Count of sub-sequences which satisfy the given condition

Given a string str consisting of digits, the task is to find the number of possible 4 digit sub-sequences which are of the form (x, x, x + 1, x + 1) where x can be from the range [0, 8]. 
Examples:

Input: str = “1122” 
Output: 1 
Only one sub-sequence is valid, i.e the entire string itself.

Input: str = “13134422” 
Output: 2 
Two Valid sub-sequences are present “1122” and “3344”. 

Approach:

• We will find out total number of possible sub-sequences for each possible x from 0 to 8.
• For each x, remove all other digits from the String, except x and x+1 as they do not affect the answer.
• Maintain a prefix Sum array to count the number of x+1 digits till i th index in the String.
• Now, for every club of digits say size K (which are x), we can choose two numbers in KC2 ways. Last two numbers can be any two numbers from all the digits (which are x+1) which follows that club of digits (count is determined using Prefix Sum Array) say size L, so there are LC2 ways to choose. Total Ways = KC2 * LC2 .
• Till, Now we can be considered x to come from the same club, but it can also be from multiple Clubs. So, we have to consider all possible pairs of clubs and multiply their size to get number of ways to choose first two numbers. For last two numbers, ways will remain same.
• In order to prevent the problem of over counting in Step 5. Only Possible way which includes the current club under consideration will be chosen as other have already been considered in calculation of previous clubs.
• Add all the ways possible for all the values of x and take Modulo.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#define ll long long int
#define MOD 1000000007
using namespace std;

// Function to return the total
// required sub-sequences
int solve(string test)
{
    int size = test.size();
    int total = 0;

    // Find ways for all values of x
    for (int i = 0; i <= 8; i++) {
        int x = i;

        // x+1
        int y = i + 1;
        string newtest;

        // Removing all unnecessary digits
        for (int j = 0; j < size; j++) {
            if (test[j] == x + 48 || test[j] == y + 48) {
                newtest += test[j];
            }
        }

        if (newtest.size() > 0) {
            int size1 = newtest.size();

            // Prefix Sum Array for X+1 digit
            int prefix[size1] = { 0 };
            for (int j = 0; j < size1; j++) {
                if (newtest[j] == y + 48) {
                    prefix[j]++;
                }
            }

            for (int j = 1; j < size1; j++) {
                prefix[j] += prefix[j - 1];
            }

            int count = 0;
            int firstcount = 0;

            // Sum of squares
            int ss = 0;

            // Previous sum of all possible pairs
            int prev = 0;

            for (int j = 0; j < size1; j++) {
                if (newtest[j] == x + 48) {
                    count++;
                    firstcount++;
                }
                else {

                    ss += count * count;

                    // To find sum of multiplication of all
                    // possible pairs
                    int pairsum
                        = (firstcount * firstcount - ss)
                          / 2;
                    int temp = pairsum;

                    // To prevent overcounting
                    pairsum -= prev;
                    prev = temp;

                    int secondway = prefix[size1 - 1];
                    if (j != 0)
                        secondway -= prefix[j - 1];

                    int answer = count * (count - 1)
                                 * secondway
                                 * (secondway - 1);
                    answer /= 4;
                    answer += (pairsum * secondway
                               * (secondway - 1))
                              / 2;

                    // Adding ways for all possible x
                    total += answer;
                    count = 0;
                }
            }
        }
    }

    return total;
}

// Driver code
int main()
{
    string test = "13134422";
    cout << solve(test) << endl;

    return 0;
}

Java

// Java Implementation of above approach
import java.io.*;

class GFG {

    // Function to return the total
    // required sub-sequences
    static int solve(String test, int MOD)
    {
        int size = test.length();
        int total = 0;

        // Find ways for all values of x
        for (int i = 0; i <= 8; i++) {
            int x = i;

            // x+1
            int y = i + 1;
            String newtest = "";

            // Removing all unnecessary digits
            for (int j = 0; j < size; j++) {
                if (test.charAt(j) == x + 48
                    || test.charAt(j) == y + 48) {
                    newtest += test.charAt(j);
                }
            }

            if (newtest.length() > 0) {
                int size1 = newtest.length();

                // Prefix Sum Array for X+1 digit
                int[] prefix = new int[size1];
                for (int j = 0; j < size1; j++) {
                    prefix[j] = 0;
                    if (newtest.charAt(j) == y + 48) {
                        prefix[j]++;
                    }
                }

                for (int j = 1; j < size1; j++) {
                    prefix[j] += prefix[j - 1];
                }

                int count = 0;
                int firstcount = 0;

                // Sum of squares
                int ss = 0;

                // Previous sum of all possible pairs
                int prev = 0;

                for (int j = 0; j < size1; j++) {
                    if (newtest.charAt(j) == x + 48) {
                        count++;
                        firstcount++;
                    }
                    else {

                        ss += count * count;

                        // To find sum of multiplication of
                        // all possible pairs
                        int pairsum
                            = (firstcount * firstcount - ss)
                              / 2;
                        int temp = pairsum;

                        // To prevent overcounting
                        pairsum -= prev;
                        prev = temp;

                        int secondway = prefix[size1 - 1];
                        if (j != 0)
                            secondway -= prefix[j - 1];

                        int answer = count * (count - 1)
                                     * secondway
                                     * (secondway - 1);
                        answer /= 4;
                        answer += (pairsum * secondway
                                   * (secondway - 1))
                                  / 2;

                        // Adding ways for all possible x
                        total += answer;
                        count = 0;
                    }
                }
            }
        }

        return total;
    }

    // Driver code
    public static void main(String[] args)
    {
        String test = "13134422";
        int MOD = 1000000007;
        System.out.println(solve(test, MOD));
    }
}

// This code is contributed by krikti..

Python3

# Python3 implementation of the approach

MOD = 1000000007

# Function to return the total
# required sub-sequences


def solve(test):

    size = len(test)
    total = 0

    # Find ways for all values of x
    for i in range(9):
        x = i

        # x+1
        y = i + 1
        newtest = ""

        # Removing all unnecessary digits
        for j in range(size):
            if (ord(test[j]) == x + 48 or ord(test[j]) == y + 48):
                newtest += test[j]

        if (len(newtest) > 0):
            size1 = len(newtest)

            # Prefix Sum Array for X+1 digit
            prefix = [0 for i in range(size1)]

            for j in range(size1):
                if (ord(newtest[j]) == y + 48):
                    prefix[j] += 1

            for j in range(1, size1):
                prefix[j] += prefix[j - 1]

            count = 0
            firstcount = 0

            # Sum of squares
            ss = 0

            # Previous sum of all possible pairs
            prev = 0

            for j in range(size1):
                if (ord(newtest[j]) == x + 48):
                    count += 1
                    firstcount += 1

                else:

                    ss += count * count

                    # To find sum of multiplication of all
                    # possible pairs
                    pairsum = (firstcount * firstcount - ss) // 2
                    temp = pairsum

                    # To prevent overcounting
                    pairsum -= prev
                    prev = temp

                    secondway = prefix[size1 - 1]
                    if (j != 0):
                        secondway -= prefix[j - 1]

                    answer = count * (count - 1) * secondway * (secondway - 1)
                    answer //= 4
                    answer += (pairsum * secondway * (secondway - 1)) // 2

                    # Adding ways for all possible x
                    total += answer
                    count = 0

    return total


# Driver code
test = "13134422"
print(solve(test))

# This code is contributed by mohit kumar 29

C#

// C# Implementation of above approach

using System;

class GFG {

    // Function to return the total
    // required sub-sequences
    static int solve(string test, int MOD)
    {
        int size = test.Length;
        int total = 0;

        // Find ways for all values of x
        for (int i = 0; i <= 8; i++) {
            int x = i;

            // x+1
            int y = i + 1;
            string newtest = "";

            // Removing all unnecessary digits
            for (int j = 0; j < size; j++) {
                if (test[j] == x + 48
                    || test[j] == y + 48) {
                    newtest += test[j];
                }
            }

            if (newtest.Length > 0) {
                int size1 = newtest.Length;

                // Prefix Sum Array for X+1 digit
                int[] prefix = new int[size1];
                for (int j = 0; j < size1; j++) {
                    prefix[j] = 0;
                    if (newtest[j] == y + 48) {
                        prefix[j]++;
                    }
                }

                for (int j = 1; j < size1; j++) {
                    prefix[j] += prefix[j - 1];
                }

                int count = 0;
                int firstcount = 0;

                // Sum of squares
                int ss = 0;

                // Previous sum of all possible pairs
                int prev = 0;

                for (int j = 0; j < size1; j++) {
                    if (newtest[j] == x + 48) {
                        count++;
                        firstcount++;
                    }
                    else {

                        ss += count * count;

                        // To find sum of multiplication of
                        // all possible pairs
                        int pairsum
                            = (firstcount * firstcount - ss)
                              / 2;
                        int temp = pairsum;

                        // To prevent overcounting
                        pairsum -= prev;
                        prev = temp;

                        int secondway = prefix[size1 - 1];
                        if (j != 0)
                            secondway -= prefix[j - 1];

                        int answer = count * (count - 1)
                                     * secondway
                                     * (secondway - 1);
                        answer /= 4;
                        answer += (pairsum * secondway
                                   * (secondway - 1))
                                  / 2;

                        // Adding ways for all possible x
                        total += answer;
                        count = 0;
                    }
                }
            }
        }

        return total;
    }

    // Driver code
    public static void Main()
    {
        string test = "13134422";
        int MOD = 1000000007;
        Console.WriteLine(solve(test, MOD));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// JavaScript Implementation of above approach

    // Function to return the total
   // required sub-sequences
    function solve(test,MOD)
    {
        let size = test.length;
    let total = 0;

    // Find ways for all values of x
    for (let i = 0; i <= 8; i++)
    {
        let x = i;

        // x+1
        let y = i + 1;
        let newtest = "";

        // Removing all unnecessary digits
        for (let j = 0; j < size; j++)
        {
            if (test[j].charCodeAt(0) == x + 48 ||
                test[j].charCodeAt(0) == y + 48)
            {
                newtest += test[j];
            }
        }

        if (newtest.length > 0) {
            let size1 = newtest.length;

            // Prefix Sum Array for X+1 digit
            let prefix = new Array(size1);
            for (let j = 0; j < size1; j++)
            {
                prefix[j] = 0;
                if (newtest[j].charCodeAt(0) == y + 48)
                {
                    prefix[j]++;
                }
            }

            for (let j = 1; j < size1; j++)
            {
                prefix[j] += prefix[j - 1];
            }

            let count = 0;
            let firstcount = 0;

            // Sum of squares
            let ss = 0;

            // Previous sum of all possible pairs
            let prev = 0;

            for (let j = 0; j < size1; j++)
            {
                if (newtest[j].charCodeAt(0) == x + 48)
                {
                    count++;
                    firstcount++;
                }
                else
                {

                    ss += count * count;

                    // To find sum of multiplication of all
                    // possible pairs
                    let pairsum =
            Math.floor((firstcount * firstcount - ss) / 2);
                    let temp = pairsum;

                    // To prevent overcounting
                    pairsum -= prev;
                    prev = temp;

                    let secondway = prefix[size1 - 1];
                    if (j != 0)
                        secondway -= prefix[j - 1];

                    let answer = count * (count - 1)
                                * secondway * (secondway - 1);
                    answer = Math.floor(answer/4);
                    answer += Math.floor((pairsum * secondway
                            * (secondway - 1)) / 2);

                    // Adding ways for all possible x
                    total += answer;
                    count = 0;
                }
            }
        }
    }

    return total;
    }

    // Driver code
    let test = "13134422";
    let MOD = 1000000007;
    document.write(solve(test,MOD));


// This code is contributed by unknown2108

</script>

Output:

2

Time complexity: O(N)
Auxiliary space: O(N)