Rearrange an array such that product of every two consecutive elements is a multiple of 4

Given an array arr[] of size N, the task is to rearrange the array elements such that for every index i(1 <= i <= N – 1), the product of arr[i] and arr[i – 1] is a multiple of 4.

Example:

Input: arr[] = {1, 10, 100}
Output: 1, 100, 10
Explanation:
1 * 100 is divisible by 4
100 * 10 is divisible by 4

Input: arr[] = {2, 7, 1, 8, 2, 8}
Output: 7, 8, 1, 8, 2, 2

Naive Approach:
The simplest approach to solve the problem is to generate all possible permutations of the array and for every permutation, check if the product of every two consecutive elements is a multiple of 4 or not.
Time complexity: O(N^2 * N!)
Auxiliary Space: O(N!)



Efficient Approach:
Follow the steps below to optimize the above approach:

  • Initialize following three variables:

    • odd = Number of odd elements.
    • four = Number of elements divisible by 4.
    • non_four = Number of even elements not divisible by 4.
  • Consider the following two cases:

    • Case 1: non_four = 0
      In this case, no even elements are present which are not divisible by 4.

      • The optimal way is to first place the “odd” elements first in the array in alternate positions.
      • Fill the vacancies with the even elements.
      • Illustration
        arr[] = {1, 1, 1, 4, 4}
        Step 0: four = 2, odd = 3
        Step 1: {1 _ 1 _ 1}
        Step 2: {1 4 1 4 1}

      • Hence, for the approach to be mathematically possible, the difference between the count of even and odd elements should not exceed 1.
    • Case 2: non_four > 0

      In this case, even elements which are not divisible by 4 exist in the array.

      • Follow the exact same strategy as depicted above by placing the elements divisible by 4 in alternate positions followed by odd elements in the vacancies.
      • Then, place all the remaining even elements at the end of the array. This is because product of two even numbers is always divisible by 4. Hence, placing the even elements towards the end of the array guarantees that product of consecutive elements among them is divisible by 4.

        Illustration:
        arr[] = {2, 7, 1, 8, 2, 8}
        Step 1: four = 2, non_four = 2, odd = 2
        Step 2: {_ 8 _ 8}
        Step 3: {1 8 7 8}
        Step 4: {1 8 7 8 2 2}

      • For this to be possible mathematically, four >= odd.

    Below is the implementation of the above approach:

    C++

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    // C++ Program to rearray array
    // elements such that the product
    // of every two consecutive
    // elements is a multiple of 4
      
    #include <bits/stdc++.h>
    using namespace std;
      
    // Function to rearrange array
    // elements such that the every
    // two consecutive elements is
    // a multiple of 4
    void Permute(vector<int>& arr,
                 int n)
    {
      
        int odd = 0, four = 0;
        int non_four = 0;
      
        vector<int> ODD, FOUR,
            NON_FOUR;
      
        for (auto x : arr) {
      
            // If element is odd
            if (x & 1) {
                odd++;
                // Odd
                ODD.push_back(x);
            }
      
            // If element is divisible
            // by 4
            else if (x % 4 == 0) {
                four++;
                // Divisible by 4
                FOUR.push_back(x);
            }
      
            // If element is not
            // divisible by 4
            else {
                non_four++;
                // Even but not divisble
                // by 4
                NON_FOUR.push_back(x);
            }
        }
      
        // Condition for rearrangement
        // to be possible
        if (non_four == 0
            && four >= odd - 1) {
      
            int x = ODD.size();
            int y = FOUR.size();
            int i;
      
            // Print ODD[i] and FOUR[i]
            // consecutively
            for (i = 0; i < x; i++) {
                cout << ODD[i] << " ";
                if (i < y)
                    cout << FOUR[i] << " ";
            }
      
            // Print the remaining
            // FOUR[i], if any
            while (i < y)
                cout << FOUR[i] << " ";
            cout << endl;
        }
      
        // Condition for rearrangement
        // to be possible
        else if (non_four > 0
                 and four >= odd) {
      
            int x = ODD.size();
            int y = FOUR.size();
            int i;
      
            // Print ODD[i] and FOUR[i]
            // consecutively
            for (i = 0; i < x; i++) {
                cout << ODD[i] << " ";
                if (i < y)
                    cout << FOUR[i] << " ";
            }
      
            // Print the remaining
            // FOUR[i], if any
            while (i < y)
                cout << FOUR[i] << " ";
      
            // Print the NON_FOUR[i]
            // elements at the end
            for (int j = 0;
                 j < (int)NON_FOUR.size();
                 j++)
                cout << NON_FOUR[j] << " ";
            cout << endl;
        }
        else
      
            // No possible configuration
            cout << "Not Possible" << endl;
    }
      
    // Driver Code
    signed main()
    {
      
        vector<int> arr = { 2, 7, 1,
                            8, 2, 8 };
        int N = sizeof(arr) / sizeof(arr);
        Permute(arr, N);
      
        return 0;
    }

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    Python3

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    # Python3 program to rearray array
    # elements such that the product
    # of every two consecutive
    # elements is a multiple of 4
      
    # Function to rearrange array
    # elements such that the every
    # two consecutive elements is
    # a multiple of 4
    def Permute(arr, n):
          
        odd = 0
        four = 0
        non_four = 0
        ODD, FOUR, NON_FOUR = [], [], []
      
        for x in arr:
      
            # If element is odd
            if (x & 1):
                odd += 1 
                  
                # Odd
                ODD.append(x)
                  
            # If element is divisible
            # by 4
            elif (x % 4 == 0):
                four += 1
                  
                # Divisible by 4
                FOUR.append(x)
      
            # If element is not
            # divisible by 4
            else:
                non_four += 1
                  
                # Even but not divisble
                # by 4
                NON_FOUR.append(x)
                  
        # Condition for rearrangement
        # to be possible
        if (non_four == 0 and four >= odd - 1):
            x = len(ODD)
            y = len(FOUR)
            i = 0
      
            # Print ODD[i] and FOUR[i]
            # consecutively
            while i < x:
                print(ODD[i], end = " ")
                if (i < y):
                    print(FOUR[i], end = " ")
      
            # Print the remaining
            # FOUR[i], if any
            while (i < y):
                print(FOUR[i], end = " ")
                i += 1
            print()
      
        # Condition for rearrangement
        # to be possible
        elif (non_four > 0 and four >= odd):
            x = len(ODD)
            y = len(FOUR)
            i = 0
      
            # Print ODD[i] and FOUR[i]
            # consecutively
            while i < x:
                print(ODD[i], end = " ")
                if (i < y):
                    print(FOUR[i], end = " ")
                i += 1
      
            # Print the remaining
            # FOUR[i], if any
            while (i < y):
                print(FOUR[i], end = " ")
                i += 1
      
            # Print the NON_FOUR[i]
            # elements at the end
            for j in NON_FOUR:
                print(j, end = " ")
        else:
      
            # No possible configuration
            print("Not Possible")
      
    # Driver Code
    if __name__ == '__main__':
      
        arr = [ 2, 7, 1, 8, 2, 8 ]
        N = len(arr)
          
        Permute(arr, N)
      
    # This code is contributed by mohit kumar 29

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    Output:

    7 8 1 8 2 2
    


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

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