I have a assingment were I need to code and decode txt files, for example: hello how are you? has to be coded as hel2o how are you? and aaaaaaaaaajkle as a10jkle.
while ( ! invoer.eof ( ) ) {
if (kar >= '0' && kar <= '9') {
counter = kar-48;
while (counter > 1){
uitvoer.put(vorigeKar);
counter--;
}
}else if (kar == '/'){
kar = invoer.get();
uitvoer.put(kar);
}else{
uitvoer.put(kar);
}
vorigeKar = kar;
kar = invoer.get ( );
}
but the problem I have is if need to decode a12bhr, the answer is aaaaaaaaaaaabhr but I can't seem to get the 12 as number without problems, I also can't use any strings or array's.
c++
I believe that you are making following mistake: imagine you give a32, then you read the character a and save it as vorigeKar (previous character, I am , Flemish so I understand Dutch :-) ).
Then you read 3, you understand that it is a number and you repeat vorigeKar three times, which leads to aaa. Then you read 2 and repeat vorigeKar two times, leading to aaaaa (five times, five equals 3 + 2).
You need to learn how to keep on reading numeric characters, and translate them into complete numbers (like 32, or 12 in your case).
Like #Dominique said in his answers, You're doing it wrong.
Let me tell you my logic, you can try it.
Pesudo Code + Logic:
Store word as a char array or string, so that it'll be easy to print at last
Loop{
Read - a //check if it's number by subtracting from '0'
Read - 1 //check if number = true. Store it in int res[] = res*10 + 1
//Also store the previous index in an index array(ie) index of char 'a' if you encounter a number first time.
Read - 2 //check if number = true. Store it in res = res*10 + 2
Read - b , h and so on till "space" character
If you encounter another number, then store it's previous character's index in index array and then store the number in a res[] array.
Now using index array you can get the index of your repeating character to be printed and print it for it's corresponding times which we have stored in the result array.
This goes for the second, third...etc:- numbers in your word till the end of the word
}
First, even though you say you can't use strings, you still need to know the basic principle behind how to turn a stream of digit characters into an integer.
Assuming the number is positive, here is a simple function that turns a series of digits into a number:
#include <iostream>
#include <cctype>
int runningTotal(char ch, int lastNum)
{
return lastNum * 10 + (ch -'0');
}
int main()
{
// As a test
char s[] = "a123b23cd1/";
int totalNumber = 0;
for (size_t i = 0; s[i] != '/'; ++i)
{
char digit = s[i]; // This is the character "read from the file"
if ( isdigit( digit) )
totalNumber = runningTotal(digit, totalNumber);
else
{
if ( totalNumber > 0 )
std::cout << totalNumber << "\n";
totalNumber = 0;
}
}
std::cout << totalNumber;
}
Output:
123
23
1
So what was done? The character array is the "file". I then loop for each character, building up the number. The runningTotal is a function that builds the integer from each digit character encountered. When a non-digit is found, we output that number and start the total from 0 again.
The code does not save the letter to "multiply" -- I leave that to you as homework. But the code above illustrates how to take digits and create the number from them. For using a file, you would simply replace the for loop with the reading of each character from the file.
I need to write a function string add(string a, string b)
where a and b are strings representing integers and the function add(a,b) returns a string
representing their sum.
Strings a and b can have a maximum of 100 characters.
I have tried different ways but failed, here is where I'm standing right now.
So I took the 2 strings and I tried adding each digit starting from last.
If in the array at [i] it's more than 10, then add 1 to [i-1], and mod it by 10 to get the last digit.
The return is empty:
string add(string a, string b){
int arrA[a.length()];
int arrB[b.length()];
string Res=" ";
//99999999 2222222
if(a.length()>=b.length()){
//i=7
for (int i=b.length();i>=0;i--){
arrA[i] = (int) (a[i]-'0') + (int) (b[i]-'0');
}
for(int i=b.length()-1;i>=1;i--)
Res[i]=arrA[i];
for(int i=a.length()-1;i>=1;i--){
if (arrA[i]>=10){
arrA[i]=arrA[i]%10;
arrA[i-1]=arrA[i-1]+1;}
}
}
else{
for (int i=a.length();i>=0;i--){
arrB[i] = (int) (a[i]-'0') + (int) (b[i]-'0');
}
for(int i=b.length()-1;i>=1;i--)
Res[i]=arrB[i];
for(int i=b.length()-1;i>=1;i--){
if (arrB[i]>=10){
arrB[i]=arrB[i]%10;
arrB[i-1]=arrB[i-1]+1;}
}
}
return Res;
}
Thank you in advance!
Think about how you would do this with pencil and paper, then write code to do the same thing.
You have two strings of digits. Start at the right, add the two digits, and if the result overflows, subtract 10 and note that you have a carry. Store the resulting digit. Move one place to the left. Repeat until done. If you run out of digits in one string, just pretend that you've got zeros for the rest of the digits.
Note that each digit in the input is the character representation of the digit. To get the numeric value, subtract '0' from each digit. Once you have the result, convert it to a character by adding '0'.
string add(string a, string b) {
int c = stoi(a) + stoi(b);
return to_string(c);
}
I am creating a small game where the user will have hints(Characters of a string) to guess the word of a string. I have the code to see each individual character of the string, but is it possible that I can see those characters printed out randomly?
string str("TEST");
for (int i = 0; i < str.size(); i++){
cout <<" "<< str[i];
output:T E S T
desired sample output: E T S T
Use random_shuffle on the string:
random_shuffle(str.begin(), str.end());
Edits:
C++11 onwards use:
auto engine = std::default_random_engine{};
shuffle ( begin(str), end(str), engine );
Use the following code to generate the letters randomly.
const int stl = str.size();
int stl2 = stl;
while (stl2 >= 0)
{
int r = rand() % stl;
if (str[r] != '0')
{
cout<<" "<<str[r];
str[r] = '0';
stl2--;
}
}
This code basically generates the random number based on the size of the String and then prints the character placed at that particular position of the string.
To avoid the reprinting of already printed character, I have converted the character printed to "0", so next time same position number is generated, it will check if the character is "0" or not.
If you need to preserve the original string, then you may copy the string to another variable and use it in the code.
Note: It is assumed that string will contain only alphabetic characters and so to prevent repetition, "0" is used. If your string may contain numbers, you may use a different character for comparison purpose
Okay so here is what I'm trying to accomplish.
First of all below table is just an example of what I created, in my assignment I'm not suppose to know any of these. Which means I don't know what they will pass and what is the length of each string.
I'm trying to accomplish one task is to get to be able to compare part of the string
//In Array `phrase` // in array `word`
"Backdoor", 0 "mark" 3 (matches "Market")
"DVD", 1 "of" 2 (matches "Get off")
"Get off", 2 "" -1 (no match)
"Market", 3 "VD" 1 (matches "DVD")
So as you can see from the above codes from the left hand side is the set of array which I store them in my class and they have upto 10 words
Here is the class definition.
class data
{
char phrase[10][40];
public:
int match(const char word[ ]);
};
so I'm using member function to access this private data.
int data::match(const char word[ ])
{
int n,
const int wordLength = strlen(word);
for (n=0 ; n <= 10; n++)
{
if (strncmp (phrase[n],word,wordLength) == 0)
{
return n;
}
}
return -1;
}
The above code that I'm trying to make it work is that it should match and and return if it found the match by returning the index n if not found should always return -1.
What happen now is always return 10.
You're almost there but your code is incomplete so I''m shootin in the dark on a few things.
You may have one too many variables representing an index. Unless n and i are different you should only use one. Also try to use more descriptive names, pos seems to represent the length of the text you are searching.
for (n=0 ; n <= searchLength ; n++)
Since the length of word never changes you don't need to call strlen every time. Create a variable to store the length in before the for loop.
const int wordLength = strlen(word);
I'm assuming the text you are searching is stored in a char array. This means you'll need to pass a pointer to the first element stored at n.
if (strncmp (&phrase[n],word,wordLength) == 0)
In the end you have something that looks like the following:
char word[256] = "there";
char phrase[256] = "hello there hippie!";
const int wordLength = strlen(word);
const int searchLength = strlen(phrase);
for (int n = 0; n <= searchLength; n++)
{
// or phrase + n
if (strncmp(&phrase[n], word, wordLength) == 0)
{
return n;
}
}
return -1;
Note: The final example is now complete to the point of returning a match.
I'm puzzled about your problem. There are some cases unclear. For eaxmple abcdefg --- abcde Match "abcde"? how many words match? any other examples, abcdefg --- dcb Match "c"?and abcdefg --- aoodeoofoo Match "a" or "adef"? if you want to find the first matched word, it's OK and very simple. But if you are to find the longest and discontinuous string, it is a big question. I think you should have a research about LCS problem (Longest Common Subsequence)
For example :
String is : abcd
shortest palindrome is abcdcba is the solution
longer palindrome can be : abcddcba
another example:
String : aaaab
shortest palindrome is aaaabaaaa
longer palindrome can be aaaaabbaaaa
Restrictions : you can only add characters in the end.
Just append the reverse of initial substrings of the string, from shortest to longest, to the string until you have a palindrome. e.g., for "acbab", try appending "a" which yields "acbaba", which is not a palindrome, then try appending "ac" reversed, yielding "acbabca" which is a palindrome.
Update: Note that you don't have to actually do the append. You know that the substring matches since you just reversed it. So all you have to do is check whether the remainder of the string is a palindrome, and if so append the reverse of the substring. Which is what Ptival wrote symbolically, so he should probably get the credit for the answer. Example: for "acbab", find the longest suffix that is a palindrome; that is "bab". Then append the remainder, "ac", in reverse: ac bab ca.
My guess for the logic:
Say you string is [a1...an] (list of characters a1 to an)
Find the smallest i such that [ai...an] is a palindrome.
The smallest palindrome is [a1 ... a(i-1)] ++ [ai ... an] ++ [a(i-1) ... a1]
where ++ denotes string concatenation.
Some pseudo code, to leave at least a bit of work on you:
def shortPalindrome(s):
for i in range(len(s)):
pal = s + reverse(s[0:i])
if isPalindrome(pal):
return pal
error()
Python code, should be easy to convert to C:
for i in range(1, len(a)):
if a[i:] == a[i:][::-1]:
break
print a + a[0:i][::-1]
I was also asked the same question recently, and here is what I wrote for my interview:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int isPalin ( char *str ) {
int i, len=strlen(str);
for (i=0; i<len/2; ++i)
if (str[i]!=str[len-i-1])
break;
return i==len/2;
}
int main(int argc, char *argv[]) {
if (argc!=2)
puts("Usage: small_palin <string>");
else {
char *str = argv[1];
int i=0, j, len=strlen(str);
while ( !isPalin(str+i) )
++i;
char *palin = malloc(len+1+i);
*(palin+len+1+i) = '\0';
strcpy(palin,str);
for (i=i-1, j=0; i>=0; --i, ++j)
*(palin+len+j) = str[i];
puts(palin);
}
return 0;
}
I feel that the program would have been more structured had I written an strrev() function and checked palindrome using strcmp(). This would enable me to reverse the starting characters of the source string and directly copy it using strcpy().
The reson why I went with this solution is that before this question I was asked to check for palindrome and I already had that isPalin() in paper. Kind of felt using existing code would be better !!
From the examples you shown looks like the longest palindrome is the original string concatenated with its reverse, and the shortest is the original string concatenated with its reverse except for the first character. But I'm pretty sure you want something more complex. Perhaps you can give better examples?
if string is made of k chars, I think you should add to this string the reversed (k-1) chars...
Below is my answer for another case: shortest palindrome by attaching characters to the front. So your task is to understand the algorithm and modify it appropriately.
Basically, it states that from a string s find the shortest palindrome by adding some characters to the front of s.
If you have never tried to solve this problem, I suggest that you solve it, and it will help you improve your problem solving skill.
After solving it, I kept looking for better solutions. I stumbled upon another programmer's solution. It is in python, and really neat. It is really interesting, but later I found out it was wrong.
class Solution:
# #param {string} s
# #return {string}
def shortestPalindrome(self, s):
A=s+s[::-1]
cont=[0]
for i in range(1,len(A)):
index=cont[i-1]
while(index>0 and A[index]!=A[i]):
index=cont[index-1]
cont.append(index+(1 if A[index]==A[i] else 0))
print cont[-1]
return s[cont[-1]:][::-1]+s
I myself looked at the Solution and saw it's interesting idea. At first, the algorithm concatenates the string and its reversed version. Then the following steps are similar to the steps for building KMP-table (or failure function) using in KMP algorithm. Why does this procedure work?
If you know KMP text searching algorithm, you will know its "lookup table" and steps to build it. Right now, I just show one important use of the table: it can show you the longest prefix of a string s that is also suffix of s (but not s itself). For example, "abcdabc" has the longest prefix which is also a suffix: "abc" (not "abcdabc" since this is the entire string!!!). To make it fun, we call this prefix is "happy substring" of s. So the happy substring of "aaaaaaaaaa" (10 a's ) is "aaaaaaaaa" (9 a's).
Now we go back and see how finding happy sub string of s can help solve the shortest palindrome problem.
Suppose that q is the shortest string added to the front of s to make the string qs is a palindrome. We can see that obviously length(q) < length(s) since ss is also a palindrome. Since qs is a palindrome, qs must end with q, or s = p+q where p is a sub string of s. Easily we see that p is also a palindrome. Therefore, in order to have shortest qs, q needs to be shortest. In turn, p is the longest palindromic sub string of s.
We call s' and q' are the reversed strings of s and q respectively. We see that s = pq, s' = q'p since p is a palindrome. So ss' = pqq'p . Now we need to find the longest p. Eureka! This also means that p is a happy sub string of the string ss'. That's how the above algorithm works!!!
However, after some thought, the above algorithm has some loophole. p is not a happy sub string of ss'! In fact, p is the longest prefix that is also a suffix of ss', but the prefix and suffix must not overlap each other. So let's make it more fun, we call "extremely happy sub string" of a string s is the longest sub string of s that is a prefix and also a suffix and this prefix and suffix must not overlap. On the other word, the "extremely happy sub string" of s must have length less than or equal half length of s.
So it turns out the "happy sub string" of ss' is not always "extremely happy sub string" of ss'. We can easily construct an example: s = "aabba". ss'="aabbaabbaa". The happy sub string of "aabbaabbaa" is "aabbaa", while the extremely happy sub string of "aabbaabbaa" is "aa". Bang!
Hence, the correct solution should be as following, based on the observation that length(p) <= length(ss')/2.
class Solution:
# #param {string} s
# #return {string}
def shortestPalindrome(self, s):
A=s+s[::-1]
cont=[0]
for i in range(1,len(A)):
index=cont[i-1]
while(index>0):
if(A[index]==A[i]):
if index < len(s):
break
index=cont[index-1]
cont.append(index+(1 if A[index]==A[i] else 0))
print cont[-1]
return s[cont[-1]:][::-1]+s
Hooray!
As you can see, algorithms are interesting!
The link to the article I wrote here
It looks like the solutions outlined here are O(N^2) (for each suffix X of the reversed string S, find if S + X is a palindrome).
I believe there is a linear, i.e O(N) solution for this problem. Consider the following statement: the only time where you would append less characters than S.Length - 1 is when the string already contains a partial palindrome, so it will be in the form of NNNNNPPPPPP, where PPPPP represent a palindrome. This means that if we can find the largest trailing palindrome, we can solve it linearly by concatenating the reverse of NNNNN to the end.
Finally, there exists a famous algorithm (Manacher, 1975) that finds the longest (and in fact, all) of the palindromes contained in a string (there is a good explanation here). It can be easily modified to return the longest trailing palidrome, thus giving a linear solution for this problem.
If anyone is interested, here is the full code for a mirror problem (append characters at the beginning):
using System.Text;
// Via http://articles.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html
class Manacher
{
// Transform S into T.
// For example, S = "abba", T = "^#a#b#b#a#$".
// ^ and $ signs are sentinels appended to each end to avoid bounds checking
private static string PreProcess(string s)
{
StringBuilder builder = new StringBuilder();
int n = s.Length;
if (n == 0) return "^$";
builder.Append('^');
for (int i = 0; i < n; i++)
{
builder.Append('#');
builder.Append(s[i]);
}
builder.Append('#');
builder.Append('$');
return builder.ToString();
}
// Modified to return only the longest palindrome that *starts* the string
public static string LongestPalindrome(string s)
{
string T = PreProcess(s);
int n = T.Length;
int[] P = new int[n];
int C = 0, R = 0;
for (int i = 1; i < n - 1; i++)
{
int i_mirror = 2 * C - i; // equals to i' = C - (i-C)
P[i] = (R > i) ? Math.Min(R - i, P[i_mirror]) : 0;
// Attempt to expand palindrome centered at i
while (T[i + 1 + P[i]] == T[i - 1 - P[i]])
P[i]++;
// If palindrome centered at i expand past R,
// adjust center based on expanded palindrome.
if (i + P[i] > R)
{
C = i;
R = i + P[i];
}
}
// Find the maximum element in P.
int maxLen = 0;
int centerIndex = 0;
for (int i = 1; i < n - 1; i++)
{
if (P[i] > maxLen
&& i - 1 == P[i] /* the && part forces to only consider palindromes that start at the beginning*/)
{
maxLen = P[i];
centerIndex = i;
}
}
return s.Substring((centerIndex - 1 - maxLen) / 2, maxLen);
}
}
public class Solution {
public string Reverse(string s)
{
StringBuilder result = new StringBuilder();
for (int i = s.Length - 1; i >= 0; i--)
{
result.Append(s[i]);
}
return result.ToString();
}
public string ShortestPalindrome(string s)
{
string palindrome = Manacher.LongestPalindrome(s);
string part = s.Substring(palindrome.Length);
return Reverse(part) + palindrome + part;
}
}
using System;
using System.Collections.Generic;
using System.Linq;
public class Test
{
public static void shortPalindrome(string [] words){
List<string> container = new List<string>(); //List of Palindromes
foreach (string word in words )
{
char[] chararray = word.ToCharArray();
Array.Reverse(chararray);
string newText = new string(chararray);
if (word == newText) container.Add(word);
}
string shortPal=container.ElementAt(0);
for(int i=0; i<container.Count; i++)
{
if(container[i].Length < shortPal.Length){
shortPal = container[i];
}
}
Console.WriteLine(" The Shortest Palindrome is {0}",shortPal);
}
public static void Main()
{
string[] word = new string[5] {"noon", "racecar","redivider", "sun", "loss"};
shortPalindrome(word);
}
}
Shortest palindrome -
Reverse iterate from last positon + 1 to beginning
Push_back the elements
#include <iostream>
#include <string>
using namespace std ;
int main()
{
string str = "abcd" ;
string shortStr = str ;
for( string::reverse_iterator it = str.rbegin()+1; it != str.rend() ; ++it )
{
shortStr.push_back(*it) ;
}
cout << shortStr << "\n" ;
}
And longer palindrome can be any longer.
Ex: abcd
Longer Palindrome - abcddcba, abcdddcba, ...