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| Original file line number | Diff line number | Diff line change |
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| """ | ||
| A Krishnamurthy number is a number whose sum total of the factorials of each digit is equal to the number itself. | ||
| Here's what I mean by that: | ||
| "145" is a Krishnamurthy Number because, | ||
| 1! + 4! + 5! = 1 + 24 + 120 = 145 | ||
| "40585" is also a Krishnamurthy Number. | ||
| 4! + 0! + 5! + 8! + 5! = 40585 | ||
| "357" or "25965" is NOT a Krishnamurthy Number | ||
| 3! + 5! + 7! = 6 + 120 + 5040 != 357 | ||
| The following function will check if a number is a Krishnamurthy Number or not and return a boolean value. | ||
| """ | ||
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| def find_factorial(n): | ||
| fact = 1 | ||
| while n != 0: | ||
| fact *= n | ||
| n -= 1 | ||
| return fact | ||
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| def krishnamurthy_number(n): | ||
| if n == 0: | ||
| return False | ||
| sum_of_digits = 0 # will hold sum of FACTORIAL of digits | ||
| temp = n | ||
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| while temp != 0: | ||
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| # get the factorial of of the last digit of n and add it to sum_of_digits | ||
| sum_of_digits += find_factorial(temp % 10) | ||
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| # replace value of temp by temp/10 | ||
| # i.e. will remove the last digit from temp | ||
| temp //= 10 | ||
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| # returns True if number is krishnamurthy | ||
| return (sum_of_digits == n) |
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| @@ -0,0 +1,36 @@ | ||
| """Magic Number | ||
| A number is said to be a magic number, | ||
| if the sum of its digits are calculated till a single digit recursively | ||
| by adding the sum of the digits after every addition. | ||
| If the single digit comes out to be 1,then the number is a magic number. | ||
| Example: | ||
| Number = 50113 => 5+0+1+1+3=10 => 1+0=1 [This is a Magic Number] | ||
| Number = 1234 => 1+2+3+4=10 => 1+0=1 [This is a Magic Number] | ||
| Number = 199 => 1+9+9=19 => 1+9=10 => 1+0=1 [This is a Magic Number] | ||
| Number = 111 => 1+1+1=3 [This is NOT a Magic Number] | ||
| The following function checks for Magic numbers and returns a Boolean accordingly. | ||
| """ | ||
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| def magic_number(n): | ||
| total_sum = 0 | ||
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| # will end when n becomes 0 | ||
| # AND | ||
| # sum becomes single digit. | ||
| while n > 0 or total_sum > 9: | ||
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| # when n becomes 0 but we have a total_sum, | ||
| # we update the value of n with the value of the sum digits | ||
| if n == 0: | ||
| n = total_sum # only when sum of digits isn't single digit | ||
| total_sum = 0 | ||
| total_sum += n % 10 | ||
| n //= 10 | ||
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| # Return true if sum becomes 1 | ||
| return total_sum == 1 | ||
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,34 @@ | ||
| # extended_gcd(a, b) modified from | ||
| # https://github.com/keon/algorithms/blob/master/algorithms/maths/extended_gcd.py | ||
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| def extended_gcd(a: int, b: int) -> [int, int, int]: | ||
| """Extended GCD algorithm. | ||
| Return s, t, g | ||
| such that a * s + b * t = GCD(a, b) | ||
| and s and t are co-prime. | ||
| """ | ||
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| old_s, s = 1, 0 | ||
| old_t, t = 0, 1 | ||
| old_r, r = a, b | ||
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| while r != 0: | ||
| quotient = old_r // r | ||
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| old_r, r = r, old_r - quotient * r | ||
| old_s, s = s, old_s - quotient * s | ||
| old_t, t = t, old_t - quotient * t | ||
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| return old_s, old_t, old_r | ||
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| def modular_inverse(a: int, m: int) -> int: | ||
| """ | ||
| Returns x such that a * x = 1 (mod m) | ||
| a and m must be coprime | ||
| """ | ||
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| s, t, g = extended_gcd(a, m) | ||
| if g != 1: | ||
| raise ValueError("a and m must be coprime") | ||
| return s % m |
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,45 @@ | ||
| def power(a: int, n: int, r: int = None): | ||
| """ | ||
| Iterative version of binary exponentiation | ||
| Calculate a ^ n | ||
| if r is specified, return the result modulo r | ||
| Time Complexity : O(log(n)) | ||
| Space Complexity : O(1) | ||
| """ | ||
| ans = 1 | ||
| while n: | ||
| if n & 1: | ||
| ans = ans * a | ||
| a = a * a | ||
| if r: | ||
| ans %= r | ||
| a %= r | ||
| n >>= 1 | ||
| return ans | ||
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| def power_recur(a: int, n: int, r: int = None): | ||
| """ | ||
| Recursive version of binary exponentiation | ||
| Calculate a ^ n | ||
| if r is specified, return the result modulo r | ||
| Time Complexity : O(log(n)) | ||
| Space Complexity : O(log(n)) | ||
| """ | ||
| if n == 0: | ||
| ans = 1 | ||
| elif n == 1: | ||
| ans = a | ||
| else: | ||
| ans = power_recur(a, n // 2, r) | ||
| ans = ans * ans | ||
| if n % 2: | ||
| ans = ans * a | ||
| if r: | ||
| ans %= r | ||
| return ans | ||
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| Original file line number | Diff line number | Diff line change |
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| @@ -1 +0,0 @@ | ||
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| def multiply(matA: list, matB: list) -> list: | ||
| """ | ||
| Multiplies two square matrices matA and matB od size n x n | ||
| Time Complexity: O(n^3) | ||
| """ | ||
| n = len(matA) | ||
| matC = [[0 for i in range(n)] for j in range(n)] | ||
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| for i in range(n): | ||
| for j in range(n): | ||
| for k in range(n): | ||
| matC[i][j] += matA[i][k] * matB[k][j] | ||
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| return matC | ||
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| def identity(n: int) -> list: | ||
| """ | ||
| Returns the Identity matrix of size n x n | ||
| Time Complecity: O(n^2) | ||
| """ | ||
| I = [[0 for i in range(n)] for j in range(n)] | ||
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| for i in range(n): | ||
| I[i][i] = 1 | ||
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| return I | ||
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| def matrix_exponentiation(mat: list, n: int) -> list: | ||
| """ | ||
| Calculates mat^n by repeated squaring | ||
| Time Complexity: O(d^3 log(n)) | ||
| d: dimesion of the square matrix mat | ||
| n: power the matrix is raised to | ||
| """ | ||
| if n == 0: | ||
| return identity(len(mat)) | ||
| elif n % 2 == 1: | ||
| return multiply(matrix_exponentiation(mat, n - 1), mat) | ||
| else: | ||
| tmp = matrix_exponentiation(mat, n // 2) | ||
| return multiply(tmp, tmp) |
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