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Subtraction of two large numbers using 10’s complement

  • Difficulty Level : Medium
  • Last Updated : 24 Sep, 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 10’s complement.

Example: 

Input: N = 10, str1 = “3434243434”, M = 14, str2 = “22324365765767” 
Output: 22320931522333

Input: N = 20, str1 = “12345334233242431433”, M = 20, str2 = “12345334233242431432” 
Output:

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 + (10’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 10’s complement of str2.
  • Now, add 10’s complement of str2 to str1.
  • If any carry is generated, then drop the carry.
  • 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 calculate the
// subtraction of two large number
// using 10's complement
#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 10's
// complement of given number
string complement10(string v)
{
    // Stores the complement
    string complement = "";
 
    // Calculate 9's complement
    for (int i = 0; i < v.size(); i++) {
 
        // Subtract every bit from 9
        complement += '9' - v[i] + '0';
    }
 
    // Add 1 to 9's complement
    // to find 10's complement
    complement = sumBig(complement, "1");
    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 aren't 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, complement10(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;
 
        // Erase first bit
        it = c.begin();
 
        c.erase(it);
 
        // Trim zeros at the beginning
        it = c.begin();
 
        while (*it == '0')
            c.erase(it);
 
        return c;
    }
 
    // If both lengths are equal
    else {
        return complement10(c);
    }
}
 
// Driver Code
int main()
{
 
    string str1 = "12345334233242431433";
    string str2 = "12345334233242431432";
 
    cout << subtract(str1, str2) << endl;
 
    return 0;
}
Output: 
1

 

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


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