Subtraction of two large numbers using 9's compliment

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 compliment of str2.
Now, add 9’s compliment of str2 to str1.
If any carry is generated, insert at the beginning of str1.
If no carry is generated, then the compliment of str1 is the final answer.

Below is the implementation of the above approach:

C++

// C++ Program to implement 
// the above appraoch 
#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 compliment9(string v) 
{ 
    // Stores the compliment 
    string compliment = ""; 
  
    for (int i = 0; i < v.size(); i++) { 
  
        // Subtract every bit from 9 
        compliment += '9' - v[i] + '0'; 
    } 
  
    // Return the result 
    return compliment; 
} 
  
// 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, compliment9(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 compliment9(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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My Personal Notes

Subtraction of two large numbers using 9’s compliment

C++

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