Count sub-arrays which have elements less than or equal to X

Given an array of n elements and an integer X. Count the number of sub-arrays of this array which have all elements less than or equal to X.

Examples:

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach : A simple approach uses two nested loops for generating all sub-arrays of the given an array and a loop to check whether all elements of a sub-array is less than or equal to X or not.

Time Complexity: O(n*n*n)

Efficient Approach: An efficient approach is to observe that we just want the count of those sub-arrays which have all elements less than or equal to X. We can create a binary array of 0s and 1s corresponding to the original array. If an element in the original is less than or equal to X, then the corresponding element in the binary array will be 1 otherwise 0. Now, our problem reduces to count the number of sub-arrays in this binary array which has all 1s. We can also see that for an array which has all 1s all of its sub-arrays will have only 1s and the total number of sub-arrays will be len*(len+1)/2. For example, {1, 1, 1, 1} will have 10 sub-arrays.

Below is the complete algorithm to solve the above problem:

• Create a corresponding binary array of the original array as described above.
• Initialize a counter variable to 0 and start traversing the binary array keeping track of the lengths of sub-arrays which has all 1s
• We can easily calculate the number of sub-arrays of an array which has all 1s by using the formula n*(n+1)/2, where n is the length of the array with all 1s.
• Calculate the length of every sub-array which has all 1s and increment the count variable by length*(length+1)/2. We can do this in O(n) time complexity

Below is the implementation of above approach:

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);     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.

PHP



Output:

6

Time Complexity: O(n), where n is the number of elements in the array.
Auxiliary Space: O(n).

Another Method: We can improve the above solution without using extra space keeping the time complexity O(n). Instead of marking elements as 0 and 1 we can keep track of start and end of each such region and update the count whenever the region ends.

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);         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

Output:

6

Time Complexity: O(n), where n is the number of elements in the array.
Auxiliary Space: O(1).

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