Leetcode Ex. 204: Count Primes
Question:
I am learning Python using LeetCode problems and came across the Count Primes problem.
I have created a solution, however, the program returned ‘Time Limit Exceeded’ when submitted.
I am not sure why this happened.
Why did this happen and how could I improve my solution to make it more time efficient?
Brief:
Count the number of prime numbers less than a non-negative number, n.
Example:
Input: 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
My Solution:
class Solution:
def isPrime(self,n: int) -> int:
isPrime = False
count = 0
for i in range (1,n+1): # Iterates through integers (acting as divisors) from 1 to n
if n % i == 0: # aka if n is divisible by i
count = count + 1
if count == 2: # if n only divisible by itself and 1
isPrime = True
return isPrime
def countPrimes(self,n):
totalPrimes = 0
for i in range(1,n): # runs from 1 to n-1
answer = self.isPrime(i)
if answer == True:
totalPrimes = totalPrimes + 1
return totalPrimes
Answers:
There are various ways to count primes in a certain range (but very inefficient and time consuming for large numbers). You seem to have the right idea, we will first need a function that correctly determines if n is prime or not (Some primality tests that are fast like Fermat and Miller-Rabin give false positives). Trial division is the easiest to implement (it is only practical for small numbers however). For each input n we will check to see if there is a prime p < sqrt(n) that divides n. If so, n is composite, otherwise, n is prime. Here is my solution:
import math
def is_prime(n):
trial_divide=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
if (n < 32):
if n in trial_divide:
return(1)
else:
return(0)
if (n < 1001):
for p in trial_divide:
if (n%p==0):
return(0)
return(1)
if (n > 1000):
f = round(math.sqrt(n)+1)
trial_divide=count_primes(f)
for p in trial_divide:
if (n%p==0):
return(0)
return(1)
We haven’t defined count_primes yet so inputs above 1000 will cause an error. Here is the prime-counting function (actually, it returns the primes less than or equal to n):
def count_primes(n):
primes=[]
for i in range(n+1):
if is_prime(i):
primes.append(i)
return(primes)
Putting two and two together, we can call the count_primes function with something like 2000:
print(count_primes(2000))
which should return the primes less than 2000:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971,977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999]
To check how many there are, call
print(len(count_primes(2000)))
which gives 303 primes in this range.
def count_primes(num):
c, z = 0, 0
pnc = lambda l: 1 if (len([z + 1 for i in range(1, l + 1) if l % i == 0]) == 2) else 0
for n in range(0, num+1):
c += pnc(n)
return c
I am learning Python using LeetCode problems and came across the Count Primes problem.
I have created a solution, however, the program returned ‘Time Limit Exceeded’ when submitted.
I am not sure why this happened.
Why did this happen and how could I improve my solution to make it more time efficient?
Brief:
Count the number of prime numbers less than a non-negative number, n.
Example:
Input: 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
My Solution:
class Solution:
def isPrime(self,n: int) -> int:
isPrime = False
count = 0
for i in range (1,n+1): # Iterates through integers (acting as divisors) from 1 to n
if n % i == 0: # aka if n is divisible by i
count = count + 1
if count == 2: # if n only divisible by itself and 1
isPrime = True
return isPrime
def countPrimes(self,n):
totalPrimes = 0
for i in range(1,n): # runs from 1 to n-1
answer = self.isPrime(i)
if answer == True:
totalPrimes = totalPrimes + 1
return totalPrimes
There are various ways to count primes in a certain range (but very inefficient and time consuming for large numbers). You seem to have the right idea, we will first need a function that correctly determines if n is prime or not (Some primality tests that are fast like Fermat and Miller-Rabin give false positives). Trial division is the easiest to implement (it is only practical for small numbers however). For each input n we will check to see if there is a prime p < sqrt(n) that divides n. If so, n is composite, otherwise, n is prime. Here is my solution:
import math
def is_prime(n):
trial_divide=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
if (n < 32):
if n in trial_divide:
return(1)
else:
return(0)
if (n < 1001):
for p in trial_divide:
if (n%p==0):
return(0)
return(1)
if (n > 1000):
f = round(math.sqrt(n)+1)
trial_divide=count_primes(f)
for p in trial_divide:
if (n%p==0):
return(0)
return(1)
We haven’t defined count_primes yet so inputs above 1000 will cause an error. Here is the prime-counting function (actually, it returns the primes less than or equal to n):
def count_primes(n):
primes=[]
for i in range(n+1):
if is_prime(i):
primes.append(i)
return(primes)
Putting two and two together, we can call the count_primes function with something like 2000:
print(count_primes(2000))
which should return the primes less than 2000:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971,977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999]
To check how many there are, call
print(len(count_primes(2000)))
which gives 303 primes in this range.
def count_primes(num):
c, z = 0, 0
pnc = lambda l: 1 if (len([z + 1 for i in range(1, l + 1) if l % i == 0]) == 2) else 0
for n in range(0, num+1):
c += pnc(n)
return c