Count subarrays made up of single-digit integers only

Given an array arr[] consisting of N positive integers, the task is to count subarrays consisting of single-digit elements only.

Examples:

Input: arr[] = {0, 1, 14, 2, 5}
Output: 6
Explanation: All subarrays made of only single digit numbers are {{0}, {1}, {2}, {5}, {0, 1}, {2, 5}}. Therefore, the total count of subarrays is 6.

Input: arr[] ={12, 5, 14, 17}
Output: 1
Explanation: All subarrays made of only single digit numbers are {5}.
Therefore, the total count of subarrays is 1.

Naive Approach: The simplest approach is to traverse the array and generate all possible subarrays. For each subarray, check if all integers in it are single-digit integers or not. 

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to find the size of each block of contiguous single-digit integers and increment the count by total subarrays of that length. Follow the steps below to solve the problem:

• Initialize a variable, say res = 0 and c = 0, to store the total count of subarrays and the total count of single-digit integers in a subarray.
• Traverse the array and perform the following operations:
  • If arr[i] < 10, increment the count of c by one and count of res by c.
  • Otherwise, assign c = 0.
• Finally, print the total count of single-digit integer subarrays.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count of subarrays made
// up of single digit integers only
int singleDigitSubarrayCount(int arr[],
                             int N)
{
    // Stores count of subarrays
    int res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        if (arr[i] <= 9) {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }

        else {

            // Assign count = 0
            count = 0;
        }
    }

    cout << res;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 0, 1, 14, 2, 5 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
    singleDigitSubarrayCount(arr, N);

    return 0;
}

Java

// Java program for the above approach
class GFG
{

// Function to count of subarrays made
// up of single digit integers only
static void singleDigitSubarrayCount(int arr[],
                             int N)
{

    // Stores count of subarrays
    int res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;

    // Traverse the array
    for (int i = 0; i < N; i++)
    {
        if (arr[i] <= 9)
        {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }

        else
        {

            // Assign count = 0
            count = 0;
        }
    }
    System.out.print(res);
}

// Driver Code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 0, 1, 14, 2, 5 };

    // Size of the array
    int N = arr.length;
    singleDigitSubarrayCount(arr, N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to count of subarrays made
# up of single digit integers only
def singleDigitSubarrayCount(arr, N):

    # Stores count of subarrays
    res = 0

    # Stores the count of consecutive
    # single digit numbers in the array
    count = 0

    # Traverse the array
    for i in range(N):
        if (arr[i] <= 9):

            # Increment size of block by 1
            count += 1

            # Increment res by count
            res += count
        else:
            # Assign count = 0
            count = 0
    print (res)

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [0, 1, 14, 2, 5]

    # Size of the array
    N = len(arr)
    singleDigitSubarrayCount(arr, N)

    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG{

// Function to count of subarrays made
// up of single digit integers only
static void singleDigitSubarrayCount(int[] arr,
                             int N)
{

    // Stores count of subarrays
    int res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;

    // Traverse the array
    for (int i = 0; i < N; i++)
    {
        if (arr[i] <= 9)
        {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }
        else
        {

            // Assign count = 0
            count = 0;
        }
    }
    Console.Write(res);
}

// Driver Code
public static void Main(string[] args)
{
    // Given array
    int[] arr = { 0, 1, 14, 2, 5 };

    // Size of the array
    int N = arr.Length;
    singleDigitSubarrayCount(arr, N);
}
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// Javascript program for the above approach

// Function to count of subarrays made
// up of single digit integers only
function singleDigitSubarrayCount(arr, N)
{

    // Stores count of subarrays
    let res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    let count = 0;

    // Traverse the array
    for(let i = 0; i < N; i++)
    {
        if (arr[i] <= 9)
        {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }
        else
        {

            // Assign count = 0
            count = 0;
        }
    }
    document.write(res);
}

// Driver Code

// Given array
let arr = [ 0, 1, 14, 2, 5 ];

// Size of the array
let N = arr.length;

singleDigitSubarrayCount(arr, N);

// This code is contributed by Manoj.

</script>

Output:

6

Time Complexity: O(N)
Auxiliary Space: O(1)