habina

def uniqueString(str):
	"Return True if the string has all unique characters."
	allChar = []
	for i in range(256):
		allChar.append(True)
	for char in str:
		if(allChar[ord(char)]):
			allChar[ord(char)] = False
		else:
			return False

habina

#include <stdio.h>
void swap(char* a, char* b);
void reverse(char* str);

int main(){
    char strArray[] = "1point3acres";
    printf("Before Reverse: %s\n", strArray);
    reverse(strArray);
    printf("After Reverse: %s\n", strArray);
}

habina

def permutationString(firstStr, secondStr):
	"Return True if the second string is a permutation of the other"
	
	"""If the length of two strings are different,
	directly return false
	"""
	if(len(firstStr) != len(secondStr)):
		return False

	"Change all characters to lowercase"

habina

def replaceSpace(string):
	"""Replace all spaces in a string with '%20'"""

	"Split string by space, save into a list"
	newList = string.split()

	"str.join concatenate list at the end"
	newString = "%20".join(newList)

	return newString

habina

def stringCompression(originalString):
	"""Compress string by using the counts of repeated characters
	Input: aabcccccaaa 
	Output: a2b1c5a3
	If the length of string bigger than the compressed string
	Then return the original string
	"""

	if(originalString == ""):
		return originalString

habina

def getIndex(x, y, width):
	"""
	Return the index in 1D array
	"""
	index = x + y * width
	return index

def printMatrix(matrix, width):
	"""
	Print Matrix row by row

habina

def getIndex(x, y, width):
	return x + y * width

def printMatrix(matrix, m, n):
	"""
	Print Matrix row by row
	"""
	for i in range(m*n):
		print(matrix[i], end="\t")
		if((i + 1) % n == 0):

habina

"""
 ============================================================================
 Question        :Write code to remove duplicates from an unsorted linked list.
                  FOLLOW UP
                  How would you solve this problem if a temporary buffer is not allowed?
 Solution        : iterate through the whole, add char to a set
                   if char already existed in set, skip this node
 Time Complexity : Average Case: O(1), Worst Case: O(N)
 Space Complexity: O(N)
 Gist Link       : https://gist.github.com/habina/3896531754d629b2032f

habina

"""
 ============================================================================
 Question        : 2.2 Implement an algorithm to find the kth to last element of a singly linked list.
 Solution        : Faster pointer move forward Kth node first, then move forward to the end alone with slow pointer,
                   until faster pointer reaches the end, return slower pointer
 Time Complexity : O(N)
 Space Complexity: O(1)
 Gist Link       : https://gist.github.com/habina/f521508cef301ac0d18d
 ============================================================================
"""

habina

"""
 ============================================================================
 Question        : Implement an algorithm to delete a node in the middle of a singly linked list, given only access to that node.
                   EXAMPLE
                   Input: the node c from the linked list a->b->c->d->e
                   Result: nothing isreturned, but the new linked list looks like a- >b- >d->e
 Solution        : Copy the following pointer content, and relink the next to the next next pointer
 Time Complexity : O(1)
 Space Complexity: O(1)
 Gist Link       : https://gist.github.com/habina/1effe686ec65dbadf1ec