通过循环移动段或用最大值替换前缀来最小化使所有 Array 元素相等的步骤

给定一个由N个正整数组成的数组arr[] ，任务是通过对数组执行任意次数（可能为 0）的以下操作，打印使数组中所有元素相等所需的最小步数。

• 操作 1：选择任何前缀arr[1…k]使得 max (arr[1], arr[2], …, arr[k]) = arr[k] 并设置arr[i] = arr[k]对于[1,k−1]范围内的所有i 。
• 操作2：选择任何段[L，R]并将该段循环向左移动，保持其他元素不变。

例子：

Input: arr[] = {1, 2, 12, 20, 18, 19}
Output: 2
Explanation: It can be solved in two steps:
In the first step, apply operation 2 by choosing L = 4 and R = 6 yields the sequence {1, 2, 12, 18, 19, 20}. 
In the second step, apply operation 1 on the prefix arr{1, 2, 12, 18, 19, 20} which converts arr into {20, 20, 20, 20, 20} where all elements are equal.

Input: arr[] = {2, 2, 2, 2}
Output: 0
Explanation: the elements in the array are all same.

方法：给定的问题可以通过使用以下观察来解决：

• 如果所有元素都相同，则不需要进行任何操作。
• 如果数组的最大值最后是，那么只对整个数组执行操作 1 会使所有元素相等。
• 如果最大值在最后一个位置之前的某个位置，则说 j。然后在段 [j, N] 中执行操作 2，然后对前缀 [1, N] 执行操作 1，使所有元素相等。

因此，根据上述观察，答案可以是 0、1 或 2。请按照下面提到的步骤进行操作。

• 迭代数组并从数组中找到最大值。
• 如果数组的所有元素都相同，则返回 0。
• 如果最大值出现在数组的末尾，则答案为 1，如观察结果所示。
• 如果这两个都不是，那么答案是 2。

下面是上述方法的实现：

C++

// C++ code to implement the above approach
#include 
using namespace std;
 
// Function to print the minimum number
// of steps required to make all
// the elements are equal
void solve(int arr[], int N)
{
    // Initially assume all elements
    // are equal
    bool isEqual = true;
 
    // For finding the largest element
    int maxi = arr[0];
 
    // Loop to iterate through the array
    for (int i = 1; i < N; i++) {
 
        // If any element is not equal
        // than previous element, mark
        // it as false
        if (arr[i] != arr[i - 1]) {
            isEqual = false;
        }
 
        // Storing greater element
        maxi = max(maxi, arr[i]);
    }
 
    // If all the elements are equal
    if (isEqual) {
        cout << "0";
    }
 
    // If max element is found
    // at the last index
    else if (maxi == arr[N - 1]) {
        cout << "1";
    }
 
    // If max element is found
    // other than the last index
    else {
        cout << "2";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 12, 20, 18, 19 };
    int N = sizeof(arr) / sizeof(arr[0]);
    solve(arr, N);
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  // Function to print the minimum number
  // of steps required to make all
  // the elements are equal
  static void solve(int arr[], int N)
  {
 
    // Initially assume all elements
    // are equal
    boolean isEqual = true;
 
    // For finding the largest element
    int maxi = arr[0];
 
    // Loop to iterate through the array
    for (int i = 1; i < N; i++) {
 
      // If any element is not equal
      // than previous element, mark
      // it as false
      if (arr[i] != arr[i - 1]) {
        isEqual = false;
      }
 
      // Storing greater element
      maxi = Math.max(maxi, arr[i]);
    }
 
    // If all the elements are equal
    if (isEqual) {
      System.out.println("0");
    }
 
    // If max element is found
    // at the last index
    else if (maxi == arr[N - 1]) {
      System.out.println("1");
    }
 
    // If max element is found
    // other than the last index
    else {
      System.out.println("2");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 1, 2, 12, 20, 18, 19 };
    int N = arr.length;
    solve(arr, N);
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python3 code to implement the above approach
   
# Function to print the minimum number
# of steps required to make all
# the elements are equal
def solve(arr, N):
     
    # Initially assume all elements
    # are equal
    isEqual = True
   
    # For finding the largest element
    maxi = arr[0]
   
    # Loop to iterate through the array
    for i in range(1, N):
   
        # If any element is not equal
        # than previous element, mark
        # it as false
        if (arr[i] != arr[i - 1]):
            isEqual = False
             
        # Storing greater element
        maxi = max(maxi, arr[i])
   
    # If all the elements are equal
    if (isEqual):
        print("0")
     
    # If max element is found
    # at the last index
    elif (maxi == arr[N - 1]):
        print("1")
     
    # If max element is found
    # other than the last index
    else:
        print("2")
     
# Driver Code
if __name__=="__main__":
   
    arr = [ 1, 2, 12, 20, 18, 19 ]
    N = len(arr)
     
    solve(arr, N)
 
# This code is contributed by Akash Jha

C#

// C# code to implement the above approach
 
using System;
 
class GFG {
 
  // Function to print the minimum number
  // of steps required to make all
  // the elements are equal
  static void solve(int[] arr, int N)
  {
 
    // Initially assume all elements
    // are equal
    bool isEqual = true;
 
    // For finding the largest element
    int maxi = arr[0];
 
    // Loop to iterate through the array
    for (int i = 1; i < N; i++) {
 
      // If any element is not equal
      // than previous element, mark
      // it as false
      if (arr[i] != arr[i - 1]) {
        isEqual = false;
      }
 
      // Storing greater element
      maxi = Math.Max(maxi, arr[i]);
    }
 
    // If all the elements are equal
    if (isEqual) {
      Console.WriteLine("0");
    }
 
    // If max element is found
    // at the last index
    else if (maxi == arr[N - 1]) {
      Console.WriteLine("1");
    }
 
    // If max element is found
    // other than the last index
    else {
      Console.WriteLine("2");
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = { 1, 2, 12, 20, 18, 19 };
    int N = arr.Length;
    solve(arr, N);
  }
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

// Javascript code to implement the above approach
 
// Function to print the minimum number
// of steps required to make all
// the elements are equal
function solve(arr, N)
{
 
    // Initially assume all elements
    // are equal
    let isEqual = true;
 
    // For finding the largest element
    let maxi = arr[0];
 
    // Loop to iterate through the array
    for (let i = 1; i < N; i++) {
 
        // If any element is not equal
        // than previous element, mark
        // it as false
        if (arr[i] != arr[i - 1]) {
            isEqual = false;
        }
 
        // Storing greater element
        maxi = Math.max(maxi, arr[i]);
    }
 
    // If all the elements are equal
    if (isEqual) {
        document.write("0");
    }
 
    // If max element is found
    // at the last index
    else if (maxi == arr[N - 1]) {
        document.write("1");
    }
 
    // If max element is found
    // other than the last index
    else {
        document.write("2");
    }
}
 
// Driver Code
 
let arr = [1, 2, 12, 20, 18, 19];
let N = arr.length;
solve(arr, N);
 
// This code is contributed by saurabh_jaiswal.

输出

2

时间复杂度： O(N)
辅助空间： O(1)