# Modulo of a large Binary String

Given a large binary string str and an integer K, the task is to find the value of str % K.

Examples:

Input: str = “1101”, K = 45
Output: 13
decimal(1101) % 45 = 13 % 45 = 13

Input: str = “11010101”, K = 112
Output: 101
decimal(11010101) % 112 = 213 % 112 = 101

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

Approach: It is known that (str % K) where str is a binary string can be written as ((str[n – 1] * 20) + (str[n – 2] * 21) + … + (str * 2n – 1)) % K which in turn can be written as (((str[n – 1] * 20) % K) + ((str[n – 2] * 21) % K) + … + ((str * 2n – 1)) % K) % K. This can be used to find the required answer without actually converting the given binary string to its decimal equivalent.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to return the value of (str % k)  ` `int` `getMod(string str, ``int` `n, ``int` `k)  ` `{  ` ` `  `    ``// pwrTwo[i] will store ((2^i) % k)  ` `    ``int` `pwrTwo[n];  ` `    ``pwrTwo = 1 % k;  ` `    ``for` `(``int` `i = 1; i < n; i++)  ` `    ``{  ` `        ``pwrTwo[i] = pwrTwo[i - 1] * (2 % k);  ` `        ``pwrTwo[i] %= k;  ` `    ``}  ` ` `  `    ``// To store the result  ` `    ``int` `res = 0;  ` `    ``int` `i = 0, j = n - 1;  ` `    ``while` `(i < n)  ` `    ``{  ` ` `  `        ``// If current bit is 1  ` `        ``if` `(str[j] == ``'1'``)  ` `        ``{  ` ` `  `            ``// Add the current power of 2  ` `            ``res += (pwrTwo[i]);  ` `            ``res %= k;  ` `        ``}  ` `        ``i++;  ` `        ``j--;  ` `    ``}  ` `    ``return` `res;  ` `}  ` ` `  `// Driver code  ` `int` `main()  ` `{  ` `    ``string str = ``"1101"``;  ` `    ``int` `n = str.length();  ` `    ``int` `k = 45;  ` ` `  `    ``cout << getMod(str, n, k) << endl;  ` `}  ` ` `  `// This code is contributed by ashutosh450 `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to return the value of (str % k) ` `    ``static` `int` `getMod(String str, ``int` `n, ``int` `k) ` `    ``{ ` ` `  `        ``// pwrTwo[i] will store ((2^i) % k) ` `        ``int` `pwrTwo[] = ``new` `int``[n]; ` `        ``pwrTwo[``0``] = ``1` `% k; ` `        ``for` `(``int` `i = ``1``; i < n; i++) { ` `            ``pwrTwo[i] = pwrTwo[i - ``1``] * (``2` `% k); ` `            ``pwrTwo[i] %= k; ` `        ``} ` ` `  `        ``// To store the result ` `        ``int` `res = ``0``; ` `        ``int` `i = ``0``, j = n - ``1``; ` `        ``while` `(i < n) { ` ` `  `            ``// If current bit is 1 ` `            ``if` `(str.charAt(j) == ``'1'``) { ` ` `  `                ``// Add the current power of 2 ` `                ``res += (pwrTwo[i]); ` `                ``res %= k; ` `            ``} ` `            ``i++; ` `            ``j--; ` `        ``} ` ` `  `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``String str = ``"1101"``; ` `        ``int` `n = str.length(); ` `        ``int` `k = ``45``; ` ` `  `        ``System.out.print(getMod(str, n, k)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the value of (str % k)  ` `def` `getMod(_str, n, k) :  ` ` `  `    ``# pwrTwo[i] will store ((2^i) % k)  ` `    ``pwrTwo ``=` `[``0``] ``*` `n  ` `    ``pwrTwo[``0``] ``=` `1` `%` `k  ` `    ``for` `i ``in` `range``(``1``, n):  ` `        ``pwrTwo[i] ``=` `pwrTwo[i ``-` `1``] ``*` `(``2` `%` `k)  ` `        ``pwrTwo[i] ``%``=` `k  ` ` `  `    ``# To store the result  ` `    ``res ``=` `0` `    ``i ``=` `0` `    ``j ``=` `n ``-` `1` `    ``while` `(i < n) :  ` ` `  `        ``# If current bit is 1  ` `        ``if` `(_str[j] ``=``=` `'1'``) :  ` ` `  `            ``# Add the current power of 2  ` `            ``res ``+``=` `(pwrTwo[i])  ` `            ``res ``%``=` `k  ` `             `  `        ``i ``+``=` `1` `        ``j ``-``=` `1` ` `  `    ``return` `res  ` ` `  `# Driver code  ` `_str ``=` `"1101"` `n ``=` `len``(_str)  ` `k ``=` `45` ` `  `print``(getMod(_str, n, k))  ` ` `  `# This code is contributed by ` `# divyamohan123 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the value of (str % k)  ` `    ``static` `int` `getMod(``string` `str, ``int` `n, ``int` `k)  ` `    ``{  ` `        ``int` `i; ` `         `  `        ``// pwrTwo[i] will store ((2^i) % k)  ` `        ``int` `[]pwrTwo = ``new` `int``[n];  ` `         `  `        ``pwrTwo = 1 % k;  ` `         `  `        ``for` `(i = 1; i < n; i++)  ` `        ``{  ` `            ``pwrTwo[i] = pwrTwo[i - 1] * (2 % k);  ` `            ``pwrTwo[i] %= k;  ` `        ``}  ` ` `  `        ``// To store the result  ` `        ``int` `res = 0;  ` `        ``i = 0; ` `        ``int` `j = n - 1;  ` `        ``while` `(i < n) ` `        ``{  ` ` `  `            ``// If current bit is 1  ` `            ``if` `(str[j] == ``'1'``)  ` `            ``{  ` ` `  `                ``// Add the current power of 2  ` `                ``res += (pwrTwo[i]);  ` `                ``res %= k;  ` `            ``}  ` `            ``i++;  ` `            ``j--;  ` `        ``}  ` `        ``return` `res;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``string` `str = ``"1101"``;  ` `        ``int` `n = str.Length;  ` `        ``int` `k = 45;  ` ` `  `        ``Console.Write(getMod(str, n, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```13
```

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