Find number of integers that can be obtained by replacing ? in given string for perfect divisibility by 9

Given a String S of length N. S contains 'X' at some places. Then your task is to output the number distinct of integers that can be formed by replacing 'X' from the range [0, 9] and the formed integer leaves the remainder 0 after dividing them by 9.

Note: Numbers with leading 0s are not allowed to form.

Examples:

Input: N = 2, S = XX
Output: 10
Explanation: There can be 10 possible integers, If we use the digits from the range [0, 9] at both places of X. The 10 numbers will be: {18, 27, 36, 45, 54, 63, 72, 81, 90, 99}. They all leave remainder 0 when dividing by 9.

Input: N = 2, S = 9X
Output: 2
Explanation: It can be verified that there will be only two possible such integers. Which are 90 and 99.

Approach: Implement the idea below to solve the problem

The problem is observation based. Let us observe it.

There are 3 cases to observe the problem:

• Case 1: If sum of all digit characters is divisible by 9. Then:
  • Example 1: 2X7X (Count of X = 2)
    • Possible Integers = 2070, 2079, 2178, 2277, 2376, 2475, 2574, 2673, 2772, 2871, 2970, 2979
    • Total: 12
  • Example 2: 2X7 (Count of X = 1)
    • Possible integers = 207 and 297
    • Total Possibilities = 2
  • Then the count of such integers is: (Count('X')-1) 1's followed by 2.
• Case 2: If sum of all digit characters is not divisible by 9
  • Example 1: 2X5X (Count of X = 2)
    • Possible Integers = 2052, 2151, 2250, 2259, 2358, 2457, 2556, 2655, 2754, 2853, 2952
    • Total: 11
  • Example 2: 2X5 (Count of X = 1)
    • Possible Integers = 225
    • Total possibilities = 1
  • Then the count of such possible integers is: (Count('X')) 1s.
• Case 3: If first character of string is 'X'
  • Please note that there can't be leading 0s.
  • Example 1: X27X (Count of X = 2)
    • Possible Integers = 1278, 2277, 3276, 4275, 5274, 6273, 7272, 8271, 9270, 9279
    • Total Possibilities = 10
  • Example 2: X5X (Count of X = 2)
    • Possible Integers = 153, 252, 351, 450, 459, 558, 657, 756, 855, 954
    • Total possibilities = 10
  • Then the count of such possible integers is: 1 then followed by (Count('X') - 1) 0's.

Step-by-step approach:

• Create two variables let say sum and q to store the sum of integers in s and count of X respectively.
• Run a loop on S and count X and calculate Sum
• If (First Character is X)
  • Output 1 then (q-1) times 0s
• If (Sum is divisible by 9)
  • Output 1 q times
• Else
  • Output 1 q-2 times and then 2 once.

Below is the implementation of the above approach:

#include <iostream>
#include <string>

using namespace std;

// Method to output the number of such integers
void Count_integers(int N, const string& S)
{

    // sum = To count the sum of integers except X
    // q = Count of X
    int sum = 0, q = 0;

    // Loop for initializing value of Q and Sum
    for (int i = 0; i < N; i++) {
        if (S[i] == 'X')
            q++;
        else {
            sum += (S[i] - '0');
        }
    }

    // Printing count of integers
    // according to discussed patterns
    if (S[0] == 'X') {
        cout << "1";
        for (int i = 1; i < q; i++)
            cout << "0";
    }
    else if (sum % 9 != 0) {
        for (int i = 0; i < q; i++)
            cout << "1";
    }
    else {
        for (int i = 0; i < q - 1; i++)
            cout << "1";
        cout << "2";
    }
}

// Driver Function
int main()
{
    // Inputs
    int N = 4;
    string S = "XX90";

    // Function call
    Count_integers(N, S);

    return 0;
}

// Java code to implement the approach

import java.util.*;

// Driver Class
public class GFG {

    // Driver Function
    public static void main(String[] args)
    {
        // Inputs
        int N = 4;
        String S = "XX90";

        // Function call
        Count_integers(N, S);
    }

    // Method to output the number of such integers
    public static void Count_integers(int N, String S)
    {

        // sum = To count the sum of integers except X
        // q = Count of X
        int sum = 0, q = 0;

        // Loop for initializing value of Q and Sum
        for (int i = 0; i < N; i++) {
            if (S.charAt(i) == 'X')
                q++;
            else {
                sum += (S.charAt(i) - '0');
            }
        }

        // Printing count of integers
        // according to discussed patterns
        if (S.charAt(0) == 'X') {
            System.out.print("1");
            for (int i = 1; i < q; i++)
                System.out.print("0");
        }
        else if (sum % 9 != 0) {
            for (int i = 0; i < q; i++)
                System.out.print("1");
        }
        else {
            for (int i = 0; i < q - 1; i++)
                System.out.print("1");
            System.out.print("2");
        }
    }
}

# Python code for the above approach

# Method to output the number of such integers
def Count_integers(N, S):
    # sum = To count the sum of integers except X
    # q = Count of X
    sum, q = 0, 0

    # Loop for initializing value of Q and Sum
    for i in range(N):
        if (S[i] == 'X'):
            q += 1
        else:
            sum += ord(S[i])

    # Printing count of integers
    # according to discussed patterns
    if (S[0] == 'X'):
        print("1", end="")
        for i in range(1, q):
            print("0", end="")
    elif (sum % 9 != 0):
        for i in range(q):
            print("1", end="")
    else:
        for i in range(q - 1):
            print("1", end="")
        print("2")

# Driver function
def main():
    # Inputs
    N = 4
    S = "XX90"
    # Function call
    Count_integers(N, S)


# Driver call
if __name__ == '__main__':
    main()

# This code is contributed by ragul21

using System;

// Driver Class
class GFG
{
    // Driver Function
    static void Main(string[] args)
    {
        // Inputs
        int N = 4;
        string S = "XX90";

        // Function call
        CountIntegers(N, S);
    }

    // Method to output the number of such integers
    static void CountIntegers(int N, string S)
    {
        // sum = To count the sum of integers except X
        // q = Count of X
        int sum = 0, q = 0;

        // Loop for initializing value of Q and Sum
        for (int i = 0; i < N; i++)
        {
            if (S[i] == 'X')
                q++;
            else
            {
                sum += (S[i] - '0');
            }
        }

        // Printing count of integers
        // according to discussed patterns
        if (S[0] == 'X')
        {
            Console.Write("1");
            for (int i = 1; i < q; i++)
                Console.Write("0");
        }
        else if (sum % 9 != 0)
        {
            for (int i = 0; i < q; i++)
                Console.Write("1");
        }
        else
        {
            for (int i = 0; i < q - 1; i++)
                Console.Write("1");
            Console.Write("2");
        }
    }
}

// JS Implementation
// Method to output the number of such integers
function countIntegers(N, S) {
    // sum = To count the sum of integers except X
    // q = Count of X
    let sum = 0, q = 0;

    // Loop for initializing value of Q and Sum
    for (let i = 0; i < N; i++) {
        if (S[i] === 'X') {
            q++;
        } else {
            sum += parseInt(S[i]); // Convert character to integer and add to sum
        }
    }

    // Printing count of integers according to discussed patterns
    if (S[0] === 'X') {
        process.stdout.write("1");
        for (let i = 1; i < q; i++) {
            process.stdout.write("0");
        }
    } else if (sum % 9 !== 0) {
        for (let i = 0; i < q; i++) {
            process.stdout.write("1");
        }
    } else {
        for (let i = 0; i < q - 1; i++) {
            process.stdout.write("1");
        }
        process.stdout.write("2");
    }
}

// Driver Function

// Inputs
const N = 4;
const S = "XX90";

// Function call
countIntegers(N, S);

// This code is contributed by Sakshi

10

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