1.Implement an algorithm to determine if a string has all unique characters What if you can not use additional data structures?

The length of ACSII code of a character is 8, so we can build a array, the length is 260, to represent the hash table of all characters.

bool allCharUnique(string s){

int a[260];

memset(a,0,sizeof a);

for(int i=0;i<s.length();i++){


else return false;


return true;


2.Write code to reverse a C-Style String (C-String means that “abcd” is represented as five characters, including the null character )

consider the null character. we traverse the string to find the last index of valid char.

void reverseCStyleString(char* s){

int i=0,j=0;

char *cur = s;







3.Design an algorithm and write code to remove the duplicate characters in a string without using any additional buffer NOTE: One or two additional variables are fine An extra copy of the array
is not


Write the test cases for this method

Here is two idea. One is O(n^2), by loop all characters of the string, and each char, loop all characters whose index is bigger than current char. if
these two char is equal, then erase the char whose index is bigger.

Other idea is build a map array, map[8], represents the hash table of all characters. Following is this code:

void removeDuplicateChar(string s){


int map[]={0,0,0,0,0,0,0,0};

int ascii,index,pos;

int cur=0;

for(int i=0;i<s.length();i++){

ascii = (int)s[i];

index = ascii/32;

pos = ascii%32;

int res = map[index]&(1<<pos);









Test case: NULL , empty string, all duplicates and others.

4.Write a method to decide if two strings are anagrams or not.

One idea: sort these two strings, then look whether of these two strings are equal. O(nlogn)

Other idea: map the first string by unordered_map<char,int>. then decide the map 0f the second string is equal to the first. O(n). Following is this code:

bool isAnagrams(string a,string b){

if(a==NULL||b==NULL)return false;

if(a.length()==0&&b.length()==0)return true;

if(a.length()!=b.length())return false;

unordered_map<char,int> map;

for(int i=0;i<a.length();i++){


else map[a[i]]++;


for(int i=0;i<b.length();i++){

if(map.count(b[i])==0)return false;



if(map[b[i]]<0)return false;



return true;


5.Write a method to replace all spaces in a string with "%20".

"%20" is a string with length of 3. each space is 1 length. So the string's length is extend to s.length()+spaceCount*2. First enlarge the string,whose length
is s.length()+spaceCount*2, then we copy the original string to this new string. When we come across a space, the new string is insert '0' '2' '%'.

void replaceAllSpaces(string s){


int spaceCount=0,preLenth=s.length();

for(int i=0;i<s.length();i++){

if(s[i]==' ')spaceCount++;


for(int i=0;i<2*spaceCount;i++)s+=' ';

int cur=s.length()-1,j=preLenth-1;


if(s[j]==' '){











6.Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees Can you do this in place?

ClockWise rotate: first we turn over all pixels diagonal. then reverse each line.

other idea: rotate each layer such as loop diagram. left layer ->top layer,bottom layer -> left layer, right layer->bottom layer, top layer->right layer.

void clockwiseRotate(vector<vector<int> > matrix){


int n=matrix.size();

for(int i=0;i<n;i++){

for(int j=i;j<n;j++){




for(int i=0;i<n;i++){





7.Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column is set to 0.

we build two record array(row,column) to track in which row/column should be set to 0.

void setMatrix(vector<vector<int> > matrix){


int *row = new int[matrix.size()];

int *column = new int[matrix[0].size()];

memset(row,0,sizeof row);memset(column,0,sizeof column);

for(int i=0;i<matrix.size();i++){

for(int j=0;j<matrix[i].size();j++){






for(int i=0;i<matrix.size();i++){

for(int j=0;j<matrix[i].size();j++){






8.Assume you have a method isSubstring which checks if one word is a substring of another Given two strings,s1and s2,write code to check if s2 is a rotation of s1using only one call to isSubstring
(i e , “waterbottle” is a rotation of “erbottlewat”).

first, make sure s1.length() equals to s2.length(). if(s1.length()!=s2.length())return false;

Then, concatenate s1 with itself and look s2 is or not the substring of s1s1.

bool isRotation(string s1,string s2){

if(s1==NULL||s2==NULL)return false;

int len = s1.length();


string s1s1 = s1+s1;

return isSubstring(s1s1,s2);


return false;



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