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# Add n binary strings

Given n binary strings, return their sum (also a binary string).

Examples:

Input:  arr[] = ["11", "1"]
Output: "100"

Input : arr[] = ["1", "10", "11"]
Output : "110"

### Approach 1:

Algorithm

1. Initialize the ‘result’ as an empty string.
2. Traverse the input from i = 0 to n-1.
3. For each i, add arr[i] to the ‘result’. How to add ‘result’ and arr[i]? Start from the last characters of the two strings and compute the digit sum one by one. If the sum becomes more than 1, then store carry for the next digit. Make this sum as the ‘result’.
4. The value of ‘result’ after traversing the entire input is the final answer.

Implementation:

## C++

 // C++ program to add n binary strings #include using namespace std;   // This function adds two binary strings and return // result as a third string string addBinaryUtil(string a, string b) {     string result = ""; // Initialize result     int s = 0; // Initialize digit sum       // Traverse both strings starting from last     // characters     int i = a.size() - 1, j = b.size() - 1;     while (i >= 0 || j >= 0 || s == 1) {           // Compute sum of last digits and carry         s += ((i >= 0) ? a[i] - '0' : 0);         s += ((j >= 0) ? b[j] - '0' : 0);           // If current digit sum is 1 or 3,         // add 1 to result         result = char(s % 2 + '0') + result;           // Compute carry         s /= 2;           // Move to next digits         i--;         j--;     }     return result; }   // function to add n binary strings string addBinary(string arr[], int n) {     string result = "";     for (int i = 0; i < n; i++)         result = addBinaryUtil(result, arr[i]);     return result; }   // Driver program int main() {     string arr[] = { "1", "10", "11" };     int n = sizeof(arr) / sizeof(arr[0]);     cout << addBinary(arr, n) << endl;     return 0; }

## Java

 // Java program to add n binary strings class GFG {       // This function adds two binary     // strings and return result as     // a third string     static String addBinaryUtil(String a, String b)     {         String result = ""; // Initialize result         int s = 0; // Initialize digit sum           // Traverse both strings starting         // from last characters         int i = a.length() - 1, j = b.length() - 1;         while (i >= 0 || j >= 0 || s == 1)         {               // Compute sum of last digits and carry             s += ((i >= 0) ? a.charAt(i) - '0' : 0);             s += ((j >= 0) ? b.charAt(j) - '0' : 0);               // If current digit sum is 1 or 3,             // add 1 to result             result = s % 2 + result;               // Compute carry             s /= 2;               // Move to next digits             i--;             j--;         }         return result;     }       // function to add n binary strings     static String addBinary(String arr[], int n)     {         String result = "";         for (int i = 0; i < n; i++)         {             result = addBinaryUtil(result, arr[i]);         }         return result;     }       // Driver code     public static void main(String[] args)     {         String arr[] = {"1", "10", "11"};         int n = arr.length;         System.out.println(addBinary(arr, n));     } }   // This code is contributed by Rajput-JI

## Python3

 # Python3 program to add n binary strings   # This function adds two binary strings and # return result as a third string def addBinaryUtil(a, b):           result = ""; # Initialize result     s = 0;       # Initialize digit sum       # Traverse both strings     # starting from last characters     i = len(a) - 1;     j = len(b) - 1;     while (i >= 0 or j >= 0 or s == 1):           # Compute sum of last digits and carry         s += (ord(a[i]) - ord('0')) if(i >= 0) else 0;         s += (ord(b[j]) - ord('0')) if(j >= 0) else 0;           # If current digit sum is 1 or 3,         # add 1 to result         result = chr(s % 2 + ord('0')) + result;           # Compute carry         s //= 2;           # Move to next digits         i -= 1;         j -= 1;       return result;   # function to add n binary strings def addBinary(arr, n):     result = "";     for i in range(n):         result = addBinaryUtil(result, arr[i]);     return result;   # Driver code arr = ["1", "10", "11"]; n = len(arr); print(addBinary(arr, n));       # This code is contributed by mits

## C#

 // C# program to add n binary strings using System;   class GFG {           // This function adds two binary     // strings and return result as     // a third string     static String addBinaryUtil(String a,                                 String b)     {         // Initialize result         String result = "";                   // Initialize digit sum         int s = 0;           // Traverse both strings starting         // from last characters         int i = a.Length - 1, j = b.Length - 1;         while (i >= 0 || j >= 0 || s == 1)         {               // Compute sum of last digits and carry             s += ((i >= 0) ? a[i] - '0' : 0);             s += ((j >= 0) ? b[j] - '0' : 0);               // If current digit sum is 1 or 3,             // add 1 to result             result = s % 2 + result;               // Compute carry             s /= 2;               // Move to next digits             i--;             j--;         }         return result;     }       // function to add n binary strings     static String addBinary(String []arr, int n)     {         String result = "";         for (int i = 0; i < n; i++)         {             result = addBinaryUtil(result, arr[i]);         }         return result;     }       // Driver code     public static void Main(String[] args)     {         String []arr = {"1", "10", "11"};         int n = arr.Length;         Console.WriteLine(addBinary(arr, n));     } }   // This code is contributed by 29AjayKumar

## PHP

 = 0 || \$j >= 0 || \$s == 1)     {           // Compute sum of last digits and carry         \$s += ((\$i >= 0) ? ord(\$a[\$i]) - ord('0') : 0);         \$s += ((\$j >= 0) ? ord(\$b[\$j]) - ord('0') : 0);           // If current digit sum is 1 or 3,         // add 1 to result         \$result = chr(\$s % 2 + ord('0')).\$result;           // Compute carry         \$s =(int)(\$s/2);           // Move to next digits         \$i--;         \$j--;     }     return \$result; }   // function to add n binary strings function addBinary(\$arr, \$n) {     \$result = "";     for (\$i = 0; \$i < \$n; \$i++)         \$result = addBinaryUtil(\$result, \$arr[\$i]);     return \$result; }   // Driver code     \$arr = array( "1", "10", "11" );     \$n = count(\$arr);     echo addBinary(\$arr, \$n)."\n";       // This code is contributed by mits ?>

## Javascript



Output

110

Complexity Analysis:

• Time complexity: O(n)
• Auxiliary Space: O(n)

### Approach 2:

By converting the binary strings to decimal numbers.

Here’s a step-by-step explanation of the approach:

1. Initialize a variable sum to store the sum of the decimal numbers converted from the binary strings.
2. Loop through each binary string in the array:
1. Initialize a variable num to store the decimal number converted from the current binary string.
2. Loop through each character in the current binary string from right to left:
3. If the current character is ‘1’, add 2 raised to the power of its position from right to num.
4. Add num to sum.
3.  Initialize an empty string result to store the binary representation of sum.
4. While sum is greater than 0:
1. If sum is even, prepend “0” to result, otherwise prepend “1”.
2. Divide sum by 2.
5. Return result.

## C++

 #include #include #include   using namespace std;   string addBinary(string arr[], int n) {     int sum = 0;     for (int i = 0; i < n; i++)     {         int num = 0;         for (int j = arr[i].size() - 1; j >= 0; j--)             if (arr[i][j] == '1')                 num += pow(2, arr[i].size() - j - 1);         sum += num;     }     string result = "";     while (sum > 0)     {         result = (sum % 2 == 0 ? "0" : "1") + result;         sum /= 2;     }     return result; }   int main() {     string arr[] = { "1", "10", "11" };     int n = sizeof(arr) / sizeof(arr[0]);     cout << addBinary(arr, n) << endl;     return 0; }

Output

110

Time Complexity: O(N*K)
Auxiliary Space: O(1)

Explanation:

The time complexity of this approach is O(nk), where n is the number of binary strings and k is the maximum length of a binary string in the array. This is because we need to loop through each binary string and each character in the binary strings.

The auxiliary space complexity of this approach is O(1), as we only use a constant amount of extra space to store variables such as sum, num, and result.

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