计算元素小于或等于 X 的子数组

给定一个包含 n 个元素的数组和一个整数 X。计算该数组中所有元素小于或等于 X 的子数组的数量。
例子：

Input : arr[] = {1, 5, 7, 8, 2, 3, 9}  
            X = 6
Output : 6
Explanation : Sub-arrays are {1}, {5}, {2}, {3},
{1, 5}, {2, 3}

Input : arr[] =  {1, 10, 12, 4, 5, 3, 2, 7}   
            X = 9
Output : 16

朴素方法：一种简单的方法是使用两个嵌套循环来生成给定数组的所有子数组，并使用一个循环来检查子数组的所有元素是否小于或等于 X。
时间复杂度：O(n*n*n)
有效的方法：一种有效的方法是观察我们只想要那些所有元素小于或等于 X 的子数组的计数。我们可以创建一个对应于原始数组的 0 和 1 的二进制数组。如果原始元素中的某个元素小于或等于 X，则二进制数组中的对应元素将为 1，否则为 0。现在，我们的问题归结为计算这个全为 1 的二进制数组中的子数组的数量。我们还可以看到，对于一个全为 1 的数组，它的所有子数组都只有 1，子数组的总数将为 len*(len+1)/2。例如，{1, 1, 1, 1} 将有 10 个子数组。
下面是解决上述问题的完整算法：

• 如上所述创建原始数组的相应二进制数组。
• 将计数器变量初始化为 0 并开始遍历二进制数组，跟踪全为 1 的子数组的长度
• 我们可以使用公式 n*(n+1)/2 轻松计算全为 1 的数组的子数组的数量，其中 n 是全为 1 的数组的长度。
• 计算每个全为 1 的子数组的长度，并将计数变量增加长度*（长度+1）/2。我们可以在 O(n) 时间复杂度内做到这一点

以下是上述方法的实现：

C++

// C++ program to count all sub-arrays which
// has all elements less than or equal to X
#include 
using namespace std;
 
// function to count all sub-arrays which
// has all elements less than or equal to X
int countSubArrays(int arr[], int n, int x)
{
    // variable to keep track of length of
    // subarrays with all 1s
    int len = 0;
 
    // variable to keep track of all subarrays
    int count = 0;
 
    // binary array of same size
    int binaryArr[n];
 
    // creating binary array
    for (int i = 0; i < n; i++) {
        if (arr[i] <= x)
            binaryArr[i] = 1;
        else
            binaryArr[i] = 0;
    }
 
    // start traversing the binary array
    for (int i = 0; i < n; i++) {
 
        // once we find the first 1, keep checking
        // for number of consecutive 1s
        if (binaryArr[i] == 1) {
            int j;
 
            for (j = i + 1; j < n; j++)
                if (binaryArr[j] != 1)
                    break;
 
            // calculate length of the subarray
            // with all 1s
            len = j - i;
 
            // increment count
            count += (len) * (len + 1) / 2;
 
            // initialize i to j
            i = j;
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 5, 7, 8, 2, 3, 9 };
    int x = 6;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countSubArrays(arr, n, x);
    return 0;
}

Java

// Java program to count all sub-arrays which
// has all elements less than or equal to X
import java.io.*;
 
class GFG {
     
    // function to count all sub-arrays which
    // has all elements less than or equal to X
    static int countSubArrays(int arr[], int n, int x)
    {
         
        // variable to keep track of length of
        // subarrays with all 1s
        int len = 0;
     
        // variable to keep track of all subarrays
        int count = 0;
     
        // binary array of same size
        int binaryArr[] = new int[n];
     
        // creating binary array
        for (int i = 0; i < n; i++) {
            if (arr[i] <= x)
                binaryArr[i] = 1;
            else
                binaryArr[i] = 0;
        }
     
        // start traversing the binary array
        for (int i = 0; i < n; i++) {
     
            // once we find the first 1, keep checking
            // for number of consecutive 1s
            if (binaryArr[i] == 1) {
                int j;
     
                for (j = i + 1; j < n; j++)
                    if (binaryArr[j] != 1)
                        break;
     
                // calculate length of the subarray
                // with all 1s
                len = j - i;
     
                // increment count
                count += (len) * (len + 1) / 2;
     
                // initialize i to j
                i = j;
            }
        }
     
        return count;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 5, 7, 8, 2, 3, 9 };
        int x = 6;
        int n = arr.length;
         
        System.out.println(countSubArrays(arr, n, x));
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# python 3 program to count all sub-arrays which
# has all elements less than or equal to X
 
# function to count all sub-arrays which
# has all elements less than or equal to X
def countSubArrays(arr, n, x):
     
    # variable to keep track of length
    # of subarrays with all 1s
    len = 0
 
    # variable to keep track of
    # all subarrays
    count = 0
 
    # binary array of same size
    binaryArr = [0 for i in range(n)]
 
    # creating binary array
    for i in range(0, n, 1):
        if (arr[i] <= x):
            binaryArr[i] = 1
        else:
            binaryArr[i] = 0
 
    # start traversing the binary array
    for i in range(0, n, 1):
         
        # once we find the first 1,
        # keep checking for number
        # of consecutive 1s
        if (binaryArr[i] == 1):
            for j in range(i + 1, n, 1):
                if (binaryArr[j] != 1):
                    break
 
            # calculate length of the
            # subarray with all 1s
            len = j - i
 
            # increment count
            count += (len) * (int)((len + 1) / 2)
 
            # initialize i to j
            i = j
 
    return count
 
# Driver code
if __name__ == '__main__':
    arr = [1, 5, 7, 8, 2, 3, 9]
    x = 6
    n = len(arr)
    print(int(countSubArrays(arr, n, x)))
     
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to count all sub-arrays which
// has all elements less than or equal to X1
using System;
 
class GFG {
     
    // function to count all sub-arrays which
    // has all elements less than or equal
    // to X
    static int countSubArrays(int []arr,
                                int n, int x)
    {
         
        // variable to keep track of length
        // of subarrays with all 1s
        int len = 0;
     
        // variable to keep track of all
        // subarrays
        int count = 0;
     
        // binary array of same size
        int []binaryArr = new int[n];
     
        // creating binary array
        for (int i = 0; i < n; i++) {
            if (arr[i] <= x)
                binaryArr[i] = 1;
            else
                binaryArr[i] = 0;
        }
     
        // start traversing the binary array
        for (int i = 0; i < n; i++) {
     
            // once we find the first 1, keep
            // checking for number of
            // consecutive 1s
            if (binaryArr[i] == 1) {
                int j;
     
                for (j = i + 1; j< n; j++)
                    if (binaryArr[j] != 1)
                        break;
     
                // calculate length of the
                // subarray with all 1s
                len = j - i;
     
                // increment count
                count += (len) * (len + 1) / 2;
     
                // initialize i to j
                i = j;
            }
        }
     
        return count;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 1, 5, 7, 8, 2, 3, 9 };
        int x = 6;
        int n = arr.Length;
         
        Console.WriteLine(
                    countSubArrays(arr, n, x));
    }
}
 
// This code is contributed by Sam007.

Javascript

// javascript program to count all sub-arrays which
// has all elements less than or equal to X
 
    // function to count all sub-arrays which
    // has all elements less than or equal to X
    function countSubArrays(arr , n , x) {
 
        // variable to keep track of length of
        // subarrays with all 1s
        var len = 0;
 
        // variable to keep track of all subarrays
        var count = 0;
 
        // binary array of same size
        var binaryArr = Array(n).fill(0);
 
        // creating binary array
        for (i = 0; i < n; i++) {
            if (arr[i] <= x)
                binaryArr[i] = 1;
            else
                binaryArr[i] = 0;
        }
 
        // start traversing the binary array
        for (i = 0; i < n; i++) {
 
            // once we find the first 1, keep checking
            // for number of consecutive 1s
            if (binaryArr[i] == 1) {
                var j;
 
                for (j = i + 1; j < n; j++)
                    if (binaryArr[j] != 1)
                        break;
 
                // calculate length of the subarray
                // with all 1s
                len = j - i;
 
                // increment count
                count += (len) * (len + 1) / 2;
 
                // initialize i to j
                i = j;
            }
        }
 
        return count;
    }
 
    // Driver code
        var arr = [ 1, 5, 7, 8, 2, 3, 9 ];
        var x = 6;
        var n = arr.length;
 
        document.write(countSubArrays(arr, n, x));
 
// This code is contributed by Rajput-Ji

输出：

6

时间复杂度：O(n)，其中 n 是数组中元素的数量。
辅助空间：O(n)。
另一种方法：我们可以在不使用额外空间的情况下改进上述解决方案，保持时间复杂度 O(n)。我们可以跟踪每个此类区域的开始和结束，并在区域结束时更新计数，而不是将元素标记为 0 和 1。

C++

// C++ program to count all sub-arrays which
// has all elements less than or equal to X
 
#include
using namespace std;
 
int countSubArrays(int arr[], int x, int n )
    {
        int count = 0;
        int start = -1, end = -1;
     
        for(int i = 0; i < n; i++)
        {
            if(arr[i] < x)
            {
                if(start == -1)
                {
                     
                    //create a new subArray
                    start = i;
                    end = i;
                }
                else
                {
                     
                    // append to existing subarray
                    end=i;
                }
            }
            else
            {
                if(start != -1 && end != -1)
                {
                     
                    // given start and end calculate
                    // all subarrays within this range
                    int length = end - start + 1;
                    count = count + ((length * (length + 1)) / 2);
                }
                 
                start = -1;
                end = -1;
            }
 
        }
         
        if(start != -1 && end != -1)
        {
             
            // given start and end calculate all
            // subarrays within this range
            int length = end - start + 1;
            count = count + ((length * (length + 1)) / 2);
        }
         
        return count;
    }
     
    // Driver code
int main()
    {
        int arr[] = { 1, 5, 7, 8, 2, 3, 9 };
        int x = 6;
        int n = sizeof(arr) / sizeof(arr[0]);
        cout<< countSubArrays(arr, x, n);
         
//This code is contributed by  29AjayKumar
 
    }

Java

// Java program to count all sub-arrays which
// has all elements less than or equal to X
 
public class GFG {
 
    public static int countSubArrays(int arr[], int x)
    {
        int count = 0;
        int start = -1, end = -1;
         
        for(int i = 0; i < arr.length; i++)
        {
            if(arr[i] < x)
            {
                if(start == -1)
                {
                     
                    //create a new subArray
                    start = i;
                    end = i;
                }
                else
                {
                     
                    // append to existing subarray
                    end=i;
                }
            }
            else
            {
                if(start != -1 && end != -1)
                {
                     
                    // given start and end calculate
                    // all subarrays within this range
                    int length = end - start + 1;
                    count = count + ((length * (length + 1)) / 2);
                }
                 
                start = -1;
                end = -1;
            }
 
        }
         
        if(start != -1 && end != -1)
        {
             
            // given start and end calculate all
            // subarrays within this range
            int length = end - start + 1;
            count = count + ((length * (length + 1)) / 2);
        }
         
        return count;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 5, 7, 8, 2, 3, 9 };
        int x = 6;
        System.out.println(countSubArrays(arr, x));
 
    }
}

Python3

# Python3 program to count all sub-arrays which
# has all elements less than or equal to X
 
def countSubArrays(arr, x, n ):
    count = 0;
    start = -1; end = -1;
 
    for i in range(n):
        if(arr[i] < x):
            if(start == -1):
 
                # create a new subArray
                start = i;
                end = i;
            else:
                 
                # append to existing subarray
                end = i;
        else:
            if(start != -1 and end != -1):
 
                # given start and end calculate
                # all subarrays within this range
                length = end - start + 1;
                count = count + ((length *
                                 (length + 1)) / 2);
            start = -1;
            end = -1;
 
    if(start != -1 and end != -1):
 
        # given start and end calculate all
        # subarrays within this range
        length = end - start + 1;
        count = count + ((length *
                         (length + 1)) / 2);
 
    return count;
 
# Driver code
arr = [ 1, 5, 7, 8, 2, 3, 9 ];
x = 6;
n = len(arr);
print(countSubArrays(arr, x, n));
 
# This code is contributed
# by PrinciRaj1992

C＃

// C# program to count all sub-arrays which
// has all elements less than or equal to X
using System;
 
class GFG
{
 
    public static int countSubArrays(int []arr, int x)
    {
        int count = 0;
        int start = -1, end = -1;
         
        for(int i = 0; i < arr.Length; i++)
        {
            if(arr[i] < x)
            {
                if(start == -1)
                {
                     
                    //create a new subArray
                    start = i;
                    end = i;
                }
                else
                {
                     
                    // append to existing subarray
                    end=i;
                }
            }
            else
            {
                if(start != -1 && end != -1)
                {
                     
                    // given start and end calculate
                    // all subarrays within this range
                    int length = end - start + 1;
                    count = count + ((length * (length + 1)) / 2);
                }
                 
                start = -1;
                end = -1;
            }
 
        }
         
        if(start != -1 && end != -1)
        {
             
            // given start and end calculate all
            // subarrays within this range
            int length = end - start + 1;
            count = count + ((length * (length + 1)) / 2);
        }
         
        return count;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 1, 5, 7, 8, 2, 3, 9 };
        int x = 6;
        Console.WriteLine(countSubArrays(arr, x));
    }
}
 
// This code contributed by Rajput-Ji

Javascript

    // Javascript program to count all sub-arrays which
    // has all elements less than or equal to X
     
    function countSubArrays(arr, x)
    {
        let count = 0;
        let start = -1, end = -1;
          
        for(let i = 0; i < arr.length; i++)
        {
            if(arr[i] < x)
            {
                if(start == -1)
                {
                      
                    //create a new subArray
                    start = i;
                    end = i;
                }
                else
                {
                      
                    // append to existing subarray
                    end=i;
                }
            }
            else
            {
                if(start != -1 && end != -1)
                {
                      
                    // given start and end calculate
                    // all subarrays within this range
                    let length = end - start + 1;
                    count = count + parseInt((length * (length + 1)) / 2, 10);
                }
                  
                start = -1;
                end = -1;
            }
  
        }
          
        if(start != -1 && end != -1)
        {
              
            // given start and end calculate all
            // subarrays within this range
            let length = end - start + 1;
            count = count + parseInt((length * (length + 1)) / 2, 10);
        }
          
        return count;
    }
     
    let arr = [ 1, 5, 7, 8, 2, 3, 9 ];
    let x = 6;
    document.write(countSubArrays(arr, x));
     

输出：

6

时间复杂度： O(n)，其中 n 是数组中元素的数量。
辅助空间： O(1)。