How can I reduce the time complexity of the following loop?

Question

for(i=0;i<N-2;i++)
       count=(count*10)%M;

Here, N can be upto 10^18 and M is (10^9 +7). Since this loop takes O(n) time to execute, I get TLE in my code. Any way to reduce the time complexity?

This is still a programming question (not off-topic in my opinion) but help you will find in math. — Yunnosch, Apr 11 '18 at 06:55
what is the initial value of count? Here M is constant i...N is also known. Now the whole things depend on the initial value of count. — MD Ruhul Amin, Apr 11 '18 at 06:55
@ruhul initial value of count is a one digit integer. N is a variable 2<=N<=10^18 — Utkarsh Pandey, Apr 11 '18 at 06:57
https://en.wikipedia.org/wiki/Exponentiation#Efficient_computation_with_integer_exponents — Oliver Charlesworth, Apr 11 '18 at 07:23

Abhishek Keshri · Accepted Answer · 2018-04-11T07:45:01.383

3

The question is basically:

(count*a^b)%mod = ((count%mod)*((a^b)%mod))%mod

a = 10, b = 10^18

You can find ((a^b)%mod) using:

long long power(long long x, long long y, long long p)
{
    long long res = 1;      // Initialize result

    x = x % p;  // Update x if it is more than or 
                // equal to p

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;

        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;  
    }
    return res;
}

Time Complexity of the power function is O(log y).

In your case count is a 1-digit number so we can simply multiply this with (count%mod), and finally take mod of the result. If count is a big number too, and can cause overflow then we can do:

long long mulmod(long long a, long long b, long long mod)
{
    long long res = 0; // Initialize result
    a = a % mod;
    while (b > 0)
    {
        // If b is odd, add 'a' to result
        if (b % 2 == 1)
            res = (res + a) % mod;

        // Multiply 'a' with 2
        a = (a * 2) % mod;

        // Divide b by 2
        b /= 2;
    }

    // Return result
    return res % mod;
}

edited Apr 11 '18 at 07:45

answered Apr 11 '18 at 07:33

Abhishek Keshri

3,074
14
31

https://stackoverflow.com/questions/20971888/modular-multiplication-of-large-numbers-in-c?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa – Abhishek Keshri Apr 11 '18 at 07:52
To multiply a\*b, first calculate a\*b/2 then add it twice. For calculating a\*b/2 calculate a*b/4 and so on. That is why. I use this in online coding, so it is correct and verified a lot of times. – Abhishek Keshri Apr 11 '18 at 08:18
The original problem calculates "x * 10^(N-1) % M". Since N could be very large, say 10^18, b in "((a^b)%mod)" could be as large as 10^(10^18-1), which can't be saved by a `long long` int. – Stan Apr 11 '18 at 08:28
That's why I wrote power function. – Abhishek Keshri Apr 11 '18 at 08:30
Yeah!!!!!!!!!!! For multiplication in case count is also a large number. – Abhishek Keshri Apr 11 '18 at 08:37
@AbhishekKeshri Sorry just not patient enough to read the details :P maybe neither do other people ... – Stan Apr 11 '18 at 08:40
No problem, hope you understood finally :) – Abhishek Keshri Apr 11 '18 at 08:41
`Any way to reduce the time complexity?` does your answer, answer this question? O(n) to O(1) ? – MD Ruhul Amin Apr 11 '18 at 08:54
It has reduced it to O(log n). Have mentioned it in the answer. See that. – Abhishek Keshri Apr 11 '18 at 08:56
e.g. - a^2 = (a^(2/2))*(a^(2/2)) but a^3 = a*(a^(2/2))*(a^(2/2)) – Abhishek Keshri Apr 11 '18 at 09:16
@UtkarshPandey He has implemented the fast exponentiation algorithm described in the page linked by Oliver Charlesworth: https://en.wikipedia.org/wiki/Exponentiation#Efficient_computation_with_integer_exponents – Klas Lindbäck Apr 11 '18 at 09:21

How can I reduce the time complexity of the following loop?

1 Answers1