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# 1. Method using Backtracking

def all_permutations(arr, l):
    r = len(string) - 1 # right pointer
    
    if l == r:
        print("".join(arr))
    
    for i in range(l,r+1):

pratik7368patil

# here is the solution for given problem

def count(string):
    vowel = 0    # vowel counter
    cons = 0     # consonants counter
    temp = string.lower()   # convert string to lower 
    v = ["a","e","i","o","u"]   # defining vowels
    for i in temp:
        if i in v:        # check for vowel if found increment vowel counter
            vowel += 1

pratik7368patil

# this problem can be solve using lot's of methods

# 1. Native Method 
# By visiting each character in string

def count_char(string, char):
    count = 0     # declearing veriable to count char
    
    for i in string:
        if i == char:

pratik7368patil

# if you know what is palindrome means and how to reverse a string
# then you know answer

# 1. Method using string slicing
def check_palindrome(string):
    # you can not compair case sensitive strings
    string = string.lower()
    if string == string[::-1]:
        return True
    else:

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# this problem can be solve using Hashmap with O(N) time

def check_anagram(s1, s2):
    if len(s1) != len(s2): # check for base case
        return False
    counter = [0]*26        # each character of string is between a-z
    n = ord("a")            # integer value of "a" that is 97
    for i in range(len(s1)):
        counter[ord(s1[i]) - n] += 1  # update counter at each character
        counter[ord(s2[i]) - n] -= 1  # decrease counter at each character

pratik7368patil

# This problem can be solve with O(N) time 
# but it require some extra space
# to make it more simple we can use collection module
# you can also manually add records in Hashmap or Dictionary if you don't want to use module

from collections import Counter

def find_duplicate(s):
    d = Counter(s)   # count each character ----O(N)
    

pratik7368patil

# it's very easy to remove white spaces from strings 
# it can be done using many methods but here we will disscuss some efficient one

# 1. Using replace()
s = "  h a c k you r co de " # this function doesn't modify original string 
new_s = s.replace(" ","")    # here new_s holds new string with no blank spaces at all.
print(new_s)
# the output will be "hackyourcode"

# 2. Using join() and split()

pratik7368patil

# we easily solve this using binary search algorithm with O(logN) time.

def find_peak(arr):
    l,r = 0,len(arr) - 1
    while l < r:
        m = (l+r)//2   # median of an array
        if arr[m] < arr[m+1]:   # looking for the peak element
            l = m + 1
        else:
            r = m

pratik7368patil

# searching for element in rotated sorted array
# this can be solve using binary search algorithm which takes O(logn) time to perform
# idea is if arr[mid] < arr[-1] then right side is sorted else left side is sorted

def search(arr, target):
    start, end = 0, len(arr) - 1
        
    while start <= end:
        mid = (start + end) // 2
        if arr[mid] == target:

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# our task is to find minimum and maximum number from array in one just one go
# the complexity of this program is O(n)

def min_max(arr):
    maximum = arr[0]   # assign random value from array as max
    minimum = arr[0]   # assign random value from array as min 
    
    for num in arr:
        if num > maximum:
            maximum = num