Minimize shifting half of each element to either sides to reduce Array

Given an array arr[] of size N, the task is to find minimum number of steps required to reduce all Array elements to 0 except the first and last, where in each step:

• Choose any index i (0<i<N-1) and decrement the element at that index (arr[i]) by 2
• Then increment any two elements, arr[j], where 0<=j<i and arr[k], where i<k<n, by 1.

If it is not possible to reduce the given array as desired, return -1.

Input: arr[] = {2, 3, 3, 4, 7}
Output: 6
Explanation: Consider the below steps as how the reduction will be done

steps           array
0              {2, 3, 3, 4, 7}         original array
1              {2, 4, 1, 4, 8}         deduct 2 from a[2] add it to a[1] and a[4]
2             {3, 2, 2, 4, 8}          deduct 2 from a[1] add it to a[0] and a[2]
3             {4, 0, 2, 4, 9}          deduct 2 from a[1] add it to a[0] and a[4]
4             {5, 0, 0, 4, 10}       deduct 2 from a[2] add it to a[0] and a[4]
5             {6, 0, 0, 2, 11}       deduct 2 from a[3] add it to a[0] and a[4]
6            {7, 0, 0, 0, 12}        deduct 2 from a[3] add it to a[0] and a[4]

Input: arr[] = {3,2,1,3,5,4}
Output: 7

Approach:

Intuition: The above problem can be solved with the help of below observations:

• Case 1: A[i] is even
  • As we always subtract 2 from a[i], this means a[i] must be an even number, else after subtracting 2 certain number of times, 1 will remain on which no operation can be performed any further.
• Case 2: A[i] is odd
  • So to make every element 0, we need to have even numbers, and if we do not have even numbers then we have to make it even.
  • When we subtract 2 from a[i], we select i such that arr[i] is even and arr[j] and/or arr[k] are odd.
  • So adding 1 to them will make then an even number.
• Case 3: If there is no odd element
  • If we do not have any odd number, then we can choose the first and last elements, and directly add 1 to a[0] and a[n-1] as we don’t have to reduce these elements.

Illustration 1:

Suppose we have {3,2,1,3,5,4}, 

Step 1: There is an odd element in the array. So choose the 1st odd element in index range (0, N-1), i.e. a[2] = 1
              We take 2 from a[1] add 1 to a[0] and 1 to a[2].
              So the array becomes {4, 0, 2, 3, 5, 4}, and our 1 becomes 2, which is now an even number. 

Step 2: Now consider next odd number in range (0, N-1), i.e. a[3] = 3
              We take 2 from a[2] add 1 to a[0] and 1 to a[2]. 

             So the array becomes {5, 0, 0, 4, 5, 4}, and our 3 becomes 4, which is now  an even number. 

Step 3: Now consider next odd number in range (0, N-1), i.e. a[4] = 5
              We take 2 from a[3] add 1 to a[0] and 1 to a[4].

             So the array becomes {6, 0, 0, 2, 6, 4}, and our 5 becomes 6, which is now  an even number. 

Step 4 – 7: We take 2 from each even numbers i.e, 2 and 6 and shift it to a[0] and a[n-1]
              As at each step we can reduce by 2, So we will require 1 step for element 2 and 3 steps for element 6.

Therefore the total operations become 7 and finally the array becomes {10, 0, 0, 0, 0, 8}.

Illustration 2:

Now if we only have 3 elements and the middle value is an odd number,
Then answer is not possible because the only possible selection is j=0,i=1 and k=2.

So after subtracting 2 a certain number of times a[i] will become 1. 

So in this case a valid answer is not possible, and the answer will be -1.

Below is the Implementation :

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to shift the array
int stepsToRearrange(vector<int>& A, int n)
{
    // if A has 3 elements and
    // 2nd element is odd then
    // answer is not possible
    if (n == 3 && (A[1] % 2 == 1))
        return -1;

    int steps = 0, countOne = 0;

    // Select even numbers and subtract 2
    // Now for the odd number
    // we add one and make them even.
    // so total steps = (a[i]+1)/2
    for (int i = 1; i < n - 1; i++) {
        steps += (A[i] + 1) / 2;
        countOne += (A[i] == 1);
    }

    // If all numbers in index 1 to n-2 = 1
    // then answer is not possible
    if (countOne == n - 2)
        return -1;

    // return steps
    return steps;
}

// Driver Code
int main()
{
    vector<int> arr = { 3, 2, 1, 3, 5, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);

    cout << stepsToRearrange(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG {

  // Function to shift the array
  public static int stepsToRearrange(int A[], int n)
  {

    // if A has 3 elements and
    // 2nd element is odd then
    // answer is not possible
    if (n == 3 && (A[1] % 2 == 1))
      return -1;

    int steps = 0, countOne = 0;

    // Select even numbers and subtract 2
    // Now for the odd number
    // we add one and make them even.
    // so total steps = (a[i]+1)/2
    for (int i = 1; i < n - 1; i++) {
      steps += (A[i] + 1) / 2;
      if (A[i] == 1) {
        countOne += 1;
      }
    }

    // If all numbers in index 1 to n-2 = 1
    // then answer is not possible
    if (countOne == n - 2)
      return -1;

    // return steps
    return steps;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 3, 2, 1, 3, 5, 4 };
    int N = arr.length;

    System.out.print(stepsToRearrange(arr, N));
  }
}

// This code is contributed by Taranpreet

Python3

# python3 program for the above approach

# Function to shift the array
def stepsToRearrange(A, n):

    # if A has 3 elements and
    # 2nd element is odd then
    # answer is not possible
    if (n == 3 and (A[1] % 2 == 1)):
        return -1

    steps, countOne = 0, 0

    # Select even numbers and subtract 2
    # Now for the odd number
    # we add one and make them even.
    # so total steps = (a[i]+1)/2
    for i in range(1, n-1):
        steps += (A[i] + 1) // 2
        countOne += (A[i] == 1)

    # If all numbers in index 1 to n-2 = 1
    # then answer is not possible
    if (countOne == n - 2):
        return -1

    # return steps
    return steps

# Driver Code
if __name__ == "__main__":

    arr = [3, 2, 1, 3, 5, 4]
    N = len(arr)

    print(stepsToRearrange(arr, N))

# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG {

    // Function to shift the array
    static int stepsToRearrange(int[] A, int n)
    {
        // if A has 3 elements and
        // 2nd element is odd then
        // answer is not possible
        if (n == 3 && (A[1] % 2 == 1))
            return -1;

        int steps = 0, countOne = 0;

        // Select even numbers and subtract 2
        // Now for the odd number
        // we add one and make them even.
        // so total steps = (a[i]+1)/2
        for (int i = 1; i < n - 1; i++) {
            steps += (A[i] + 1) / 2;
            if (A[i] == 1) {
                countOne += 1;
            }
        }

        // If all numbers in index 1 to n-2 = 1
        // then answer is not possible
        if (countOne == n - 2)
            return -1;

        // return steps
        return steps;
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 3, 2, 1, 3, 5, 4 };
        int N = arr.Length;

        Console.Write(stepsToRearrange(arr, N));
    }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript program for the above approach

// Function to shift the array
function stepsToRearrange(A, n)
{
    // if A has 3 elements and
    // 2nd element is odd then
    // answer is not possible
    if (n == 3 && (A[1] % 2 == 1))
        return -1;

    let steps = 0, countOne = 0;

    // Select even numbers and subtract 2
    // Now for the odd number
    // we add one and make them even.
    // so total steps = (a[i]+1)/2
    for (let i = 1; i < n - 1; i++) {
        steps += Math.floor((A[i] + 1) / 2);
        countOne += (A[i] == 1);
    }

    // If all numbers in index 1 to n-2 = 1
    // then answer is not possible
    if (countOne == n - 2)
        return -1;

    // return steps
    return steps;
}

// Driver Code
let arr = [ 3, 2, 1, 3, 5, 4 ];
let N = arr.length;

document.write(stepsToRearrange(arr, N));

// This code is contributed by Samim Hossain Mondal.
</script>

Output

7

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