Min and Max of 2 numbers

Problem: Find minimum and maximum of given 2 numbers.

Solution in Python:

def max(n, m):
    """
    Author: Mayur P Srivastava
    """

    if n >= m:
        return n
    return m

def min(n, m):
    """
    Author: Mayur P Srivastava
    """

    if n < m:
        return n
    return m

Concepts Learned: Logical conditions.

Cricket net run rate

Problem: Calculate net run rate for a cricket match, given runs scored by team A (batted first), overs played by team A, runs scored by team B, overs played by team B, winning team name, whether team A was bowled out, whether team B was bowled out, total number of overs in one innings.

Solution in Python:


from __future__ import division

import math

def net_run_rate(runsA, oversA, runsB, oversB,
                 winning_team,
                 bowled_outA=False,
                 bowled_outB=False,
                 total_overs=50,
                 eps=1e-8):
    """
    Author: Mayur P Srivastava
    Reference: http://en.wikipedia.org/wiki/Net_run_rate
    """

    assert winning_team in ['AB', 'A', 'B']

    if winning_team == 'AB':
        return 0.0, 0.0

    oversA = parse_overs(oversA)
    oversB = parse_overs(oversB)

    if abs(oversA - total_overs) > eps:
        bowled_outA = True

    if abs(oversB - total_overs) > eps and winning_team == 'A':
        bowled_outB = True

    if bowled_outA:
        oversA = total_overs
    if bowled_outB:
        oversB = total_overs

    rrA = runsA / oversA
    rrB = runsB / oversB
    
    if winning_team == 'A':
        nrrA = rrA - rrB
        nrrB = -nrrA
    else:
        nrrB = rrB - rrA
        nrrA = -nrrB

    return (nrrA, nrrB), (runsA, runsB), (oversA, oversB)


def parse_overs(o):
    completed_overs = math.floor(o)

    balls = math.floor(0.5 + 10 * (o - completed_overs))
    assert balls >= 0 and balls < 6
    
    return completed_overs + balls / 6.0

Concepts Learned: Maths

Cricket run rate

Problem: Compute current run rate and required run rate for a cricket game. Given: current score, target score, number of overs bowled, total number of overs.

Solution in Python:


from __future__ import division

import math

def calculate_run_rates(current_score, target_score, current_overs, total_overs):
    """

    Author: Mayur P Srivastava

    In overs, fraction part represents number of balls,
    e.g. 5.1, 5.2, 5.3, 5.4, 5.5, 6.0
    """

    current_overs = parse_overs(current_overs)
    total_overs   = parse_overs(total_overs)

    if current_overs > 0:
        current_rr = current_score / current_overs
    else:
        current_rr = 0

    remaining_overs = total_overs - current_overs
    runs_to_win     = target_score - current_score + 1

    required_rr = runs_to_win / remaining_overs


def parse_overs(o):
    completed_overs = math.floor(o)

    balls = math.floor(0.5 + 10 * (o - completed_overs))
    assert balls >= 0 and balls < 6
    
    return completed_overs + balls / 6.0

Concepts Learned: Maths

Matrix Addition

Problem: Add the given 2 matrixes.

Solution in Python:


def add(A, B):
    """
    Author: Mayur P Srivastava
    """

    m1, n1 = shape(A)
    m2, n2 = shape(B)

    if not can_add(m1, n1, m2, n2):
        return None

    C = create_matrix(m1, n2)

    for i in range(m1):
        for j in range(n2):
            C[i][j] = A[i][j] + B[i][j]

    return C

def shape(A):
    m = len(A)   
    n = 0
    for row in A:
        n2 = len(row)
        if n == 0:
            n = n2
        elif n != n2:
            assert False

    return m, n

def can_add(m1, n1, m2, n2):
    return m1 == m2 and n1 == n2

def create_matrix(m, n, value=0):
    matrix = []
    for i in range(m):
        row = []
        for j in range(n):
            row.append(value)
        matrix.append(row)
    return matrix

Concepts Learned: Nested loops and Maths

Matrix Multiplication

Problem: Multiply the given 2 matrixes.

Solution in Python:

def multiply(A, B):
    """
    Author: Mayur P Srivastava
    """


    m1, n1 = shape(A)
    m2, n2 = shape(B)

    if not can_multiply(m1, n1, m2, n2):
        return None

    C = create_matrix(m1, n2)

    for i in range(m1):
        for j in range(n2):
            c = 0
            for k in range(n1):
                c += A[i][k] * B[k][j]
            C[i][j] = c

    return C

def shape(A):
    m = len(A)   
    n = 0
    for row in A:
        n2 = len(row)
        if n == 0:
            n = n2
        elif n != n2:
            assert False

    return m, n

def can_multiply(m1, n1, m2, n2):
    return n1 == m2

def create_matrix(m, n, value=0):
    matrix = []
    for i in range(m):
        row = []
        for j in range(n):
            row.append(value)
        matrix.append(row)
    return matrix

Concepts Learned: Maths

Divisibility

Problem: Check whether a given number n is divisible by another number m.

Solution in Python:

def is_divisible(n, m):
    """
    Author: Mayur P Srivastava
    """


    if m == 0:
        return False

    return n % m == 0

Concepts Learned: Maths

Midpoint

Problem: Compute the mid-point of two 2-dimensional points.

Solution in Python:

def compute_midpoint(x1, y1, x2, y2):
    """
    Author: Mayur P Srivastava
    """

    return (x1 + x2) / 2.0, (y1 + y2) / 2.0

Concepts Learned: Maths

Check whether matrix multiplication is possible

Problem: Given dimensions of 2 matrixes, check whether they can be multiplied.

Solution in Python:

def can_multiply(m1, n1, m2, n2):
    """
    Author: Mayur P Srivastava
    """

    return n1 == m2

Concepts Learned: Maths.

Triangle

Problem: Check whether given three 2-dimensional points form a triangle.


Solution in Python:

from __future__ import division

def check_triangle(x1, y1, x2, y2, x3, y3, eps=1e-16):
    """
    Author: Mayur P Srivastava
    """

    dx12 = x1 - x2
    dy12 = y1 - y2
    dx23 = x2 - x3
    dy23 = y2 - y3

    if dx12 == 0 and dx23 == 0:
        return False
    if abs(dx12) < eps and abs(dx23) < eps:
        return False
    if dx12 == 0 or abs(dx12) < eps:
        return True
    if dx23 == 0 or abs(dx23) < eps:
        return True

    slope12 = dy12 / dx12
    slope23 = dy23 / dx23

    return not (abs(slope12 - slope23) < eps)

Concepts Learned: Maths.

Sum of digits

Problem: Compute the sum of digits for the given number as a single digit number.

Example:

sum_of_digits(5674) = 4 (5+6+7+4=22, 2+2=4)


Solution in Python:

def sum_of_digits(n):

    """
    Author: Mayur P Srivastava
    """

    while n > 9:
        n = sum_of_digits2(n)
    return n


def sum_of_digits2(n):
    sum = 0

    while n > 0:
        sum += n % 10
        n = n // 10

    return sum

Concepts Learned: Loops and numbers.

Euclidean Distance

Problem: Compute the Euclidean distance between two 2-dimensional points.

Solution in Python:

import math

def compute_euclidean_distance(x1, y1, x2, y2):
    """
    Author: Mayur P Srivastava
    """

    dx = x2 - x1
    dy = y2 - y1
    return math.sqrt(dx*dx + dy*dy)

Concepts Learned: Numbers.

Quadratic Equation

Problem: Compute the values of x in the given quadratic equations ax^2+ bx + c for given a, b, c.


Solution in Python:

from __future__ import division

import math

def solve(a, b, c, eps=1e-12):
    """
    Author: Mayur P Srivastava
    """

    if abs(a) < eps:
        return None, None

    discriminant = b * b - 4 * a * c
    if discriminant < 0:
        return None, None

    return (-b + math.sqrt(discriminant)) / (2*a), (-b - math.sqrt(discriminant)) / (2*a)

Concepts Learned: Numbers.

Reverse the given string.

Problem: Reverse the given string.

Solution in Python:


def reverse(s):

    """
    Author: Mayur P Srivastava
    """

    arr = list(s)    
    n   = len(arr)

    for i in range(n // 2):
        arr[i], arr[n-i-1] = arr[n-i-1], arr[i]

    return "".join(arr)


Concepts Learned: Single loop.

Capitalize first letter of each word in a sentence.

Problem: Capitalize first letter of each word in a sentence.

Example:


capitalize_words('This is a sample sentence')
This Is A Sample Sentence

Solution in Python:


def capitalize_words(sentence):
    """
    Author: Mayur P Srivastava
    """

    arr = list(sentence)

    start_of_word = False

    for i in range(len(arr)):
        c = arr[i]
        if c == ' ' or c == '\t' or c == '\n':
            start_of_word = True
        else:
            if start_of_word:
                ascii = ord(c)
                start_of_word = False
                if ascii >= 97 and ascii <= 122:
                    arr[i] = chr(ascii - 32)

    return "".join(arr)

Concepts Learned: Single loop.

Binary fraction to Decimal

Problem: Convert the given binary fraction to decimal.

Example:


binary_to_decimal_fraction(".1")
0.5

binary_to_decimal_fraction("0.0001100110011001")
0.0999908447265625

binary_to_decimal_fraction("0.01")
0.25

binary_to_decimal_fraction("0.0101100110011001")
0.3499908447265625

Solution in Python:

from __future__ import division

def binary_to_decimal_fraction(binary_str):
    """
    Author: Mayur P Srivastava
    """

    if binary_str[0] == '.':

        start_pos = 1

    elif binary_str[0] == '0' and binary_str[1] == '.':

        start_pos = 2

    else:

        assert False

    div     = 1

    decimal = 0

    for pos in range(start_pos, len(binary_str)):

        c = int(binary_str[pos])

        div *= 2

        decimal += c / div

    return decimal

Concepts Learned: Single loop, number system.

Decimal fraction to Binary

Problem: Convert the given fractional decimal to binary.

Example:


decimal_to_binary_fraction(.0)
0

decimal_to_binary_fraction(.1)
0.0001100110011001

decimal_to_binary_fraction(.25)
0.01

decimal_to_binary_fraction(.35)
0.0101100110011001



Solution in Python:

import math

def decimal_to_binary_fraction(n, max_digits=16, eps=1e-12):

    """
    Author: Mayur P Srivastava
    """

    assert n >= 0 and n < 1 

    if n < eps:

        return "0"

    binary_str = "0."

    num_digits = 0

    while num_digits < max_digits and abs(n) > eps:

        n = n * 2

        if n >= 1:

            binary_str += "1"

            n -= 1

        else:

            binary_str += "0"

        num_digits += 1

    return binary_str

Concepts Learned: Single loops, number system

Hexadecimal to Decimal

Problem: Convert the given hexadecimal string into decimal.

Example:


hex_to_decimal('198F')
6543

hex_to_decimal('19AF')

6575

Solution in Python:

def hex_to_decimal(hex_str):
    """
    Author: Mayur P Srivastava
    """

    decimal_number = 0

    n = len(hex_str)

    for i in range(n - 1, -1, -1):

        c = ord(hex_str[i])

        if c >= 65 and c <= 70:

            c = 10 + c - 65

        elif c >= 97 and c <= 102:

            c = 10 + c - 97

        elif c >= 48 and c <= 57:

            c = c - 48

        else:

            assert False

        decimal_number += 16**(n-i-1) * c

    return decimal_number

Concepts Learned: Single loop, number system.

Decimal to Hexadecimal

Problem: Convert the given decimal whole number to hexadecimal.

Example:


decimal_to_hex(32)
20

decimal_to_hex(31)
1F

decimal_to_hex(6543)
198F

Solution in Python:

def decimal_to_hex(n):
    """
    Author: Mayur P Srivastava
    """

    if n == 0:

        return "0";

    hex_str = "";

    while n > 0:

        c = n % 16

        n = n // 16

        if c > 9:

            c = chr(65 + c - 10)

        else:

            c = str(c)

        hex_str = c + hex_str

    return hex_str

Concepts Learned: Single loops, number systems

Palindrome Number

Problem: Check whether a whole number is palindrome or not.


Solution in Python:


import math

def check_for_palindrome_number(n):
    """
    Author: Mayur P Srivastava
    """

    assert n >= 0

    n_digits = int(math.floor(math.log10(n)))

    upper_mult = 10 ** n_digits
    lower_mult = 10

    while lower_mult <= upper_mult:
        lower = (n % lower_mult) // (lower_mult // 10)
        upper = (n // upper_mult) % 10

        if lower != upper:
            return False

        lower_mult *= 10
        upper_mult //=10

    return True

Concepts Learned: Single Loops, numbers.

Palindrome Word

Problem: Check whether a word is palindrome or not. Make sure it is not case-sensitive, i.e. AbCBa should be a palindrome.

Solution in Python:


import math

UPPER_TO_LOWER_OFFSET = ord('a') - ord('A')

def check_for_palindrome(word):
    """
    Author: Mayur P Srivastava
    """

    n = len(word)

    for i in range(int(math.floor(n // 2))):
        if not canonical_equals(word[i], word[n - i - 1]):
            return False

    return True


def canonical_equals(letter1, letter2):
    ascii1 = canonical_ascii(ord(letter1))
    ascii2 = canonical_ascii(ord(letter2))
    return ascii1 == ascii2


def canonical_ascii(ascii):
    if ascii >= 65 and ascii <= 90:
        return ascii + UPPER_TO_LOWER_OFFSET
    return ascii

Concepts Learned: Single loop.

Binary to Decimal

Problem: Convert the given binary integer to decimal.

Example:


binary_to_decimal(111)
7

binary_to_decimal(10101)
21

binary_to_decimal(111110)
62


Solution in Python:


def binary_to_decimal(binary_str):
    """
    Author: Mayur P Srivastava
    """

    decimal_number = 0

    n = len(binary_str)

    for i in range(n - 1, -1, -1):
        decimal_number += 2**(n-i-1) * int(binary_str[i])

    return decimal_number



Concepts Learned: Single loop, binary numbers.

Decimal to Binary

Problem: Convert the given decimal integer to binary.

Example:


decimal_to_binary(5)
101

decimal_to_binary(11)
1011

decimal_to_binary(35)
100011


Solution in Python:


def decimal_to_binary(n):
    """
    Author: Mayur P Srivastava
    """

    if n == 0:
        return "0";

    binary_str = "";

    while n > 0:
        bit = n % 2
        n = n // 2
        binary_str = str(bit) + binary_str

    return binary_str

Concepts Learned: Single loops, binary numbers.

Nested Loop: Trigonometric Functions

Problem: Plot the given trigonometric function.

Example:


trignometry_plot(fn=math.sin, plot_char="#")
                                        #
                                              #
                                                    #
                                                         #
                                                              #
                                                                 #
                                                                    #
                                                                     #
                                                                     #
                                                                    #
                                                                 #
                                                              #
                                                         #
                                                    #
                                              #
                                        #
                                 #
                           #
                      #
                 #
              #
           #
          #
          #
           #
              #
                 #
                      #
                           #
                                 #
                                       #

Solution in Python:


import math

def trignometry_plot(fn=math.sin, n_x_steps=30, y_orig=40,
                     y_range=30, plot_char="."):
    """
    Author: Mayur P Srivastava
    """

    pi = 3.14159

    for xi in range(n_x_steps+1):
        x = 2 * pi / n_x_steps * xi
        y = y_orig + y_range * fn(x)
        row = ""
        if y >= 0:
            for i in range(y):
                row += " "
            row += plot_char

        print row

Concepts Learned: Nested loops, trigonometric functions.