I need help solving this issue, I am expecting a number to come out but get this error instead
Line 65: Char 5: error: conflicting types for 'main' int main(int argc, char *argv[]) { ^ Line 47: Char 5: note: previous definition is here int main() ^ 1 error generated.
here is some of my code
class Solution {
public:
int value(char r){
if (r == 'I')
return 1;
if (r == 'V')
return 5;
if (r == 'X')
return 10;
if (r == 'L')
return 50;
if (r == 'C')
return 100;
if (r == 'D')
return 500;
if (r == 'M')
return 1000;
return -1;
}
int romanToInt(string& s) {
int ret = 0;
for (int i = 0; i < s.length(); i++) {
int s1 = value(s[i]);
if (i + 1 < s.length()) {
int s2 = value(s[i + 1]);
if (s1 >= s2) {
ret = ret + s1;
}
else {
ret = ret + s2 - s1;
i++;
}
}
else {
ret = ret + s1;
}
}
return ret;
}
};
int main()
{
Solution m;
string str = "III";
cout << "Integer form of Roman Numeral is " << m.romanToInt(str) << endl;
return 0;
}
I am trying to use a pointer array where it reads the line letter by letter and recognizing the value of the letter in the function value(), I think I understand that my main needs to be formatted differently in order to do this task but I am a little stuck on how to do so.
You have probably defined int main twice. Considering you have an error on line 65 while your code is less than 60 lines long I would assume there is more code than what was copied here.
Related
Code wars Question:(Isogram)
An Isogram is a word that has no repeating letters,consecutive or non-consecutive.Implement a function that determines whether a string that contains only letters is an Isogram. Assume the empty string is an Isogram.Ignore letter case.
Test Cases:
"Dermatoglyphics"-->true
"aba" --> false
"balLoon"--> false
My Code:
#include <bits/stdc++.h>
using namespace std;
bool isIsogram(string s)
{
for (int i = 0; i < s.length(); i++)
{
for (int j = i + 1; j < s.length(); j++)
{
if (s[i] == s[j] || s[i] == s[j] + 32 || s[i] == s[j] - 32)
{
return false;
}
return true;
}
}
}
int main()
{
cout << isIsogram("abd") << endl;
return 0;
}
If there is an optimized solution for this problem?
Can we reduce time complexity of this problem?
Your code is of O(N^2) Time Complexity.
Yes, there is an Optimized Way of doing it.
Here's the optimized solution: O(N)
#include <bits/stdc++.h>
using namespace std;
bool Isogram(string s)
{
int N = 125;
bool check[N];
for (int i = 0; i < 125; i++)
{
check[i] = false;
}
for (int i = 0; i < s.size(); i++)
{
if (check[s[i]] != false)
{
return false;
}
check[s[i]] = true;
if (s[i] + 32 < N)
check[s[i] + 32] = true;
if (s[i] - 32 >= 0)
check[s[i] - 32] = true;
}
return true;
}
int main()
{
string s;
cin >> s;
if (Isogram(s) == true)
cout << "The word " << s << " is an Isogram" << endl;
else
cout << "The word " << s << " is not an Isogram" << endl;
return 0;
}
We made an array of 125 elements because the ASCII value of lowercase particles lies in the range from 65 to 90, and for uppercase, it is 97 to 122.
And we are adding 32 and subtracting 32 to cover the uppercase alphabet (if the element is in lowercase) and vice versa.
Hope you get it...
I am doing the CS50 course and am on week 2. One of the problems of week 2 is called "Caesar". Essentially you have to write code which cyphers text by shifting letters that use the users inputted preferred number. After running my code I keep getting this error
"error: implicitly declaring library function 'strlen' with
type 'unsigned long (const char *)'
[-Werror,-Wimplicit-function-declaration]
for(i = 0, l = strlen(text); i < n; i++)"
This is the code:
int main(int argc, string argv[])
{
string n = argv[1];
int y = argc;
int key = get_int("./caesar ");//getting the number from the user
int k = (key);//assigning key a variable name.
string text = get_string("plaintext: ");//letting the user input their text.
if (key < 1)//Trying to make limit for acceptable input.
{
printf("ERROR");
return 1;
}
int l;
int i;
//for loop containing the encipher process
for(i = 0, l = strlen(text); i < n; i++)
{
if(isalpha(i))
{
if (isupper[i])
{
printf("ciphertext: %c",(text[i] + k)%26 + 65);
}
else (islower[i])
{
printf("ciphertext: %c",(text[i] + k)%26 + 65);
}
}
}
printf("ciphertext: %c", d || c);
return;
int checking_key(int y,string n)
int num = argc;
string key = y;
int num_key = atoi(key);
if(argc != 2)
{
return 0;
}
else
{
if (num_key > 0)
{
return num_key;
}
else
{
return 0;
}
}
}
From man strlen:
Synopsis
#include <string.h>
size_t strlen(const char *s);
Just like one needs to "include" cs50.h to use any of the get_* functions, string.h must be "include"d to access its functions, eg strlen.
Additionally (per comments):
The "ordered comparison" in the compile error
ordered comparison between pointer and integer ('int' and 'string' (aka 'char *')) [-Werror] for(i = 0, l = strlen(text); i < n; i++)
is i < n. Error says one of them is an int and one of them is a string.
On closer inspection this program is a long way from a clean compile. Recommend you follow along with the spec and "approach this problem one step at a time"
I just want to know that for finding the Krishnamurthy number, we have to first find the factorial of the digits, then the addition of those numbers. (like, 1!+4!+5! = 145).
So, below is my code, and I have applied a factorial function over there. But the output is not coming in favor (145 is not a Kri...).
#include <stdio.h>
#include <stdlib.h>
void main()
{
int digit,factorial = 1, temp, input, sum = 0;
printf("Enter a Number:\n");
scanf("%d",&input);
int Factorial(int digit){
factorial = factorial*digit;
return 0;
}
temp = input;
while(temp>0){
digit = temp%10;
temp = temp/10;
sum = sum + Factorial(digit);
}
if(sum==input){
printf("%d is a Krishnamurthy Number",input);
}
else{
printf("%d is not a Krishnamurthy Number",input);
}
}
Have I done anything wrong in logic, or function declaration or definition? Please help.
your factorial function is not performing correctly. factorial means, multiplication of all digits starting from n downto 1 -
(n-1) * (n-2) * ... * (n)
but your function is not giving the desired result you want.
int Factorial(int digit){
factorial = factorial*digit;
return 0;
}
you need to change that function to get the factorial value of a number, you can either iterate a loop downto one or use a recursive approach to get the factorial.
int Factorial(int digit){
int result = 1;
for(int i=n; i>=1; i--){
result *= i;
}
return result;
}
or
int Factorial(int digit) {
if(n <= 1) return digit;
return digit * Factorial(digit-1);
}
you can follow the thread to understand the depth of recursive function mentioned above.
#include<stdio.h>
int main(int argc, char* argv[], char* envp[])
{
int sum = 0,
int a,
int p = 0,
int d,
int i,
int fact;
//code
printf("Enter a number: ");
scanf("%d", &a);
p = a;
while (a > 0)
{
fact = 1;
d = a % 10;
a /= 10;
for (i = d; i >= 1; i--)
{
fact *= i;
}
sum += fact;
}
if (sum == p)
printf("It is a Krishnamurthy number.\n");
else
printf("It is not a Krishnamurthy number.\n");
printf("\n\n");
return(0);
}
I'm posting this although much has already been posted about this question. I didn't want to post as an answer since it's not working. The answer to this post (Finding the rank of the Given string in list of all possible permutations with Duplicates) did not work for me.
So I tried this (which is a compilation of code I've plagiarized and my attempt to deal with repetitions). The non-repeating cases work fine. BOOKKEEPER generates 83863, not the desired 10743.
(The factorial function and letter counter array 'repeats' are working correctly. I didn't post to save space.)
while (pointer != length)
{
if (sortedWordChars[pointer] != wordArray[pointer])
{
// Swap the current character with the one after that
char temp = sortedWordChars[pointer];
sortedWordChars[pointer] = sortedWordChars[next];
sortedWordChars[next] = temp;
next++;
//For each position check how many characters left have duplicates,
//and use the logic that if you need to permute n things and if 'a' things
//are similar the number of permutations is n!/a!
int ct = repeats[(sortedWordChars[pointer]-64)];
// Increment the rank
if (ct>1) { //repeats?
System.out.println("repeating " + (sortedWordChars[pointer]-64));
//In case of repetition of any character use: (n-1)!/(times)!
//e.g. if there is 1 character which is repeating twice,
//x* (n-1)!/2!
int dividend = getFactorialIter(length - pointer - 1);
int divisor = getFactorialIter(ct);
int quo = dividend/divisor;
rank += quo;
} else {
rank += getFactorialIter(length - pointer - 1);
}
} else
{
pointer++;
next = pointer + 1;
}
}
Note: this answer is for 1-based rankings, as specified implicitly by example. Here's some Python that works at least for the two examples provided. The key fact is that suffixperms * ctr[y] // ctr[x] is the number of permutations whose first letter is y of the length-(i + 1) suffix of perm.
from collections import Counter
def rankperm(perm):
rank = 1
suffixperms = 1
ctr = Counter()
for i in range(len(perm)):
x = perm[((len(perm) - 1) - i)]
ctr[x] += 1
for y in ctr:
if (y < x):
rank += ((suffixperms * ctr[y]) // ctr[x])
suffixperms = ((suffixperms * (i + 1)) // ctr[x])
return rank
print(rankperm('QUESTION'))
print(rankperm('BOOKKEEPER'))
Java version:
public static long rankPerm(String perm) {
long rank = 1;
long suffixPermCount = 1;
java.util.Map<Character, Integer> charCounts =
new java.util.HashMap<Character, Integer>();
for (int i = perm.length() - 1; i > -1; i--) {
char x = perm.charAt(i);
int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
charCounts.put(x, xCount);
for (java.util.Map.Entry<Character, Integer> e : charCounts.entrySet()) {
if (e.getKey() < x) {
rank += suffixPermCount * e.getValue() / xCount;
}
}
suffixPermCount *= perm.length() - i;
suffixPermCount /= xCount;
}
return rank;
}
Unranking permutations:
from collections import Counter
def unrankperm(letters, rank):
ctr = Counter()
permcount = 1
for i in range(len(letters)):
x = letters[i]
ctr[x] += 1
permcount = (permcount * (i + 1)) // ctr[x]
# ctr is the histogram of letters
# permcount is the number of distinct perms of letters
perm = []
for i in range(len(letters)):
for x in sorted(ctr.keys()):
# suffixcount is the number of distinct perms that begin with x
suffixcount = permcount * ctr[x] // (len(letters) - i)
if rank <= suffixcount:
perm.append(x)
permcount = suffixcount
ctr[x] -= 1
if ctr[x] == 0:
del ctr[x]
break
rank -= suffixcount
return ''.join(perm)
If we use mathematics, the complexity will come down and will be able to find rank quicker. This will be particularly helpful for large strings.
(more details can be found here)
Suggest to programmatically define the approach shown here (screenshot attached below) given below)
I would say David post (the accepted answer) is super cool. However, I would like to improve it further for speed. The inner loop is trying to find inverse order pairs, and for each such inverse order, it tries to contribute to the increment of rank. If we use an ordered map structure (binary search tree or BST) in that place, we can simply do an inorder traversal from the first node (left-bottom) until it reaches the current character in the BST, rather than traversal for the whole map(BST). In C++, std::map is a perfect one for BST implementation. The following code reduces the necessary iterations in loop and removes the if check.
long long rankofword(string s)
{
long long rank = 1;
long long suffixPermCount = 1;
map<char, int> m;
int size = s.size();
for (int i = size - 1; i > -1; i--)
{
char x = s[i];
m[x]++;
for (auto it = m.begin(); it != m.find(x); it++)
rank += suffixPermCount * it->second / m[x];
suffixPermCount *= (size - i);
suffixPermCount /= m[x];
}
return rank;
}
#Dvaid Einstat, this was really helpful. It took me a WHILE to figure out what you were doing as I am still learning my first language(C#). I translated it into C# and figured that I'd give that solution as well since this listing helped me so much!
Thanks!
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
namespace CsharpVersion
{
class Program
{
//Takes in the word and checks to make sure that the word
//is between 1 and 25 charaters inclusive and only
//letters are used
static string readWord(string prompt, int high)
{
Regex rgx = new Regex("^[a-zA-Z]+$");
string word;
string result;
do
{
Console.WriteLine(prompt);
word = Console.ReadLine();
} while (word == "" | word.Length > high | rgx.IsMatch(word) == false);
result = word.ToUpper();
return result;
}
//Creates a sorted dictionary containing distinct letters
//initialized with 0 frequency
static SortedDictionary<char,int> Counter(string word)
{
char[] wordArray = word.ToCharArray();
int len = word.Length;
SortedDictionary<char,int> count = new SortedDictionary<char,int>();
foreach(char c in word)
{
if(count.ContainsKey(c))
{
}
else
{
count.Add(c, 0);
}
}
return count;
}
//Creates a factorial function
static int Factorial(int n)
{
if (n <= 1)
{
return 1;
}
else
{
return n * Factorial(n - 1);
}
}
//Ranks the word input if there are no repeated charaters
//in the word
static Int64 rankWord(char[] wordArray)
{
int n = wordArray.Length;
Int64 rank = 1;
//loops through the array of letters
for (int i = 0; i < n-1; i++)
{
int x=0;
//loops all letters after i and compares them for factorial calculation
for (int j = i+1; j<n ; j++)
{
if (wordArray[i] > wordArray[j])
{
x++;
}
}
rank = rank + x * (Factorial(n - i - 1));
}
return rank;
}
//Ranks the word input if there are repeated charaters
//in the word
static Int64 rankPerm(String word)
{
Int64 rank = 1;
Int64 suffixPermCount = 1;
SortedDictionary<char, int> counter = Counter(word);
for (int i = word.Length - 1; i > -1; i--)
{
char x = Convert.ToChar(word.Substring(i,1));
int xCount;
if(counter[x] != 0)
{
xCount = counter[x] + 1;
}
else
{
xCount = 1;
}
counter[x] = xCount;
foreach (KeyValuePair<char,int> e in counter)
{
if (e.Key < x)
{
rank += suffixPermCount * e.Value / xCount;
}
}
suffixPermCount *= word.Length - i;
suffixPermCount /= xCount;
}
return rank;
}
static void Main(string[] args)
{
Console.WriteLine("Type Exit to end the program.");
string prompt = "Please enter a word using only letters:";
const int MAX_VALUE = 25;
Int64 rank = new Int64();
string theWord;
do
{
theWord = readWord(prompt, MAX_VALUE);
char[] wordLetters = theWord.ToCharArray();
Array.Sort(wordLetters);
bool duplicate = false;
for(int i = 0; i< theWord.Length - 1; i++)
{
if(wordLetters[i] < wordLetters[i+1])
{
duplicate = true;
}
}
if(duplicate)
{
SortedDictionary<char, int> counter = Counter(theWord);
rank = rankPerm(theWord);
Console.WriteLine("\n" + theWord + " = " + rank);
}
else
{
char[] letters = theWord.ToCharArray();
rank = rankWord(letters);
Console.WriteLine("\n" + theWord + " = " + rank);
}
} while (theWord != "EXIT");
Console.WriteLine("\nPress enter to escape..");
Console.Read();
}
}
}
If there are k distinct characters, the i^th character repeated n_i times, then the total number of permutations is given by
(n_1 + n_2 + ..+ n_k)!
------------------------------------------------
n_1! n_2! ... n_k!
which is the multinomial coefficient.
Now we can use this to compute the rank of a given permutation as follows:
Consider the first character(leftmost). say it was the r^th one in the sorted order of characters.
Now if you replace the first character by any of the 1,2,3,..,(r-1)^th character and consider all possible permutations, each of these permutations will precede the given permutation. The total number can be computed using the above formula.
Once you compute the number for the first character, fix the first character, and repeat the same with the second character and so on.
Here's the C++ implementation to your question
#include<iostream>
using namespace std;
int fact(int f) {
if (f == 0) return 1;
if (f <= 2) return f;
return (f * fact(f - 1));
}
int solve(string s,int n) {
int ans = 1;
int arr[26] = {0};
int len = n - 1;
for (int i = 0; i < n; i++) {
s[i] = toupper(s[i]);
arr[s[i] - 'A']++;
}
for(int i = 0; i < n; i++) {
int temp = 0;
int x = 1;
char c = s[i];
for(int j = 0; j < c - 'A'; j++) temp += arr[j];
for (int j = 0; j < 26; j++) x = (x * fact(arr[j]));
arr[c - 'A']--;
ans = ans + (temp * ((fact(len)) / x));
len--;
}
return ans;
}
int main() {
int i,n;
string s;
cin>>s;
n=s.size();
cout << solve(s,n);
return 0;
}
Java version of unrank for a String:
public static String unrankperm(String letters, int rank) {
Map<Character, Integer> charCounts = new java.util.HashMap<>();
int permcount = 1;
for(int i = 0; i < letters.length(); i++) {
char x = letters.charAt(i);
int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
charCounts.put(x, xCount);
permcount = (permcount * (i + 1)) / xCount;
}
// charCounts is the histogram of letters
// permcount is the number of distinct perms of letters
StringBuilder perm = new StringBuilder();
for(int i = 0; i < letters.length(); i++) {
List<Character> sorted = new ArrayList<>(charCounts.keySet());
Collections.sort(sorted);
for(Character x : sorted) {
// suffixcount is the number of distinct perms that begin with x
Integer frequency = charCounts.get(x);
int suffixcount = permcount * frequency / (letters.length() - i);
if (rank <= suffixcount) {
perm.append(x);
permcount = suffixcount;
if(frequency == 1) {
charCounts.remove(x);
} else {
charCounts.put(x, frequency - 1);
}
break;
}
rank -= suffixcount;
}
}
return perm.toString();
}
See also n-th-permutation-algorithm-for-use-in-brute-force-bin-packaging-parallelization.
Given 2 strings like bangalore and blr, return whether one appears as a subsequence of the other. The above case returns true whereas bangalore and brl returns false.
Greedy strategy should work for this problem.
Find the first letter of the suspected substring (blr) in the big string (*b*angalore)
Find the second letter starting at the index of the first letter plus one (anga*l*ore)
Find the third letter starting at the index of the second letter plus one (o*r*e)
Continue until you can no longer find the next letter of blr in the string (no match), or you run out of letters in the subsequence (you have a match).
Here is a sample code in C++:
#include <iostream>
#include <string>
using namespace std;
int main() {
string txt = "quick brown fox jumps over the lazy dog";
string s = "brownfoxzdog";
int pos = -1;
bool ok = true;
for (int i = 0 ; ok && i != s.size() ; i++) {
ok = (pos = txt.find(s[i], pos+1)) != string::npos;
}
cerr << (ok ? "Found" : "Not found") << endl;
return 0;
}
// Solution 1
public static boolean isSubSequence(String str1, String str2) {
int i = 0;
int j = 0;
while (i < str1.length() && j < str2.length()) {
if (str1.charAt(i) == str2.charAt(j)) {
i++;
j++;
} else {
i++;
}
}
return j == str2.length();
}
// Solution 2 using String indexOf method
public static boolean isSubSequenceUsingIndexOf(String mainStr, String subStr) {
int i = 0;
int index = 0;
while(i<subStr.length()) {
char c = subStr.charAt(i);
if((index = mainStr.indexOf(c, index)) == -1) {
return false;
}
i++;
}
return true;
}
O(N) solution, where N is the length of the substring.
bool subsequence( string s1, string s2 ){
int n1 = s1.length();
int n2 = s2.length();
if( n1 > n2 ){
return false;
}
int i = 0;
int j = 0;
while( i < n1 && j < n2 ){
if( s1[i] == s2[j] ){
i++;
}
j++;
}
return i == n1;
}
Find the length of the longest common subsequence. If it is equal to the length of small string, return true