Divide a big number into two parts that differ by k

Given a big positive number N. The task is divide N into two number ‘A’ and ‘B’ such that difference between them is K (1 <= K <= 10¹⁰⁰) i.e A – B = K.

Examples:

Input : N = 10, K = 2
Output : A = 6 B = 4

Input : N = 20, K = 4
Output : A = 12 B = 8

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Let the two required number be ‘A’ and ‘B’. So, we know sum of ‘A’ and ‘B’ will end upto N.
So, one equation became,
A + B = N
And also, we want the difference between ‘A’ and ‘B’ equals to ‘K’.
So, another equation becomes,
A – B = K
On adding both the equation, we get
2*A = N + K

So, A = (N + K)/2
Then we can find B by, B = (N – A)

Now, to handle the Big Integer, we have store Integers in character array and define a function to perform the operation on them.

Below is C implementation of this approach:

// C program to Divide a Big 
// Number into two parts 
#include <stdio.h> 
#include <string.h> 
#define MAX 100 
  
// Function to adds two Numbers 
// represnted as array of character. 
void add(char v1[], char v2[]) 
{ 
    int i, d, c = 0; 
  
    // length of string 
    int l1 = strlen(v1); 
    int l2 = strlen(v2); 
  
    // initializing extra character 
    // position to 0 
    for (i = l1; i < l2; i++) 
        v1[i] = '0'; 
  
    for (i = l2; i < l1; i++) 
        v2[i] = '0'; 
  
    // Adding each element of character 
    // and storing the carry. 
    for (i = 0; i < l1 || i < l2; i++) { 
        d = (v1[i] - '0') + (v2[i] - '0') + c; 
        c = d / 10; 
        d %= 10; 
        v1[i] = '0' + d; 
    } 
  
    // If remainder remains. 
    while (c) { 
        v1[i] = '0' + (c % 10); 
        c /= 10; 
        i++; 
    } 
  
    v1[i] = '\0'; 
    v2[l2] = '\0'; 
} 
  
// Function to subtracts two numbers 
// represnted by string. 
void subs(char v1[], char v2[]) 
{ 
    int i, d, c = 0; 
  
    // Finding the length of the string. 
    int l1 = strlen(v1); 
    int l2 = strlen(v2); 
  
    // initializing extra character position to 0. 
    for (i = l2; i < l1; i++) 
        v2[i] = '0'; 
  
    // Substrating each element of character. 
    for (i = 0; i < l1; i++) { 
        d = (v1[i] - '0' - c) - (v2[i] - '0'); 
  
        if (d < 0) { 
            d += 10; 
            c = 1; 
        } 
        else
            c = 0; 
  
        v1[i] = '0' + d; 
    } 
  
    v2[l2] = '\0'; 
    i = l1 - 1; 
  
    while (i > 0 && v1[i] == '0') 
        i--; 
  
    v1[i + 1] = '\0'; 
} 
  
// Function divides a number represented by 
// character array a constant. 
int divi(char v[], int q) 
{ 
    int i, l = strlen(v); 
  
    int c = 0, d; 
  
    // Dividing each character element by constant. 
    for (i = l - 1; i >= 0; i--) { 
        d = c * 10 + (v[i] - '0'); 
        c = d % q; 
        d /= q; 
        v[i] = '0' + d; 
    } 
  
    i = l - 1; 
  
    while (i > 0 && v[i] == '0') 
        i--; 
  
    v[i + 1] = '\0'; 
    return c; 
} 
  
// Function to reverses the character array. 
void rev(char v[]) 
{ 
    int l = strlen(v); 
    int i; 
    char cc; 
  
    // Reversing the array. 
    for (i = 0; i < l - 1 - i; i++) { 
        cc = v[i]; 
        v[i] = v[l - 1 - i]; 
        v[l - i - 1] = cc; 
    } 
} 
  
// Wrapper Function 
void divideWithDiffK(char a[], char k[]) 
{ 
  
    // Reversing the character array. 
    rev(a); 
    rev(k); 
  
    // Adding the each element of both array 
    // and storing the sum in array a[]. 
    add(a, k); 
  
    // Dividing the array a[] by 2. 
    divi(a, 2); 
  
    // Reversing the character array to get output. 
    rev(a); 
    printf("%s ", a); 
  
    // Substracting each element of array 
    // i.e calculating a = a - b 
    rev(a); 
    subs(a, k); 
  
    // Reversing the character array to get output. 
    rev(a); 
    printf("%s", a); 
} 
  
// Driven Program 
int main() 
{ 
    char a[MAX] = "100", k[MAX] = "20"; 
    divideWithDiffK(a, k); 
    return 0; 
} 

Output:

60 40

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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