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Find the missing digit in given product of large positive integers

  • Difficulty Level : Hard
  • Last Updated : 14 Sep, 2021

Given two large integers in form of strings A and B and their product also in form of string C such that one digit of the product is replaced with X, the task is to find the replaced digit in the product C.

Examples:

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Input: A = 51840, B = 273581, C = 1418243×040
Output: 9
Explanation:
The product of integer A and B  is 51840 * 273581 = 14182439040. On comparing it with C, it can be concluded that the replaced digit was 9. Therefore, print 9.

Input: A = 123456789, B = 987654321, C = 12193263111×635269
Output: 2



Naive Approach: The simplest approach to solve the given problem is to find the product of two large integers A and B using the approach discussed in this article and then comparing it with C to find the resultant missing digit X.

Time Complexity: O((log10A)*(log10B))
Auxiliary Space: O(log10A + log10B)

Efficient Approach: The above approach can also be optimized by using the following observations:

Suppose N is an integer such that N = amam-1am-2 …….a2a1a0 where ax represents the xth digit. Now, N can be represented as:
=> N = am * 10m + am-1 * 10m-1 + ………… + a1 * 10 + a0

Performing the modulo operartion with 11 in the above equation:

=> N (mod 11) = am * 10m + am-1 * 10m-1 + ………… + a1 * 10 + a0 (mod 11)
=> N (mod 11) = am(-1)m + am-1(-1)m-1 + ……….. + a1(-1) + a0   (mod 11)   [Since 10 ≡ -1 (mod 11)]
=> N (mod 11) = T (mod 11) where T = a0 a1 + a2 …… + (-1)mam

Therefore, from the above equation, A * B = C can be converted into (A % 11) * (B % 11) = (C % 11) where the left-hand side of the equation is a constant value and the right-hand side will be an equation in 1 variable X which can be solved to find the value of X. There might be the possibility of X can have a negative value after performing modulo with 11, for that case consider the positive value of X.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the replaced digit
// in the product of a*b
int findMissingDigit(string a, string b,
                     string c)
{
    // Keeps track of the sign of the
    // current digit
    int w = 1;
 
    // Stores the value of a % 11
    int a_mod_11 = 0;
 
    // Find the value of a mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = a.size() - 1; i >= 0; i--) {
        a_mod_11 = (a_mod_11 + w * (a[i] - '0')) % 11;
        w = w * -1;
    }
 
    // Stores the value of b % 11
    int b_mod_11 = 0;
    w = 1;
 
    // Find the value of b mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = b.size() - 1;
         i >= 0; i--) {
 
        b_mod_11 = (b_mod_11
                    + w * (b[i] - '0'))
                   % 11;
        w = w * -1;
    }
 
    // Stores the value of c % 11
    int c_mod_11 = 0;
 
    // Keeps track of the sign of x
    bool xSignIsPositive = true;
    w = 1;
 
    for (int i = c.size() - 1; i >= 0; i--) {
 
        // If the current digit is the
        // missing digit, then keep
        // the track of its sign
        if (c[i] == 'x') {
            xSignIsPositive = (w == 1);
        }
        else {
            c_mod_11 = (c_mod_11
                        + w * (c[i] - '0'))
                       % 11;
        }
        w = w * -1;
    }
 
    // Find the value of x using
    // the derived equation
    int x = ((a_mod_11 * b_mod_11)
             - c_mod_11)
            % 11;
 
    // Check if x has a negative sign
    if (!xSignIsPositive) {
        x = -x;
    }
 
    // Return positive equivaluent
    // of x mod 11
    return (x % 11 + 11) % 11;
}
 
// Driver Code
int main()
{
    string A = "123456789";
    string B = "987654321";
    string C = "12193263111x635269";
    cout << findMissingDigit(A, B, C);
 
    return 0;
}

Java




// Java program for the above approach
class GFG {
 
    // Function to find the replaced digit
    // in the product of a*b
    public static int findMissingDigit(String a, String b, String c)
    {
       
        // Keeps track of the sign of the
        // current digit
        int w = 1;
 
        // Stores the value of a % 11
        int a_mod_11 = 0;
 
        // Find the value of a mod 11 for
        // large value of a as per the
        // derived formula
        for (int i = a.length() - 1; i >= 0; i--) {
            a_mod_11 = (a_mod_11 + w * (a.charAt(i) - '0')) % 11;
            w = w * -1;
        }
 
        // Stores the value of b % 11
        int b_mod_11 = 0;
        w = 1;
 
        // Find the value of b mod 11 for
        // large value of a as per the
        // derived formula
        for (int i = b.length() - 1; i >= 0; i--) {
 
            b_mod_11 = (b_mod_11 + w * (b.charAt(i) - '0')) % 11;
            w = w * -1;
        }
 
        // Stores the value of c % 11
        int c_mod_11 = 0;
 
        // Keeps track of the sign of x
        boolean xSignIsPositive = true;
        w = 1;
 
        for (int i = c.length() - 1; i >= 0; i--) {
 
            // If the current digit is the
            // missing digit, then keep
            // the track of its sign
            if (c.charAt(i) == 'x') {
                xSignIsPositive = (w == 1);
            } else {
                c_mod_11 = (c_mod_11 + w * (c.charAt(i) - '0')) % 11;
            }
            w = w * -1;
        }
 
        // Find the value of x using
        // the derived equation
        int x = ((a_mod_11 * b_mod_11) - c_mod_11) % 11;
 
        // Check if x has a negative sign
        if (!xSignIsPositive) {
            x = -x;
        }
 
        // Return positive equivaluent
        // of x mod 11
        return (x % 11 + 11) % 11;
    }
 
    // Driver Code
    public static void main(String args[]) {
        String A = "123456789";
        String B = "987654321";
        String C = "12193263111x635269";
        System.out.println(findMissingDigit(A, B, C));
 
    }
}
 
// This code is contributed by saurabh_jaiswal.

Python3




# Python3 Program to implement the above approach
 
# Function to find the replaced digit
# in the product of a*b
def findMissingDigit(a, b, c):
   
    # Keeps track of the sign of the
    # current digit
    w = 1
 
    # Stores the value of a % 11
    a_mod_11 = 0
 
    # Find the value of a mod 11 for
    # large value of a as per the
    # derived formula
    for i in range(len(a) - 1, -1, -1):
        a_mod_11 = (a_mod_11 + w * (ord(a[i]) - ord('0'))) % 11
        w = w * -1
 
    # Stores the value of b % 11
    b_mod_11 = 0
    w = 1
 
    # Find the value of b mod 11 for
    # large value of a as per the
    # derived formula
    for i in range(len(b) - 1, -1, -1):
        b_mod_11 = (b_mod_11 + w * (ord(b[i]) - ord('0'))) % 11
        w = w * -1
 
    # Stores the value of c % 11
    c_mod_11 = 0
 
    # Keeps track of the sign of x
    xSignIsPositive = True
    w = 1
 
    for i in range(len(c) - 1, -1, -1):
        # If the current digit is the
        # missing digit, then keep
        # the track of its sign
        if (c[i] == 'x'):
            xSignIsPositive = (w == 1)
        else:
            c_mod_11 = (c_mod_11 + w * (ord(c[i]) - ord('0'))) % 11
        w = w * -1
 
    # Find the value of x using
    # the derived equation
    x = ((a_mod_11 * b_mod_11) - c_mod_11) % 11
 
    # Check if x has a negative sign
    if (not xSignIsPositive):
        x = -x
 
    # Return positive equivaluent
    # of x mod 11
    return (x % 11 + 11) % 11
 
A = "123456789"
B = "987654321"
C = "12193263111x635269"
print(findMissingDigit(A, B, C))
 
# This code is contributed by divyeshrabadiya07.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the replaced digit
// in the product of a*b
static int findMissingDigit(string a, string b,
                     string c)
{
   
    // Keeps track of the sign of the
    // current digit
    int w = 1;
 
    // Stores the value of a % 11
    int a_mod_11 = 0;
 
    // Find the value of a mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = a.Length - 1; i >= 0; i--) {
        a_mod_11 = (a_mod_11 + w * ((int)a[i] - 48)) % 11;
        w = w * -1;
    }
 
    // Stores the value of b % 11
    int b_mod_11 = 0;
    w = 1;
 
    // Find the value of b mod 11 for
    // large value of a as per the
    // derived formula
    for (int i = b.Length - 1;
         i >= 0; i--) {
 
        b_mod_11 = (b_mod_11
                    + w * ((int)b[i] - 48))
                   % 11;
        w = w * -1;
    }
 
    // Stores the value of c % 11
    int c_mod_11 = 0;
 
    // Keeps track of the sign of x
    bool xSignIsPositive = true;
    w = 1;
 
    for (int i = c.Length - 1; i >= 0; i--) {
 
        // If the current digit is the
        // missing digit, then keep
        // the track of its sign
        if (c[i] == 'x') {
            xSignIsPositive = (w == 1);
        }
        else {
            c_mod_11 = (c_mod_11
                        + w * ((int)c[i] - '0'))
                       % 11;
        }
        w = w * -1;
    }
 
    // Find the value of x using
    // the derived equation
    int x = ((a_mod_11 * b_mod_11)
             - c_mod_11)
            % 11;
 
    // Check if x has a negative sign
    if (xSignIsPositive == false) {
        x = -x;
    }
 
    // Return positive equivaluent
    // of x mod 11
    return (x % 11 + 11) % 11;
}
 
// Driver Code
public static void Main()
{
    string A = "123456789";
    string B = "987654321";
    string C = "12193263111x635269";
    Console.Write(findMissingDigit(A, B, C));
}
}
 
// This code is contributed by ipg2016107.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the replaced digit
        // in the product of a*b
        function findMissingDigit(a, b,
            c)
        {
            // Keeps track of the sign of the
            // current digit
            let w = 1;
 
            // Stores the value of a % 11
            let a_mod_11 = 0;
 
            // Find the value of a mod 11 for
            // large value of a as per the
            // derived formula
            for (let i = a.length - 1; i >= 0; i--) {
                a_mod_11 = (a_mod_11 + w * (a[i].charCodeAt(0) - '0'.charCodeAt(0))) % 11;
                w = w * -1;
            }
 
            // Stores the value of b % 11
            let b_mod_11 = 0;
            w = 1;
 
            // Find the value of b mod 11 for
            // large value of a as per the
            // derived formula
            for (let i = b.length - 1;
                i >= 0; i--) {
 
                b_mod_11 = (b_mod_11
                    + w * (b[i].charCodeAt(0) - '0'.charCodeAt(0)))
                    % 11;
                w = w * -1;
            }
 
            // Stores the value of c % 11
            let c_mod_11 = 0;
 
            // Keeps track of the sign of x
            let xSignIsPositive = true;
            w = 1;
 
            for (let i = c.length - 1; i >= 0; i--) {
 
                // If the current digit is the
                // missing digit, then keep
                // the track of its sign
                if (c[i] == 'x') {
                    xSignIsPositive = (w == 1);
                }
                else {
                    c_mod_11 = (c_mod_11
                        + w * (c[i].charCodeAt(0) - '0'.charCodeAt(0)))
                        % 11;
                }
                w = w * -1;
            }
 
            // Find the value of x using
            // the derived equation
            let x = ((a_mod_11 * b_mod_11)
                - c_mod_11)
                % 11;
 
            // Check if x has a negative sign
            if (!xSignIsPositive) {
                x = -x;
            }
 
            // Return positive equivaluent
            // of x mod 11
            return (x % 11 + 11) % 11;
        }
 
        // Driver Code
        let A = "123456789";
        let B = "987654321";
        let C = "12193263111x635269";
        document.write(findMissingDigit(A, B, C));
 
// This code is contributed by Potta Lokesh
</script>
Output: 
2

 

Time Complexity: O(log10A + log10B)
Auxiliary Space: O(1)




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