通过将相邻对替换为其乘积可获得的最小数组

给定大小为N的数组arr [] ，任务是打印给定数组可以通过执行以下操作而减小到的最小大小：

• 删除任意两个相邻的元素，例如arr [i]和arr [i + 1]，并在数组中的该位置插入单个元素arr [i] * arr [i + 1] 。
• 如果所有数组元素都相等，则打印数组的大小。

例子：

Input: arr[] = {1, 7, 7, 1, 7, 1} 
Output: 1 
Explanation: 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {7, 7, 1, 7, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {49, 1, 7, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {49, 7, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {343, 1} 
Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {343} 

Input: arr[] = {2, 2, 2, 2} 
Output: 4 

方法：该方法基于以下想法：如果所有数组元素都相同，则无法在给定数组上执行给定的操作。否则，在每种情况下，都可以将数组大小减小为1。以下是步骤：

1. 遍历给定数组arr [] 。
2. 如果数组的所有元素都相同，则将N打印为所需答案。
3. 否则，请始终选择能够提供最大乘积以将arr []的大小减小为1的相邻元素。因此，最小可能的大小将为1 。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to minimize the size of
// array by performing given operations
int minLength(int arr[], int N)
{
 
    for (int i = 1; i < N; i++) {
 
        // If all array elements
        // are not same
        if (arr[0] != arr[i]) {
            return 1;
        }
    }
 
    // If all array elements
    // are same
    return N;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 1, 3, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << minLength(arr, N);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
 
class GFG{
     
// Function to minimize the size of
// array by performing given operations
static int minLength(int arr[], int N)
{
    for(int i = 1; i < N; i++)
    {
         
        // If all array elements
        // are not same
        if (arr[0] != arr[i])
        {
            return 1;
        }
    }
     
    // If all array elements
    // are same
    return N;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 2, 1, 3, 1 };
    int N = arr.length;
 
    // Function call
    System.out.print(minLength(arr, N));
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 implementation of the above approach
 
# Function to minimize the size of
# array by performing given operations
def minLength(arr, N):
     
    for i in range(1, N):
         
        # If all array elements
        # are not same
        if (arr[0] != arr[i]):
            return 1
             
    # If all array elements
    # are same
    return N
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 2, 1, 3, 1 ]
    N = len(arr)
     
    # Function call
    print(minLength(arr, N))
 
# This code is contributed by akhilsaini

C#

// C# implementation of the above approach
using System;
 
class GFG{
     
// Function to minimize the size of
// array by performing given operations
static int minLength(int[] arr, int N)
{
    for(int i = 1; i < N; i++)
    {
         
        // If all array elements
        // are not same
        if (arr[0] != arr[i])
        {
            return 1;
        }
    }
 
    // If all array elements
    // are same
    return N;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 2, 1, 3, 1 };
    int N = arr.Length;
 
    // Function call
    Console.Write(minLength(arr, N));
}
}
 
// This code is contributed by akhilsaini

输出：

1

时间复杂度： O(N)
辅助空间： O(1)