包含给定Q元素X的数组中的最大间隔

给定一个由N个元素组成的数组arr []和形式为[X]的Q个查询。对于每个查询，任务是找到这样的阵列的最大区间[L，R]，在间隔的最大元素是ARR [X]，使得1≤大号≤X≤R上。
注意：数组具有从1开始的索引。

例子：

Input: N = 5, arr[] = {2, 1, 2, 3, 2}, Q = 3, query[] = {1, 2, 4} 
Output: 
[1, 3] 
[2, 2] 
[1, 5] 
Explanation : 
In 1st query, x = 1, so arr[x] = 2 and answer is L = 1 and R = 3. here, we can see that max(arr[1], arr[2], arr[3]) = arr[x], which is the maximum intervals. 
In 2nd query, x = 2, so arr[x] = 1 and since it is the smallest element of the array, so the interval contains only one element, thus the range is [2, 2]. 
In 3rd query, x = 4, so arr[x] = 4, which is maximum element of the arr[], so the answer is whole array, L = 1 and R = N.

Input: N = 4, arr[] = { 1, 2, 2, 4}, Q = 2, query[] = {1, 2} 
Output: 
[1, 1] 
[1, 3] 
Explanation: 
In 1st query, x = 1, so arr[x] = 1 and since it is the smallest element of the array, so the interval contains only one element, thus the range is [1, 1]. 
In 2nd query, x = 2, so arr[x] = 2 and answer is L = 1 and R = 3. here, we can see that max(arr[1], arr[2], arr[3]) = arr[x] = arr[2] = 2, which is the maximum intervals. 

方法：这个想法是为arr []中的每个值K从1到N预先计算最大间隔。步骤如下：

1. 对于arr []中的每个元素K ，固定元素K的索引，然后找到我们可以将间隔扩展到左右的程度。
2. 减少左迭代器直到arr [left]≤K ，类似地增加右迭代器直到arr [right]≤K 。
3. left和right的最终值代表该间隔的开始索引和结束索引，分别存储在arrL []和arrR []中。
4. 在我们预先计算了每个值的间隔范围之后。然后，对于每个查询，我们需要打印arr [x]的间隔范围，即arrL [arr [x]]和arrR [arr [x]] 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to precompute the interval
// for each query
void utilLargestInterval(int arr[],
                         int arrL[],
                         int arrR[],
                         int N)
{
 
    // For every values [1, N] find
    // the longest intervals
    for (int maxValue = 1;
         maxValue <= N; maxValue++) {
 
        int lastIndex = 0;
 
        // Iterate the array arr[]
        for (int i = 1; i <= N; i++) {
 
            if (lastIndex >= i
                || arr[i] != maxValue)
                continue;
            int left = i, right = i;
 
            // Shift the left pointers
            while (left > 0
                   && arr[left] <= maxValue)
                left--;
 
            // Shift the right pointers
            while (right <= N
                   && arr[right] <= maxValue)
                right++;
 
            left++, right--;
            lastIndex = right;
 
            // Store the range of interval
            // in arrL[] and arrR[].
            for (int j = left; j <= right; j++) {
 
                if (arr[j] == maxValue) {
                    arrL[j] = left;
                    arrR[j] = right;
                }
            }
        }
    }
}
 
// Function to find the largest interval
// for each query in Q[]
void largestInterval(
    int arr[], int query[], int N, int Q)
{
 
    // To store the L and R of X
    int arrL[N + 1], arrR[N + 1];
 
    // Function Call
    utilLargestInterval(arr, arrL,
                        arrR, N);
 
    // Iterate to find ranges for each query
    for (int i = 0; i < Q; i++) {
 
        cout << "[" << arrL[query[i]]
             << ", " << arrR[query[i]]
             << "]\n";
    }
}
 
// Driver Code
int main()
{
    int N = 5, Q = 3;
 
    // Given array arr[]
    int arr[N + 1] = { 0, 2, 1, 2, 3, 2 };
 
    // Given Queries
    int query[Q] = { 1, 2, 4 };
 
    // Function Call
    largestInterval(arr, query, N, Q);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to precompute the interval
// for each query
static void utilLargestInterval(int arr[],
                                int arrL[],
                                int arrR[],
                                int N)
{
 
    // For every values [1, N] find
    // the longest intervals
    for(int maxValue = 1;
            maxValue <= N; maxValue++)
    {
       int lastIndex = 0;
        
       // Iterate the array arr[]
       for(int i = 1; i <= N; i++)
       {
          if (lastIndex >= i ||
                 arr[i] != maxValue)
              continue;
          int left = i, right = i;
           
          // Shift the left pointers
          while (left > 0 &&
                 arr[left] <= maxValue)
              left--;
           
          // Shift the right pointers
          while (right <= N &&
                 arr[right] <= maxValue)
              right++;
           
          left++;
          right--;
          lastIndex = right;
           
          // Store the range of interval
          // in arrL[] and arrR[].
          for(int j = left; j <= right; j++)
          {
             if (arr[j] == maxValue)
             {
                 arrL[j] = left;
                 arrR[j] = right;
             }
          }
       }
    }
}
 
// Function to find the largest interval
// for each query in Q[]
static void largestInterval(int arr[],
                            int query[],
                            int N, int Q)
{
     
    // To store the L and R of X
    int []arrL = new int[N + 1];
    int []arrR = new int[N + 1];
 
    // Function Call
    utilLargestInterval(arr, arrL,
                        arrR, N);
 
    // Iterate to find ranges for
    // each query
    for(int i = 0; i < Q; i++)
    {
       System.out.print("[" + arrL[query[i]] +
                       ", " + arrR[query[i]] + "]\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5, Q = 3;
 
    // Given array arr[]
    int arr[] = { 0, 2, 1, 2, 3, 2 };
 
    // Given queries
    int query[] = { 1, 2, 4 };
 
    // Function call
    largestInterval(arr, query, N, Q);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to precompute the interval
# for each query
def utilLargestInterval(arr, arrL, arrR, N):
  
    # For every values [1, N] find
    # the longest intervals
    for maxValue in range(1, N + 1):
        lastIndex = 0
  
        # Iterate the array arr[]
        for i in range(N + 1):
            if (lastIndex >= i or
                   arr[i] != maxValue):
                continue
             
            left = i
            right = i
  
            # Shift the left pointers
            while (left > 0 and
               arr[left] <= maxValue):
                left -= 1
  
            # Shift the right pointers
            while (right <= N and
               arr[right] <= maxValue):
                right += 1
  
            left += 1
            right -= 1
            lastIndex = right
  
            # Store the range of interval
            # in arrL[] and arrR[].
            for j in range(left, right + 1):
                if (arr[j] == maxValue):
                    arrL[j] = left
                    arrR[j] = right
             
# Function to find the largest interval
# for each query in Q[]
def largestInterval(arr, query, N, Q):
  
    # To store the L and R of X
    arrL = [0 for i in range(N + 1)]
    arrR = [0 for i in range(N + 1)]
  
    # Function call
    utilLargestInterval(arr, arrL, arrR, N);
  
    # Iterate to find ranges for each query
    for i in range(Q):
        print('[' + str(arrL[query[i]]) +
             ', ' + str(arrR[query[i]]) + ']')
  
# Driver code
if __name__=="__main__":
     
    N = 5
    Q = 3
  
    # Given array arr[]
    arr = [ 0, 2, 1, 2, 3, 2 ]
  
    # Given Queries
    query = [ 1, 2, 4 ]
  
    # Function call
    largestInterval(arr, query, N, Q)
 
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to precompute the interval
// for each query
static void utilLargestInterval(int []arr,
                                int []arrL,
                                int []arrR,
                                int N)
{
 
    // For every values [1, N] find
    // the longest intervals
    for(int maxValue = 1;
            maxValue <= N; maxValue++)
    {
       int lastIndex = 0;
        
       // Iterate the array []arr
       for(int i = 1; i <= N; i++)
       {
          if (lastIndex >= i ||
                 arr[i] != maxValue)
              continue;
               
          int left = i, right = i;
           
          // Shift the left pointers
          while (left > 0 &&
                 arr[left] <= maxValue)
              left--;
               
          // Shift the right pointers
          while (right <= N &&
                 arr[right] <= maxValue)
              right++;
           
          left++;
          right--;
          lastIndex = right;
           
          // Store the range of interval
          // in arrL[] and arrR[].
          for(int j = left; j <= right; j++)
          {
             if (arr[j] == maxValue)
             {
                 arrL[j] = left;
                 arrR[j] = right;
             }
          }
       }
    }
}
 
// Function to find the largest interval
// for each query in Q[]
static void largestInterval(int []arr,
                            int []query,
                            int N, int Q)
{
     
    // To store the L and R of X
    int []arrL = new int[N + 1];
    int []arrR = new int[N + 1];
 
    // Function Call
    utilLargestInterval(arr, arrL,
                        arrR, N);
 
    // Iterate to find ranges for
    // each query
    for(int i = 0; i < Q; i++)
    {
       Console.Write("[" + arrL[query[i]] +
                    ", " + arrR[query[i]] + "]\n");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 5, Q = 3;
 
    // Given array []arr
    int []arr = { 0, 2, 1, 2, 3, 2 };
 
    // Given queries
    int []query = { 1, 2, 4 };
 
    // Function call
    largestInterval(arr, query, N, Q);
}
}
 
// This code is contributed by Princi Singh

输出：

[1, 3]
[2, 2]
[1, 5]

时间复杂度： O(Q + N ² )
辅助空间： O(N)