最小化将每个元素的一半移动到两侧以减少 Array

给定一个大小为N的数组arr[] ，任务是找到将除第一个和最后一个之外的所有 Array 元素减少到 0 所需的最小步骤数，其中在每个步骤中：

• 选择任何索引 i (0(arr[i])处的元素减2
• 然后将任意两个元素arr[j]（其中 0<=j 。

如果无法根据需要减少给定数组，则返回 -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

方法：

直觉：上述问题可以借助以下观察来解决：

• 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.

插图 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}.

插图 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.

以下是实施：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to shift the array
int stepsToRearrange(vector& 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 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

// 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.

输出

7

时间复杂度：O(N)
辅助空间：O(1)