Find two numbers whose difference is given as Binary string
Last Updated :
13 Dec, 2023
Given a string S, which contains a certain number of 1s followed by a certain number of 0s. Then your task is to output the binary representation of two integers A and B by following below conditions:
- A >= B
- The number of set bits in both A and B are minimum.
- A – B = K, where K is the Decimal representation of S.
Examples:
Input: S = 11100
Output: A = 100000, B = 100
Explanation: It is visible that A > B. Binary representation of A and B is 100000 and 100 respectively, which collectively have set bits equal to 2 and A – B = 32 – 4 = 28 which is Decimal representation of S.
Input: S = 1111100000
Output: A = 10000000000, B = 100000
Explanation: It is visible that A > B. Binary representation of A and B is 10000000000 and 100000 respectively, which collectively have set bits equal to 2 and A – B = 1024 – 32 = 992 which is Decimal representation of S.
Observations:
- If string S has only one 1, then we can have A = K and B = 0
- Otherwise, for every possible S, there will exist A and B such that A – B = K and A and B both are perfect powers of 2.
- Examples:
- S = 110 then A = 8 and B = 2
- S = 11110 then A = 32, B = 2
- S = 11110000 then A = 256, B = 32
Approach: Implement the idea below to solve the problem:
The problem can be solved using the above observations. Let’s say we start moving from right to left and X is the position of the first occurrence of 1 and Y is the position of last occurrence of 1, then we always say that A = (2 ^ (Y + 1)) and B = (2 ^ X) such that A – B = K.
Steps were taken to solve the problem:
- Declare a variable let say Count to count the number of set bits in S.
- Run a loop and count the number of set bits in S.
- If (Count == 1), then A = S, B = 0
- Else
- Start moving from right to left and find the first and last occurrence of 1 in S say X and Y respectively.
- After traversing the string, make A = (2 ^ (Y + 1)) and B = (2 ^ (X + 1))
- Print A and B
Code to implement the approach:
C++
#include <bits/stdc++.h>
using namespace std;
void func(string& s)
{
int count = 0;
for (char ch : s) {
if (ch == '1') {
count++;
}
}
if (count == 0) {
cout << "0 0\n";
}
else if (count == 1) {
cout << s << " 0\n";
}
else {
string A, B;
int X = -1, Y = -1;
for (int i = s.length() - 1; i >= 0; i--) {
if (s[i] == '1') {
Y = i;
if (X == -1)
X = i;
}
}
A += '1';
for (int i = s.length() - 1; i >= Y; i--)
A += '0';
B += '1';
for (int i = s.length() - 1; i > X; i--)
B += '0';
cout << A << " " << B << "\n";
}
}
int main()
{
string S = "11110000";
func(S);
return 0;
}
|
Java
import java.util.*;
class Main {
public static void main(String[] args)
{
String S = "11110000";
func(S);
}
public static void func(String s)
{
int count = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '1') {
count++;
}
}
if (count == 0) {
System.out.println("0 0");
}
else if (count == 1) {
System.out.println(s + " " + 0);
}
else {
StringBuilder A = new StringBuilder();
StringBuilder B = new StringBuilder();
int X = -1, Y = -1;
for (int i = s.length() - 1; i >= 0; i--) {
if (s.charAt(i) == '1') {
Y = i;
if (X == -1)
X = i;
}
}
A.append("1");
for (int i = s.length() - 1; i >= Y; i--)
A.append("0");
B.append("1");
for (int i = s.length() - 1; i > X; i--)
B.append("0");
System.out.println(A + " " + B);
}
}
}
|
Python
def func(s):
count = s.count('1')
if count == 0:
print("0 0")
elif count == 1:
print(s + " 0")
else:
A, B = "", ""
X, Y = -1, -1
for i in range(len(s) - 1, -1, -1):
if s[i] == '1':
Y = i
if X == -1:
X = i
A += '1' + '0' * (len(s) - Y)
B += '1' + '0' * (len(s) - X - 1)
print(A, B)
if __name__ == "__main__":
S = "11110000"
func(S)
|
C#
using System;
class Program
{
static void Func(string s)
{
int count = 0;
foreach (char ch in s)
{
if (ch == '1')
{
count++;
}
}
if (count == 0)
{
Console.WriteLine("0 0");
}
else if (count == 1)
{
Console.WriteLine($"{s} 0");
}
else
{
string A = "", B = "";
int X = -1, Y = -1;
for (int i = s.Length - 1; i >= 0; i--)
{
if (s[i] == '1')
{
Y = i;
if (X == -1)
X = i;
}
}
A += '1';
for (int i = s.Length - 1; i >= Y; i--)
A += '0';
B += '1';
for (int i = s.Length - 1; i > X; i--)
B += '0';
Console.WriteLine($"{A} {B}");
}
}
static void Main()
{
string S = "11110000";
Func(S);
}
}
|
Javascript
function func(s) {
let count = 0;
for (let i = 0; i < s.length; i++) {
if (s[i] === '1') {
count++;
}
}
if (count === 0) {
console.log("0 0");
} else if (count === 1) {
console.log(`${s} 0`);
} else {
let A = '';
let B = '';
let X = -1;
let Y = -1;
for (let i = s.length - 1; i >= 0; i--) {
if (s[i] === '1') {
Y = i;
if (X === -1) {
X = i;
}
}
}
A += '1';
for (let i = s.length - 1; i >= Y; i--) {
A += '0';
}
B += '1';
for (let i = s.length - 1; i > X; i--) {
B += '0';
}
console.log(`${A} ${B}`);
}
}
const S = "11110000";
func(S);
|
Time Complexity: O(N), where N is the length of the string S.
Auxiliary Space: O(1)
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