dsal 包含 Q 查询的给定元素 X 的数组中的最大间隔

包含 Q 查询的给定元素 X 的数组中的最大间隔

原文:https://www . geeksforgeeks . org/包含给定元素 x 的最大数组间隔查询/

给定一个由 N 元素和 Q 查询组成的【X】数组arr[]。对于每个查询，任务是找到数组的最大区间【L，R】，使得区间中最大的元素为arr【X】，使得 1 ≤ L ≤ X ≤ R 。 注意:数组有基于 1 的索引。

示例:

输入: N = 5，arr[] = {2，1，2，3，2}，Q = 3，query[] = {1，2，4} 输出: 【1，3】【2，2】【1，5】 说明: 在第一次查询中，x = 1，所以 arr[x] = 2，答案为 L = 1，R = 3。在这里，我们可以看到 max(arr[1]，arr[2]，arr[3]) = arr[x]，这是最大间隔。在第二个查询中，x = 2，所以 arr[x] = 1，由于它是数组中最小的元素，所以区间只包含一个元素，因此范围为[2，2]。第三次查询，x = 4，所以 arr[x] = 4，这是 arr[]的最大元素，所以答案是全数组，L = 1，R = N。

输入: N = 4，arr[] = { 1，2，2，4}，Q = 2，query[] = {1，2} 输出: 【1，1】【1，3】 解释: 在第一次查询中，x = 1，所以 arr[x] = 1 并且由于它是数组中最小的元素，所以区间只包含一个元素，因此范围为[1，1]。第二次查询，x = 2，所以 arr[x] = 2，答案是 L = 1，R = 3。在这里，我们可以看到 max(arr[1]，arr[2]，arr[3]) = arr[x] = arr[2] = 2，这是最大间隔。

方法:思路是预计算arr【】中从 1 到 N 的每个值 K 的最大区间。以下是步骤:

对于arr【】中的每个元素 K ，固定元素 K 的索引，然后找出我们可以将间隔扩展到它的左右多少。
递减左迭代器直到 arr【左】≤ K ，类似地递增右迭代器直到 arr【右】≤ K 。
左右的最终值代表区间的开始和结束索引，分别存储在 arrL[] 和 arrR[] 中。
在我们预先计算了每个值的区间范围之后。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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

java 描述语言

<script>

// Javascript program for the above approach

// Function to precompute the interval
// for each query
function utilLargestInterval(arr, arrL, arrR, N)
{

    // For every values [1, N] find
    // the longest intervals
    for (var maxValue = 1;
         maxValue <= N; maxValue++) {

        var lastIndex = 0;

        // Iterate the array arr[]
        for (var i = 1; i <= N; i++) {

            if (lastIndex >= i
                || arr[i] != maxValue)
                continue;
            var 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 (var 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[]
function largestInterval( arr, query, N, Q)
{

    // To store the L and R of X
    var arrL = Array(N+1).fill(0),arrR = Array(N+1).fill(0);

    // Function Call
    utilLargestInterval(arr, arrL,
                        arrR, N);

    // Iterate to find ranges for each query
    for (var i = 0; i < Q; i++) {

        document.write( "[" + arrL[query[i]]
             + ", " + arrR[query[i]]
             + "]<br>");
    }
}

// Driver Code
var N = 5, Q = 3;

// Given array arr[]
var arr = [0, 2, 1, 2, 3, 2];

// Given Queries
var query = [1, 2, 4];

// Function Call
largestInterval(arr, query, N, Q);

// This code is contributed by itsok.
</script>

Output:

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

时间复杂度:O(Q+N²) 辅助空间: O(N)

版权属于：月萌API www.moonapi.com，转载请注明出处

本文链接：https://moonapi.com/news/36182.html

dsal 查看更多书籍

《GeeksForGeeks 数据结构和算法中文教程 2022-10-31》

分类

最近更新

dsal 包含 Q 查询的给定元素 X 的数组中的最大间隔

包含 Q 查询的给定元素 X 的数组中的最大间隔

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

留言

联系客服

数据知识

系统公告

开发文档

dsal查看更多书籍

《GeeksForGeeks 数据结构和算法中文教程 2022-10-31》

dsal 包含 Q 查询的给定元素 X 的数组中的最大间隔

包含 Q 查询的给定元素 X 的数组中的最大间隔

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

留言

联系客服

dsal 查看更多书籍