I am facing difficulty to figure out the time complexity of this below function. Please help to find how to solve this question?
int sumOfDigits(int n){
int sum;`
if(n < 10){
return n;
}
sum = (n % 10) + sumOfDigits(n / 10);
return sum;
}
The function considers each digit of a number n and returns the sum of them.
The number of iterations is ⌊log10(n)⌋ + 1 and depends on the number of digits of n. Therefore, T(n) = O(log n)
Related
I have a C code off finding large perfect numbers below,
#include <stdio.h>
int main ()
{
unsigned long long num,i,sum;
while (scanf ("%llu",&num) != EOF && num)
{
sum = 1;
for (i=2; i*i<=num; i++)
{
if (num % i == 0)
{
if (i*i == num)
sum += i;
else
sum += (i + num/i);
}
}
if (sum == num)
printf ("Perfect\n");
else if (sum > num)
printf ("Abundant\n");
else
printf ("Deficient\n");
}
return 0;
}
I tried to find whether a number is perfect, abundant or deficient. I run a loop upto the square root of numto minimize the runtime. It works fine <= 10^15 but for the larger values it takes too long time to execute.
For example,for the following input sets,
8
6
18
1000000
1000000000000000
0
this code shows the following outputs,
Deficient
Perfect
Abundant
Abundant
Abundant
But, for 10^16 it doesn't respond quickly.
So, is there any better way to find a perfect number for too long values? Or is there any better algorithm to implement here??? :)
Yes, there is a better algorithm.
Your algorithm is basically the simple one--adding up the divisors of a number to find... the sum of the divisors of a number (excluding itself). But you can use the number-theoretic formula for finding the sum of the divisors of a number (including itself). If the prime numbers dividing n are p1, p2, ..., pk and the powers of those primes in the canonical decomposition of n are a1, a2, ..., ak, then the sum of the divisors of n is
(p1**(a1+1) - 1) / (p1 - 1) * (p2**(a2+1) - 1) / (p2 - 1) * ...
* (pk**(ak+1) - 1) / (pk - 1)
You can find the prime divisors and their exponents more quickly than finding all the divisors of n. Subtract n from that expression above and you get the sum you want.
There are some tricks, of course, to find the pis and ais more efficiently: I'll leave that to you.
By the way, if your purpose is just to find the perfect numbers, as in your title, you would do better to use Euclid's formula for even prime numbers. Find the Mersenne prime numbers by examining all 2**p-1 for prime p to see if they are prime--there are shortcuts to doing this as well--then constructing a perfect number from that Mersenne prime. This would leave out any odd perfect numbers, though. If you find any, let the mathematical community know--that would make you world famous.
Of course, the fastest way of all to find perfect numbers is to use the lists already made of some of them.
It is a matter of factorization of numbers. You can read more here: https://en.wikipedia.org/wiki/Integer_factorization
Unfortunately no good news for you - the bigger the number gets, the longer it takes.
To start with your code, try not to multiply i*i each iteration.
Instead of:
for (i=2; i*i<=num; i++)
calculate square root of num first, and then compare
i <= square_root_of_num in the loop.
// Program to determine whether perfect or not
# include <bits/stdc++.h>
using namespace std;
map<long long int, int> mp; // to store prime factors and there frequency
void primeFactors(long long int n)
{
// counting the number of 2s that divide n
while (n%2 == 0)
{
mp[2] = mp[2]+1;
n = n/2;
}
long long int root = sqrt(n);
// n must be odd at this point. So we can skip every even numbers next
for (long long int i = 3; i <= root; i = i+2)
{
// While i divides n, count frequency of i prime factor and divide n
while (n%i == 0)
{
mp[i] = mp[i]+1;
n = n/i;
}
}
// This condition is to handle the case whien n is a prime number
// greater than 2
if (n > 2)
{
mp[n] = mp[n]+1;
}
}
long long int pow(long long int base, long long int exp)
{
long long int result = 1;
base = base;
while (exp>0)
{
if (exp & 1)
result = (result*base);
exp >>= 1;
base = (base*base);
}
return result;
}
int main ()
{
long long num, p, a, sum;
while (scanf ("%lld",&num) != EOF && num)
{
primeFactors(num);
sum = 1;
map<long long int, int> :: iterator i;
for(i=mp.begin(); i!=mp.end(); i++)
{
p = i->first;
a = i->second;
sum = sum*((pow(p,a+1)-1)/(p-1));
}
if (sum == 2*num)
printf ("Perfect\n");
else if (sum > num)
printf ("Abundant\n");
else
printf ("Deficient\n");
mp.clear();
}
return 0;
}
I have a Computer Science Midterm tomorrow and I need help determining the complexity of these recursive functions. I know how to solve simple cases, but I am still trying to learn how to solve these harder cases. Any help would be much appreciated and would greatly help in my studies, Thank you!
fonction F(n)
if n == 0
return 1
else
return F(n-1) * n
fonction UniqueElements(A[0..n-1])
for i=0 to i <= n-2 do
for j=i+1 to j <= n-1 do
if A[i] == A[j]
return false
return true
fonction BinRec(n)
if n == 1
return 1
else
return BinRec(floor(n/2)) + 1
For hands on learning, you can plug the functions into a program, and test their worst case scenario performance.
When trying to calculate O by hand, here are some things to remember
The +, -, *, and / offsets can be ignored. So 1 to n+5 and 1 to 5n is considered equivalent to 1 to n.
Also, Only the highest order of magnitude counts, so for O 2^n + n^2 + n, 2^n grows the fastest, so it is equivalent to O 2^n
With recursive functions, you are looking at how many times the function is called in the method (the split count) and how much it needs to be called (the depth, usually is equal to list length). So the final O will be depth_count^split_count
With loops, each nested loop multiplies to the one it's in, and sequential loops add, so (1-n){(1-n){}} (1-n){} is (n * n) + n) => n^2 + n =(only highest growth counts)> n^2
PRACTICE! You will need to practice to get the hang of the gatchas of growth rate and how control flows interact. (so do online practice quizs)
function F(n){
count++
if (n == 0)
return 1
else
return F(n-1) * n
}
function UniqueElements(A){
for (var i=0 ; i <= A.length-2; i++){
for (var j=i+1;j <= A.length-1; j++){
if (A[i] == A[j]){
return false
}
}
}
return true
}
function BinRec(n) {
count++
if (n == 1)
return 1
else
return BinRec(Math.floor(n/2)) + 1
}
count = 0;
console.log(F(10));
console.log(count);
count = 0;
console.log(UniqueElements([1,2,3,5]));
console.log(count);
count = 0;
console.log(BinRec(40));
console.log(count);
I want to solve a mathematical problem in a fastest possible way.
I have a set of natural numbers between 1 to n, for example {1,2,3,4,n=5} and I want to calculate a formula like this:
s = 1*2*3*4+1*2*3*5+1*2*4*5+1*3*4*5+2*3*4*5
as you can see, each element in the sum is a multiplications of n-1 numbers in the set. For example in (1*2*3*4), 5 is excluded and in (1*2*3*5), 4 is excluded. I know some of the multiplications are repeated, for example (1*2) is repeated in 3 of the multiplications. How can I solve this problem with least number of multiplications.
Sorry for bad English.
Thanks.
Here is a way that does not "cheat" by replacing multiplication with repeated addition or by using division. The idea is to replace your expression with
1*2*3*4 + 5*(1*2*3 + 4*(1*2 + 3*(1 + 2)))
This used 9 multiplications for the numbers 1 through 5. In general I think the multiplication count would be one less than the (n-1)th triangular number, n * (n - 1) / 2 - 1. Here is Python code that stores intermediate factorial values to reduce the number of multiplications to just 6, or in general 2 * n - 4, and the addition count to the same (but half of them are just adding 1):
def f(n):
fact = 1
term = 2
sum = 3
for j in range(2, n):
fact *= j
term = (j + 1) * sum
sum = fact + term
return sum
The only way to find which algorithm is the fastest is to code all of them in one language, and run each using a timer.
The following would be the most straightforward answer.
def f(n):
result = 0
nList = [i+1 for i in range(n)]
for i in range(len(nList)):
result += reduce(lambda x, y: x*y,(nList[:i]+nList[i+1:]))
return result
Walkthrough - use the reduce function to multiply all list's of length n-1 and add to the variable result.
If you just want to minimise the number of multiplications, you can replace all the multiplications by additions, like this:
// Compute 1*2*…*n
mult_all(n):
if n = 1
return 1
res = 0
// by adding 1*2*…*(n-1) an entirety of n times
for i = 1 to n do
res += mult_all(n-1)
return res
// Compute sum of 1*2*…*(i-1)*(i+1)*…*n
sum_of_mult_all_but_one(n):
if n = 1
return 0
// by computing 1*2*…*(n-1) + (sum 1*2*…*(i-1)*(i+1)*…*(n-1))*n
res = mult_all(n-1)
for i = 1 to n do
res += sum_of_mult_all_but_one(n-1)
return res
Here is an answer that would work with javascript. It is not the fastest way because it is not optimized, but it should work if you want to just find the answer.
function combo(n){
var mult = 1;
var sum = 0;
for (var i = 1; i <= n; i++){
mult = 1;
for (var j = 1; j<= n; j++){
if(j != i){
mult = mult*j;
}
}
sum += mult;
}
return (sum);
}
alert(combo(n));
This is a contest problem (ACM ICPC South America 2015), it was the hardest in the problem set.
Summary: Given integers N and K, count the number of sequences a of length N consisting of integers 1 ≤ ai ≤ K, subject to the condition that for any x in that sequence there has to be a pair i, j satisfying i < j and ai = x − 1 and aj = x, i.e. the last x is preceded by x − 1 at some point.
Example: for N = 1000 and K = 100 the solution should be congruent to 265428620 modulo (109 + 7). Other examples and details can be found in the problem description.
I tried everything in my knowledge, but I need pointers to know how to do it. I even printed some lists with brute force to find the pattern, but I didn't succeed.
I'm looking for an algorithm, or formula that allows me to get to the right solution for this problem. It can be any language.
EDIT:
I solved the problem using a formula I found on the internet (someone who explained this problem). However, just because I programmed it, doesn't mean I understand it, so the question remains open. My code is here (the online judge returns Accepted):
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
ll mod = 1e9+7;
ll memo[5001][5001];
ll dp(int n, int k){
// K can't be greater than N
k = min(n, k);
// if N or K is 1, it means there's only one possible list
if(n <= 1 || k <= 1) return 1;
if(memo[n][k] != -1) return memo[n][k];
ll ans1 = (n-k) * dp(n-1, k-1);
ll ans2 = k * dp(n-1, k);
memo[n][k] = ((ans1 % mod) + (ans2 % mod)) % mod;
return memo[n][k];
}
int main(){
int n, q;
for(int i=0; i<5001; i++)
fill(memo[i], memo[i]+5001, -1);
while(scanf("%d %d", &n, &q) == 2){
for(int i=0; i<q; i++){
int k;
scanf("%d", &k);
printf("%s%lld", i==0? "" : " ", dp(n, k));
}
printf("\n");
}
return 0;
}
The most important lines are the recursive call, particularly, these lines
ll ans1 = (n-k) * dp(n-1, k-1);
ll ans2 = k * dp(n-1, k);
memo[n][k] = ((ans1 % mod) + (ans2 % mod)) % mod;
Here I show the brute force algorithm for the problem in python. It works for small numbers, but for very big numbers it takes too much time. For N=1000 and K=5 it is already infeasible (Needs more than 100 years time to calculate)(In C it should also be infeasible as C is only 100 times faster than Python). So the problem actually forces you to find a shortcut.
import itertools
def checkArr(a,K):
for i in range(2,min(K+1,max(a)+1)):
if i-1 not in a:
return False
if i not in a:
return False
if a.index(i-1)>len(a)-1-a[::-1].index(i):
return False
return True
def num_sorted(N,K):
result=0
for a in itertools.product(range(1,K+1), repeat=N):
if checkArr(a,K):
result+=1
return result
num_sorted(3,10)
It returns 6 as expected.
I have the following code
public int X(int n)
{
if (n == 0)
return 0;
if (n == 1)
return 1;
else
return (X(n- 1) + X(n- 2));
}
I want to calculate the complexity of time and memory of this code
My code consists of a constant checking if (n == 0) return 0; so this will take a constant time assume c so we have either c or c or the calculation of the recursion functions which I can't calculate
Can anyone help me in this?
To calculate the value of X(n), you are calculating X(n-1) and X(n-2)
So T(n) = T(n-1) + T(n-2);
T(0) = 1
T(1) = 1
which is exponential O(2^n)
If you want detailed proof of how it will be O(2^n), check here.
Space complexity is linear.
(Just to be precise, If you consider the stack space taken for recursion, it's O(n))