数组中位置 j 的计数，使得 arr[i] 在索引范围 [i, j] 中最大，端点相同

给定一个由N个正整数组成的数组arr[] ，任务是找到所有j使得arr[j] = arr[i]以及范围[min(j, i), max(j, i)]小于或等于arr[i]其中1 ≤ i ≤ N 。

例子：

Input: arr[] = {4, 7, 7, 9, 7}, N = 5
Output: 1 2 2 1 1
Explanation:
For i = 1, j = 1 is the only element such that arr[i] = arr[j] and no element in the middle has value greater than arr[1].
For i = 2, j = 2 and 3 are the elements such that arr[i] = arr[j] and no elements in the middle has value greater than arr[2].
For i = 3, j = 2 and 3 are the elements such that arr[i] = arr[j] and no elements in the middle has value greater than arr[3].
For i = 4, j = 4 is the only element such that arr[i] = arr[j] and no element in the middle has value greater than arr[4].
For i = 5, j = 5 is the only element such that arr[i] = arr[j] and no element in the middle has value greater than arr[5].

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

方法：解决问题最简单的方法是使用两个嵌套的for循环遍历数组，找到满足arr[i] = arr[j]且[i, j]范围内没有元素大于arr的对[我] 。请按照以下步骤解决问题：

• 初始化一个数组，比如ans[] ，它存储范围[0, N-1]中所有元素的答案。
• 使用变量i在[N-1, 0]范围内迭代并执行以下步骤：
  • 使用变量j在范围[i, 0]中迭代并执行以下步骤：
    • 如果arr[j] = arr[i] ，则将ans[i]的值增加1 。
    • 如果arr[j] > arr[i] ，则终止循环。
  • 使用变量j在[i+1, N-1]范围内迭代并执行以下步骤：
    • 如果arr[j] = arr[i] ，则将ans[i]的值增加1 。
    • 如果arr[j] > arr[i] ，则终止循环。
• 完成上述步骤后，打印数组ans[]作为答案。

下面是上述方法的实现

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find j such that arr[j]=arr[i]
// and no element is greater than arr[i] in the
// range [j, i]
void findElements(int arr[], int N)
{
 
    // To Store answer
    int ans[N];
 
    // Initialize the value of ans[i] as 0
    for (int i = 0; i < N; i++) {
        ans[i] = 0;
    }
 
    // Traverse the array in reverse direction
    for (int i = N - 1; i >= 0; i--) {
 
        // Traverse in the range [i, 0]
        for (int j = i; j >= 0; j--) {
            if (arr[j] == arr[i]) {
 
                // Increment ans[i] if arr[j] =arr[i]
                ans[i]++;
            }
            else
                // Terminate the loop
                if (arr[j] > arr[i])
                break;
        }
 
        // Traverse in the range [i+1, N-1]
        for (int j = i + 1; j < N; j++) {
            // Increment ans[i] if arr[i] = arr[j]
            if (arr[j] == arr[i])
                ans[i]++;
            else if (arr[j] > arr[i]) {
                break;
            }
        }
    }
 
    for (int i = 0; i < N; i++) {
        cout << ans[i] << " ";
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 1, 2, 1, 2, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    findElements(arr, N);
}

Java

// Java program for the above approach
public class GFG
{
 
    // Function to find j such that arr[j]=arr[i]
    // and no element is greater than arr[i] in the
    // range [j, i]
    static void findElements(int arr[], int N)
    {
 
        // To Store answer
        int ans[] = new int[N];
 
        // Initialize the value of ans[i] as 0
        for (int i = 0; i < N; i++) {
            ans[i] = 0;
        }
 
        // Traverse the array in reverse direction
        for (int i = N - 1; i >= 0; i--) {
 
            // Traverse in the range [i, 0]
            for (int j = i; j >= 0; j--) {
                if (arr[j] == arr[i]) {
 
                    // Increment ans[i] if arr[j] =arr[i]
                    ans[i]++;
                }
                else
                    // Terminate the loop
                    if (arr[j] > arr[i])
                    break;
            }
 
            // Traverse in the range [i+1, N-1]
            for (int j = i + 1; j < N; j++)
            {
                // Increment ans[i] if arr[i] = arr[j]
                if (arr[j] == arr[i])
                    ans[i]++;
                else if (arr[j] > arr[i]) {
                    break;
                }
            }
        }
 
        for (int i = 0; i < N; i++) {
            System.out.print(ans[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // Given Input
        int arr[] = { 1, 2, 1, 2, 4 };
        int N = arr.length;
 
        // Function Call
        findElements(arr, N);
    }
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to find j such that arr[j]=arr[i]
# and no element is greater than arr[i] in the
# range [j, i]
def findElements(arr, N):
     
    # Initialising a list to store ans
    # Initialize the value of ans[i] as 0
    ans = [0 for i in range(N)]
 
    # Traverse the array in reverse direction
    for i in range(N - 1, -1, -1):
         
        # Traverse in the range [i, 0]
        for j in range(i, -1, -1):
            if arr[j] == arr[i]:
                 
                # Increment ans[i] if arr[j] = arr[i]
                ans[i] += 1
                 
            else:
                 
                # Terminate the loop
                if arr[j] > arr[i]:
                    break
                     
        # Traverse in the range [i+1, N-1]
        for j in range(i+1, N):
           
            # Increment ans[i] if arr[j] = arr[i]
            if arr[j] == arr[i]:
                ans[i] += 1
            else:
                if arr[j] > arr[i]:
                    break
                     
    # Print the ans
    for i in range(N):
        print(ans[i], end = " ")
         
    return
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    arr = [ 1, 2, 1, 2, 4 ]
    N = len(arr)
     
    # Function call
    findElements(arr, N)
 
# This code is contributed by MuskanKalra1

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find j such that arr[j]=arr[i]
// and no element is greater than arr[i] in the
// range [j, i]
static void findElements(int []arr, int N)
{
     
    // To Store answer
    int []ans = new int[N];
 
    // Initialize the value of ans[i] as 0
    for(int i = 0; i < N; i++)
    {
        ans[i] = 0;
    }
 
    // Traverse the array in reverse direction
    for(int i = N - 1; i >= 0; i--)
    {
 
        // Traverse in the range [i, 0]
        for(int j = i; j >= 0; j--)
        {
            if (arr[j] == arr[i])
            {
                 
                // Increment ans[i] if arr[j] =arr[i]
                ans[i]++;
            }
            else
             
                // Terminate the loop
                if (arr[j] > arr[i])
                    break;
        }
 
        // Traverse in the range [i+1, N-1]
        for(int j = i + 1; j < N; j++)
        {
             
            // Increment ans[i] if arr[i] = arr[j]
            if (arr[j] == arr[i])
                ans[i]++;
            else if (arr[j] > arr[i])
            {
                break;
            }
        }
    }
 
    for(int i = 0; i < N; i++)
    {
        Console.Write(ans[i] + " ");
    }
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    int []arr = { 1, 2, 1, 2, 4 };
    int N = arr.Length;
 
    // Function Call
    findElements(arr, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript

 
// JavaScript program for the above approach
 
 
// Function to find j such that arr[j]=arr[i]
// and no element is greater than arr[i] in the
// range [j, i]
function findElements(arr, N) {
 
    // To Store answer
    let ans = new Array(N);
 
    // Initialize the value of ans[i] as 0
    for (let i = 0; i < N; i++) {
        ans[i] = 0;
    }
 
    // Traverse the array in reverse direction
    for (let i = N - 1; i >= 0; i--) {
 
        // Traverse in the range [i, 0]
        for (let j = i; j >= 0; j--) {
            if (arr[j] == arr[i]) {
 
                // Increment ans[i] if arr[j] =arr[i]
                ans[i]++;
            }
            else
                // Terminate the loop
                if (arr[j] > arr[i])
                    break;
        }
 
        // Traverse in the range [i+1, N-1]
        for (let j = i + 1; j < N; j++) {
            // Increment ans[i] if arr[i] = arr[j]
            if (arr[j] == arr[i])
                ans[i]++;
            else if (arr[j] > arr[i]) {
                break;
            }
        }
    }
 
    for (let i = 0; i < N; i++) {
        document.write(ans[i] + " ");
    }
}
 
// Driver Code
 
// Given Input
let arr = [1, 2, 1, 2, 4];
let N = arr.length
 
// Function Call
findElements(arr, N);
 

输出

1 2 1 2 1 

时间复杂度： O(N ² )
辅助空间： O(N)