index] = -arr[index] - 1;    }     // Print the elements of the array    for (int i = 0; i < N; i++)        printf("%d ", arr[i]);} // Driver Codeint main(){    int arr[] = { -3, 0, 1 };    int N = sizeof(arr) / sizeof(arr);     // Function Call    findArrayWithMaxProduct(arr, N);     return 0;}

Java

 // Java program for the above approachimport java.io.*;class GFG{   // Function to find array with maximum  // product by changing array  // elements to (-1)arr[i] - 1  static void findArrayWithMaxProduct(int arr[], int N)  {     // Traverse the array    for (int i = 0; i < N; i++)       // Applying the operation on      // all the positive elements      if (arr[i] >= 0)      {        arr[i] = -arr[i] - 1;      }     if (N % 2 == 1)    {       // Stores maximum element in array      int max_element = -1;       // Stores index of maximum element      int index = -1;       for (int i = 0; i < N; i++)         // Check if current element        // is greater than the        // maximum element        if (Math.abs(arr[i]) > max_element) {           max_element = Math.abs(arr[i]);           // Find index of the          // maximum element          index = i;        }       // Perform the operation on the      // maximum element      arr[index] = -arr[index] - 1;    }     // Print the elements of the array    for (int i = 0; i < N; i++)      System.out.print(arr[i] + " ");  }   // Driver Code  public static void main(String[] args)  {    int arr[] = { -3, 0, 1 };    int N = arr.length;     // Function Call    findArrayWithMaxProduct(arr, N);  }} // This code is contributed by jithin.

Python3

 # Python3 program for the above approach # Function to find array with maximum# product by changing array# elements to (-1)arr[i] - 1def findArrayWithMaxProduct(arr, N):       # Traverse the array    for i in range(N):         # Applying the operation on        # all the positive elements        if (arr[i] >= 0):            arr[i] = -arr[i] - 1    if (N % 2 == 1):         # Stores maximum element in array        max_element = -1         # Stores index of maximum element        index = -1        for i in range(N):             # Check if current element            # is greater than the            # maximum element            if (abs(arr[i]) > max_element):                max_element = abs(arr[i])                 # Find index of the                # maximum element                index = i         # Perform the operation on the        # maximum element        arr[index] = -arr[index] - 1             # Print elements of the array    for i in arr:        print(i, end = " ") # Driver Codeif __name__ == '__main__':    arr = [-3, 0, 1]    N = len(arr)     # Function Call    findArrayWithMaxProduct(arr, N)     # This code is contributed by mohit kumar 29.

C#

 // C# program for the above approachusing System;class GFG{   // Function to find array with maximum  // product by changing array  // elements to (-1)arr[i] - 1  static void findArrayWithMaxProduct(int[] arr, int N)  {     // Traverse the array    for (int i = 0; i < N; i++)       // Applying the operation on      // all the positive elements      if (arr[i] >= 0)      {        arr[i] = -arr[i] - 1;      }     if (N % 2 == 1)    {       // Stores maximum element in array      int max_element = -1;       // Stores index of maximum element      int index = -1;      for (int i = 0; i < N; i++)         // Check if current element        // is greater than the        // maximum element        if (Math.Abs(arr[i]) > max_element) {           max_element = Math.Abs(arr[i]);           // Find index of the          // maximum element          index = i;        }       // Perform the operation on the      // maximum element      arr[index] = -arr[index] - 1;    }     // Print the elements of the array    for (int i = 0; i < N; i++)      Console.Write(arr[i] + " ");  } // Driver Codestatic public void Main(){    int[] arr = { -3, 0, 1 };    int N = arr.Length;     // Function Call    findArrayWithMaxProduct(arr, N);}} // This code is contributed by sanjoy_62.

Javascript


Output:
2 -1 -2

Time Complexity: O(N)
Auxiliary Space: O(1)

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