Modulo of a large Binary String
Last Updated :
10 Nov, 2022
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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 = 13Input: str = "11010101", K = 112
Output: 101
decimal(11010101) % 112 = 213 % 112 = 101
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[0] * 2n - 1)) % K which in turn can be written as (((str[n - 1] * 20) % K) + ((str[n - 2] * 21) % K) + ... + ((str[0] * 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++ implementation of the approach
#include <bits/stdc++.h>
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[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[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 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 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# 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[0] = 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
<script>
// Javascript implementation of the approach
// Function to return the value of (str % k)
function getMod(str, n, k)
{
// pwrTwo[i] will store ((2^i) % k)
var pwrTwo = Array(n);
pwrTwo[0] = 1 % k;
for (var i = 1; i < n; i++)
{
pwrTwo[i] = pwrTwo[i - 1] * (2 % k);
pwrTwo[i] %= k;
}
// To store the result
var res = 0;
var 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
var str = "1101";
var n = str.length;
var k = 45;
document.write( getMod(str, n, k));
</script>
Output:
13
Time Complexity: O(n)
Auxiliary Space: O(n)
