通过将相邻的不相等对重复替换为最大，将数组简化为单个元素

给定的阵列ARR []由N个整数的，任务是通过重复地更换任何一对连续的不相等的元素，以减少给定阵列的单个元件，说ARR [i]和常用3第[i + 1]，最大(ARR [ i]，arr [i + 1])+1 。如果可能，从可开始操作的位置打印元素的索引。否则，打印-1 。

例子：

Input: arr[] = {5, 3, 4, 4, 5} 
Output: 1 
Explanation: 
Step 1: Replace arr[1] and arr[2] with max(arr[1], arr[2])+1 = max(5, 3) + 1 = 6. Therefore, arr[] = {6, 4, 4, 5}.
Step 2: Replace arr[1] and arr[2] with max(arr[1], arr[2]) + 1 = max(6, 4) + 1 = 7. Therefore, arr[] = {7, 4, 5}.
Step 3: Replace arr[1] and arr[2] with max(arr[1], arr[2])+1 = max(7, 4) + 1 = 8. Therefore, arr[] = {8, 5}.
Step 4: Replace arr[1] and arr[2] with max(arr[1], arr[2]) + 1 = max(8, 5)+1 = 9. Therefore, arr[] = {9}.

Input: arr[] ={1, 1} 
Output: -1

天真的方法：最简单的方法是遍历给定的数组，并对每个元素arr [i] ，根据给定的约束条件开始连续执行给定的操作。如果对于任何元素，该数组变为单个元素，则打印索引i，否则打印-1 。

时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：想法是使用排序算法。请注意，如果所有元素都相同，答案将始终为-1 。否则，可以选择具有最大元素的索引以开始执行操作。请按照以下步骤解决问题：

• 创建另一个与给定数组相同的数组B []并创建一个变量save，以-1初始化以存储答案。
• 对数组B []排序。
• 使用变量i在[N – 1到0]范围内遍历数组，如果找到两个连续的不相等元素，即B [i]不等于B [i – 1] ，则将save更新为save = i 。
• 遍历数组后：
  • 如果save为-1 ，则打印-1并返回。
  • 否则，如果save等于arr [0]，而save不等于arr [1] ，则将save更新为1 。
  • 否则，如果save等于arr [N – 1]且save不等于arr [N – 2] ，则将save更新为N。
  • 否则，在[1，N – 1]范围内循环，并检查save是否等于arr [i] ，以使arr [i]不等于arr [i – 1]和arr [i + 1] ，然后更新另存为save = i + 1 。
• 完成上述步骤后，打印存储在变量save中的索引。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print the index from
// where the operation can be started
void printIndex(int arr[], int N)
{
    // Initialize B[]
    int B[N];
 
    // Initialize save
    int save = -1;
 
    // Make B[] equals to arr[]
    for (int i = 0; i < N; i++) {
        B[i] = arr[i];
    }
 
    // Sort the array B[]
    sort(B, B + N);
 
    // Traverse from N-1 to 1
    for (int i = N - 1; i >= 1; i--) {
 
        // If B[i] & B[i-1] are unequal
        if (B[i] != B[i - 1]) {
            save = B[i];
            break;
        }
    }
 
    // If all elements are same
    if (save == -1) {
        cout << -1 << endl;
        return;
    }
 
    // If arr[1] is maximum element
    if (save == arr[0]
        && save != arr[1]) {
        cout << 1;
    }
 
    // If arr[N-1] is maximum element
    else if (save == arr[N - 1]
             && save != arr[N - 2]) {
        cout << N;
    }
 
    // Find the maximum element
    for (int i = 1; i < N - 1; i++) {
 
        if (save == arr[i]
            && (save != arr[i - 1]
                || save != arr[i + 1])) {
            cout << i + 1;
            break;
        }
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 5, 3, 4, 4, 5 };
 
    // Length of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    printIndex(arr, N);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to print the index
// from where the operation can
// be started
static void printIndex(int arr[],
                       int N)
{
  // Initialize B[]
  int []B = new int[N];
 
  // Initialize save
  int save = -1;
 
  // Make B[] equals to arr[]
  for (int i = 0; i < N; i++)
  {
    B[i] = arr[i];
  }
 
  // Sort the array B[]
  Arrays.sort(B);
 
  // Traverse from N-1 to 1
  for (int i = N - 1; i >= 1; i--)
  {
    // If B[i] & B[i-1] are
    // unequal
    if (B[i] != B[i - 1])
    {
      save = B[i];
      break;
    }
  }
 
  // If all elements are same
  if (save == -1)
  {
    System.out.print(-1 + "\n");
    return;
  }
 
  // If arr[1] is maximum
  // element
  if (save == arr[0] &&
      save != arr[1])
  {
    System.out.print(1);
  }
 
  // If arr[N-1] is maximum
  // element
  else if (save == arr[N - 1] &&
           save != arr[N - 2])
  {
    System.out.print(N);
  }
 
  // Find the maximum element
  for (int i = 1; i < N - 1; i++)
  {
    if (save == arr[i] &&
       (save != arr[i - 1] ||
        save != arr[i + 1]))
    {
      System.out.print(i + 1);
      break;
    }
  }
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {5, 3, 4, 4, 5};
 
  // Length of array
  int N = arr.length;
 
  // Function Call
  printIndex(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the
# above approach
 
# Function to prthe index
# from where the operation can
# be started
def printIndex(arr, N):
     
    # Initialize B
    B = [0] * (N)
 
    # Initialize save
    save = -1
     
    # Make B equals to arr
    for i in range(N):
        B[i] = arr[i]
 
    # Sort the array B
    B = sorted(B)
 
    # Traverse from N-1 to 1
    for i in range(N - 1, 1, -1):
         
        # If B[i] & B[i-1] are
        # unequal
        if (B[i] != B[i - 1]):
            save = B[i]
            break
 
    # If all elements are same
    if (save == -1):
        print(-1 + "")
        return
 
    # If arr[1] is maximum
    # element
    if (save == arr[0] and
        save != arr[1]):
        print(1)
 
    # If arr[N-1] is maximum
    # element
    elif (save == arr[N - 1] and
          save != arr[N - 2]):
        print(N)
 
    # Find the maximum element
    for i in range(1, N - 1):
        if (save == arr[i] and
           (save != arr[i - 1] or
            save != arr[i + 1])):
            print(i + 1)
            break
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr
    arr = [ 5, 3, 4, 4, 5 ]
 
    # Length of array
    N = len(arr)
 
    # Function Call
    printIndex(arr, N)
 
# This code is contributed by Rajput-Ji

C#

// C# program for the
// above approach
using System;
 
class GFG{
 
// Function to print the index
// from where the operation can
// be started
static void printIndex(int []arr,
                       int N)
{
   
  // Initialize []B
  int []B = new int[N];
 
  // Initialize save
  int save = -1;
 
  // Make []B equals to []arr
  for(int i = 0; i < N; i++)
  {
    B[i] = arr[i];
  }
 
  // Sort the array []B
  Array.Sort(B);
 
  // Traverse from N-1 to 1
  for(int i = N - 1; i >= 1; i--)
  {
     
    // If B[i] & B[i-1] are
    // unequal
    if (B[i] != B[i - 1])
    {
      save = B[i];
      break;
    }
  }
 
  // If all elements are same
  if (save == -1)
  {
    Console.Write(-1 + "\n");
    return;
  }
 
  // If arr[1] is maximum
  // element
  if (save == arr[0] &&
      save != arr[1])
  {
    Console.Write(1);
  }
 
  // If arr[N-1] is maximum
  // element
  else if (save == arr[N - 1] &&
           save != arr[N - 2])
  {
    Console.Write(N);
  }
 
  // Find the maximum element
  for(int i = 1; i < N - 1; i++)
  {
    if (save == arr[i] &&
       (save != arr[i - 1] ||
        save != arr[i + 1]))
    {
      Console.Write(i + 1);
      break;
    }
  }
}
 
// Driver Code
public static void Main(String[] args)
{
   
  // Given array []arr
  int []arr = { 5, 3, 4, 4, 5 };
 
  // Length of array
  int N = arr.Length;
 
  // Function Call
  printIndex(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program for the above approach
 
// Function to prlet the index
// from where the operation can
// be started
function printIndex(arr, N)
{
     
    // Initialize B[]
    let B = [];
     
    // Initialize save
    let save = -1;
     
    // Make B[] equals to arr[]
    for(let i = 0; i < N; i++)
    {
        B[i] = arr[i];
    }
     
    // Sort the array B[]
    B.sort();
     
    // Traverse from N-1 to 1
    for(let i = N - 1; i >= 1; i--)
    {
         
        // If B[i] & B[i-1] are
        // unequal
        if (B[i] != B[i - 1])
        {
            save = B[i];
            break;
        }
    }
     
    // If all elements are same
    if (save == -1)
    {
        document.write(-1 + "<br/>");
        return;
    }
     
    // If arr[1] is maximum
    // element
    if (save == arr[0] &&
        save != arr[1])
    {
        document.write(1);
    }
     
    // If arr[N-1] is maximum
    // element
    else if (save == arr[N - 1] &&
            save != arr[N - 2])
    {
        document.write(N);
    }
     
    // Find the maximum element
    for (let i = 1; i < N - 1; i++)
    {
        if (save == arr[i] &&
           (save != arr[i - 1] ||
            save != arr[i + 1]))
        {
            document.write(i + 1);
            break;
        }
    }
}
 
// Driver Code
 
// Given array arr[]
let arr = [5, 3, 4, 4, 5];
 
// Length of array
let N = arr.length;
 
// Function Call
printIndex(arr, N);
 
// This code is contribute by target_2
 

输出：

1

时间复杂度： O(N)
辅助空间： O(N)