Subtraction of two large numbers using 9’s complement

Last Updated : 09 Aug, 2021

Given two strings str1 and str2 of given lengths N and M respectively, each representing a large number, the task is to subtract one from the other using 9’s complement.
Examples:

Input: N = 17, str1 = “12345678987654321”, M = 11, str2 = “22324324343”
Output: 12345656663329978
Input: N = 20, Str1 = “12345334233242431433”, M = 20, Str2 = “12345334233242431432”
Output: 1

Approach:
The basic idea is similar to Subtraction of two numbers using 2’s complement.

Subtraction of given strings can be written as
Str1 – Str2 = Str1 + (- Str2) = Str1 + (9’s complement of Str2)

Follow the steps below to solve the problem:

Compare the lengths of the two strings and store the smaller of the two in str2.
Calculate 9’s complement of str2.
Now, add 9’s complement of str2 to str1.
If any carry is generated, insert at the beginning of str1.
If no carry is generated, then the complement of str1 is the final answer.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return sum of two
// large numbers given as strings
string sumBig(string a, string b)
{
 
    // Compare their lengths
    if (a.length() > b.length())
        swap(a, b);
 
    // Stores the result
    string str = "";
 
    // Store the respective lengths
    int n1 = a.length(),
        n2 = b.length();
 
    int diff = n2 - n1;
 
    // Initialize carry
    int carry = 0;
 
    // Traverse from end of both strings
    for (int i = n1 - 1; i >= 0; i--) {
        // Compute sum of
        // current digits and carry
        int sum = ((a[i] - '0')
                   + (b[i + diff] - '0')
                   + carry);
 
        // Store the result
        str.push_back(sum % 10 + '0');
 
        // Update carry
        carry = sum / 10;
    }
 
    // Add remaining digits of str2[]
    for (int i = n2 - n1 - 1; i >= 0; i--) {
 
        int sum = ((b[i] - '0') + carry);
 
        str.push_back(sum % 10 + '0');
        carry = sum / 10;
    }
 
    // Add remaining carry
    if (carry)
        str.push_back(carry + '0');
 
    // Reverse resultant string
    reverse(str.begin(), str.end());
 
    return str;
}
 
// Function return 9's
// complement of given number
string complement9(string v)
{
    // Stores the complement
    string complement = "";
 
    for (int i = 0; i < v.size(); i++) {
 
        // Subtract every bit from 9
        complement += '9' - v[i] + '0';
    }
 
    // Return the result
    return complement;
}
 
// Function returns subtraction
// of two given numbers as strings
string subtract(string a, string b)
{
 
    // If second string is larger
    if (a.length() < b.length())
        swap(a, b);
 
    // Calculate respective lengths
    int l1 = a.length(),
        l2 = b.length();
 
    // If lengths are equal
    int diffLen = l1 - l2;
 
    for (int i = 0; i < diffLen; i++) {
 
        // Insert 0's to the beginning
        // of b to make both the lengths equal
        b = "0" + b;
    }
 
    // Add (complement of B) and A
    string c = sumBig(a, complement9(b));
 
    // If length of new string is greater
    // than length of first string,
    // than carry is generated
    if (c.length() > a.length()) {
        string::iterator it;
 
        // bit1 is the carry bit
        char bit1 = c[0];
        string bit = { bit1 };
        it = c.begin();
        c.erase(it);
        c = sumBig(c, bit);
        return c;
    }
 
    // If both lengths are equal
    else {
        return complement9(c);
    }
}
 
// Driver Code
int main()
{
 
    string str1 = "12345678987654321";
    string str2 = "22324324343";
 
    cout << subtract(str1, str2) << endl;
 
    return 0;
}

Output:

12345656663329978

Time Complexity: O(max(N, M))
Auxiliary Space: O(max(N, M))

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Subtraction of two large numbers using 9’s complement

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