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Commitd750e95

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Sri Hari: Batch-5/Neetcode-250/Added-articles (neetcode-gh#3794)
* Batch-5/Neetcode-250/Added-articles* Batch-5/Neetcode-250/Added-articles* Batch-5/Neetcode-250/Added-articles* Batch-5/Neetcode-250/Added-articles
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‎articles/accounts-merge.md

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‎articles/add-binary.md

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##1. Iteration
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::tabs-start
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```python
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classSolution:
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defaddBinary(self,a:str,b:str) ->str:
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res=""
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carry=0
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a, b= a[::-1], b[::-1]
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for iinrange(max(len(a),len(b))):
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digitA=ord(a[i])-ord("0")if i<len(a)else0
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digitB=ord(b[i])-ord("0")if i<len(b)else0
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total= digitA+ digitB+ carry
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char=str(total%2)
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res= char+ res
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carry= total//2
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if carry:
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res="1"+ res
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return res
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```
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```java
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publicclassSolution {
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publicStringaddBinary(Stringa,Stringb) {
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StringBuilder res=newStringBuilder();
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int carry=0;
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StringBuilder sa=newStringBuilder(a).reverse();
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StringBuilder sb=newStringBuilder(b).reverse();
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for (int i=0; i<Math.max(sa.length(), sb.length()); i++) {
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int digitA= i< sa.length()? sa.charAt(i)-'0':0;
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int digitB= i< sb.length()? sb.charAt(i)-'0':0;
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int total= digitA+ digitB+ carry;
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char c= (char)((total%2)+'0');
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res.append(c);
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carry= total/2;
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}
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if (carry>0) {
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res.append('1');
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}
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return res.reverse().toString();
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}
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}
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```
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```cpp
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classSolution {
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public:
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string addBinary(string a, string b) {
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string res = "";
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int carry = 0;
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reverse(a.begin(), a.end());
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reverse(b.begin(), b.end());
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for (int i = 0; i < max(a.length(), b.length()); i++) {
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int digitA = i < a.length() ? a[i] - '0' : 0;
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int digitB = i < b.length() ? b[i] - '0' : 0;
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int total = digitA + digitB + carry;
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char c = (total % 2) + '0';
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res += c;
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carry = total / 2;
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}
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if (carry) {
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res += '1';
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}
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reverse(res.begin(), res.end());
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return res;
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}
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};
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```
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```javascript
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class Solution {
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/**
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* @param {string} a
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* @param {string} b
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* @return {string}
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*/
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addBinary(a, b) {
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let res = [];
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let carry = 0;
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a = a.split("").reverse().join("");
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b = b.split("").reverse().join("");
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for (let i = 0; i < Math.max(a.length, b.length); i++) {
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const digitA = i < a.length ? a[i] - '0' : 0;
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const digitB = i < b.length ? b[i] - '0' : 0;
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const total = digitA + digitB + carry;
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const char = (total % 2).toString();
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res.push(char)
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carry = Math.floor(total / 2);
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}
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if (carry) {
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res.push('1');
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}
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res.reverse()
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return res.join('');
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(max(m, n))$
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* Space complexity: $O(m + n)$
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> Where $m$and $n$ are the lengths of the strings $a$and $b$ respectively.
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---
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##2. Iteration (Optimal)
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::tabs-start
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```python
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classSolution:
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def addBinary(self, a: str, b: str) -> str:
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res =[]
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carry = 0
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i, j = len(a) - 1, len(b) - 1
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while i >= 0 or j >= 0 or carry > 0:
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digitA = int(a[i]) if i >= 0 else 0
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digitB = int(b[j]) if j >= 0 else 0
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total = digitA + digitB + carry
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res.append(total % 2)
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carry = total // 2
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i -= 1
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j -= 1
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res.reverse()
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return ''.join(map(str, res))
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```
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```java
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public class Solution {
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public String addBinary(String a, String b) {
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StringBuilder res = new StringBuilder();
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int carry = 0;
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int i = a.length() - 1, j = b.length() - 1;
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while (i >= 0 || j >= 0 || carry > 0) {
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int digitA = i >= 0 ? a.charAt(i) - '0' : 0;
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int digitB = j >= 0 ? b.charAt(j) - '0' : 0;
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int total = digitA + digitB + carry;
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res.append(total % 2);
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carry = total / 2;
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i--;
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j--;
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}
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return res.reverse().toString();
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}
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}
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```
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```cpp
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classSolution {
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public:
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string addBinary(string a, string b) {
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string res = "";
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int carry = 0;
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int i = a.size() - 1, j = b.size() - 1;
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while (i >= 0 || j >= 0 || carry > 0) {
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int digitA = i >= 0 ? a[i] - '0' : 0;
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int digitB = j >= 0 ? b[j] - '0' : 0;
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int total = digitA + digitB + carry;
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res += (total % 2) + '0';
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carry = total / 2;
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i--;
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j--;
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}
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reverse(res.begin(), res.end());
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return res;
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}
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};
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```
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```javascript
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class Solution {
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/**
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* @param {string} a
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* @param {string} b
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* @return {string}
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*/
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addBinary(a, b) {
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let res = [];
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let carry = 0;
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let i = a.length - 1, j = b.length - 1;
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while (i >= 0 || j >= 0 || carry > 0) {
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const digitA = i >= 0 ? a[i] - "0" : 0;
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const digitB = j >= 0 ? b[j] - "0" : 0;
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const total = digitA + digitB + carry;
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res.push(total % 2);
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carry = Math.floor(total / 2);
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i--;
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j--;
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}
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res.reverse()
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return res.join('');
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(max(m, n))$
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* Space complexity: $O(max(m, n))$
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>Where $m$ and $n$ are the lengths of the strings $a$ and $b$ respectively.

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