I am working on a system in which we have to create an array and store distinct numbers that can be obtained from an integer. Give Input:-
12
and the array of distinct numbers that can be generated from this number is:-
x={1,2,12}
But there is a catch that the distinct numbers cannot contain any leading zeros. I can delete a digit of the given number such that resulting number doesn't have any leading zeros.
Given Input
102
Now, the array formed is
x={0,1,2,10,12}
I am not able to solve the problem with leading zeros. Can anyone help me with the problem?
I solved this problem via substring. I computed the substring through 2 loops and if that string's first index value is zero, I didn't add it into my output array.
#include<iostream>
#include<string>
#include<sstream>
using namespace std;
int main() {
int num;
cin >> num;
vector<int> output;
string snum = to_string(num); //converting my input number into a string
//computing substring
for (int i = 0; i < snum.length(); i++) {
for (int j = i; j < snum.length(); j++) {
//substring method
string temp = snum.substr(i, j);
//if the first index is zero I continue my loop (leading zero)
if (temp[0] == '0')
continue;
else
{
//converting the string again into a number and add it to my output array
stringstream str(temp);
int n = 0;
str >> n;
output.push_back(n);
}
}
}
output.push_back(num);
//cout << output[0] << endl;
for (int i = 1; i < output.size(); i++)
cout << output[i] << "\n";
}
There maybe other optimized way to solve this problem.
Related
I have been given some integers and I have to count the frequency of a specific digit in the number.
example input:
5
447474
228
6664
40
81
The first number says number of integers in the list. I am finding frequency of 4 in this case. I tried to change the integer to an array, but it is not working.
#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
int main() {
int n;
cin>>n;
for (int i=0; i<n; i++)
{
int x;
cin>>x;
int frequency=0;
int t=log10(x);
int arr[t];
for (i=t; i>0; i--)
{
arr[i]=x%10;
x=x/10;
}
for(int i=0; i<t; i++)
{
if(arr[i]==4)
{
frequency++;
}
}
std::cout << frequency << std::endl;
}
return 0;
}
No need to create an array, or to determine the number of digits. Just loop until the number reaches zero.
int digitCount(int n, int d) {
if(n < 0) n = -n;
int count = 0;
for(; n != 0; n /= 10)
if(n % 10 == d) count++;
return count;
}
Test:
cout << digitCount(447474, 4) << endl;
cout << digitCount(-447474, 4) << endl;
Output:
4
4
Your code uses VLAs which are not standard C++. See Why aren't variable-length arrays part of the C++ standard?.
log10(x) is not the number of digits. For example log10(1234) == 3.09131516 but it is 4 digits. Also you are accessing the array out of bounds in the first iteration of the loop: arr[t]. Valid indices in an array of size t are 0,1,2,...,t-1. Trying to access arr[t] is undefined behavior.
Actually you dont need any array. Instead of storing the digits in an array you can immediately check whether it is a 4 and count.
Even simpler would be to read the user input as a std::string:
#include <string>
#include <algorithm>
#include <iostream>
int main() {
std::string input;
std::cin >> input;
std::cout << std::count(input.begin(),input.end(),'4');
}
Perhaps you should add some checks to verify that the user input is actually a valid number. However, also when reading an int you should validate the input.
I'm creating this very simple C++ program.
the program asks the user to enter a few integers and stores them in an array.but when a specific integer(for example 50)is entered,the input is ended and then,all of the integers are displayed on the screen except for 50.
for example:
input:
1
2
88
50
output:
1
2
88
the error i'm getting is when i use cout to print the array,all of numbers are shown,including 50 and numbers i did'nt even entered.
this is my code so far:
#include<iostream>
int main() {
int num[100];
for(int i=0;i<=100;i++) {
cin >> num[i];
if (num[i]!=50) break;
}
for(int j=0;j<=100;j++) {
cout << num[j] << endl;
}
return 0;
}
Change the program the following way
#include<iostream>
int main()
{
const size_t N = 100;
int num[N];
size_t n = 0;
int value;
while ( n < N && std::cin >> value && value != 50 ) num[n++] = value;
for ( size_t i = 0; i < n; i++ ) std::cout << num[i] << std::endl;
return 0;
}
Here in the first loop variable n is used to count the actual number of entered values. And then this variable is used as the upper bound for the second loop.
As for your program then the valid range of indices for the first loop is 0-99 and you have to output only whose elements of the array that were inputed.
A do while loop is more suitable for your problem. The stop condition will check if the number fit inside the array (if k is not bigger than 100) and if number entered is 50.
#include<iostream>
using namespace std;
int main() {
int num[100];
int k = 0;
// A do while loop will be more suitable
do{
cin >> num[k++];
}while(k<100&&num[k-1]!=50);
for (int j = 0; j < k-1; j++) {
cout << num[j] << endl;
}
return 0;
}
Also, a better solution to get rid of 100 limitation is to use std::vector data structure that automatically adjust it's size, like this:
vector<int> num;
int temp;
do {
cin >> temp;
num.push_back(temp);
} while (temp != 50);
Note, you can use temp.size() to get the number of items stored.
You read up to 101 numbers, but if you enter 50 you break the loop and go for printing it. In the printing loop you go through all 101 numbers, but you actually may have not set all of them.
In the first loop count in a count variable the numbers you read until you meet 50 and in the printing loop just iterate count-1 times.
You have allocated an array of 100 integers on the stack. The values are not initialized to zero by default, so you end up having whatever was on the stack previously appear in your array.
You have also off-by-one in both of your loops, you allocated array of 100 integers so that means index range of 0-99.
As the question is tagged as C++, I would suggest that you leave the C-style array and instead use a std::vector to store the values. This makes it more flexible as you don't have to specify a fixed size (or manage memory) and you don't end up with uninitialized values.
Little example code (requires C++11 compiler):
#include <iostream>
#include <vector>
int main()
{
std::vector<int> numbers; // Store the numbers here
for(int i = 0; i < 100; ++i) // Ask a number 100 times
{
int n;
std::cin >> n;
if( n == 50 ) // Stop if user enters 50
break;
numbers.push_back(n); // Add the number to the numbers vector
}
for (auto n : numbers) // Print all the values in the numbers vector
std::cout << n << std::endl;
return 0;
}
There are just 2 changes in your code check it out :
int main()
{
int num[100],i; //initialize i outside scope to count number of inputs
for(i=0;i<100;i++) {
cin >> num[i];
if (num[i]==50) break; //break if the entered number is 50
}
for(int j=0;j<=i-1;j++)
{
cout << num[j] << endl;
}
return 0;
}
Okay, others already pointed out the two mistakes. You should use i < 100 in the loop conditions instead of i <= 100 and you have to keep track of how many elements you entered.
Now let me add an answer how I think it would be better.
#include <iostream>
#include <vector>
int main() {
std::vector<int> numbers; // Create an empty vector.
for (int temp; // a temp variable in the for loop.
numbers.size() < 100 && // check that we have less than 100 elements.
std::cin >> temp && // read in the temp variable,
// and check if the read was a success.
temp != 50) // lastly check that the value we read isn't 50.
{
numbers.push_back(temp); // Now we just add it to the vector.
}
for (int i = 0; i < numbers.size(); ++i)
std::cout << numbers[i]; // Now we just print all the elements of
// the vector. We only added correct items.
}
The above code doesn't even read anymore numbers after it found 50. And if you want to be able to enter any number of elements you just have to remove the check that we have less than 100 elements.
Now I commented the above code a bit much, if you compress it it'll reduce to just:
#include <iostream>
#include <vector>
int main() {
std::vector<int> numbers; // Create an empty vector.
for (int temp; numbers.size() < 100 && std::cin >> temp && temp != 50)
numbers.push_back(temp);
for (int i = 0; i < numbers.size(); ++i)
std::cout << numbers[i];
}
If you can use the C++11 standard it reduces to:
#include <iostream>
#include <vector>
int main() {
std::vector<int> numbers; // Create an empty vector.
for (int temp; numbers.size() < 100 && std::cin >> temp && temp != 50)
numbers.push_back(temp);
for (int element : numbers)
std::cout << element;
}
for (auto element : numbers) is new, it basically means for every int 'element' in 'numbers'.
I am coding an assignment for my class where a user will input 10 letter answers, and the program will return a grade. I recently changed my char arrays to string arrays, because I think it makes it easier to read.
I went to debug my code and am now getting the error "Deubug Assertion Failed." I do not know what this means or how to fix it.
Any help would be appreciated.
Thanks!
Below is my code:
// Lab 8
// programmed by Elijah Barron
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
//Function headers
string inputAnswers(string given);
int numCorrect(string correctAnswers, string given);
int main()
{
string correctAnswers = "BCADBADCAB";
string given;
int numRight = 0;
inputAnswers(given);
numCorrect(correctAnswers, given);
double grade = 10 * numRight;
cout << "Your quiz grade is " << grade << "%" << endl;
return 0;
}
//Get the answers
string inputAnswers(string given)
{
for (int n = 0; n < 10; n++)
{
cout << "Please enter your answer for question #" << n + 1 << " ";
cin >> given[n];
}
return given;
}
//Find if answers are correct or incorrect
int numCorrect(string correctAnswers, string given)
{
int numRight = 10;
int n = 0;
for (int n = 0; n < 10; n++);
{
if (given[n] != correctAnswers[n])
numRight -= 1;
}
return numRight;
}
The immediate issue is that given will start off as an empty string as you haven't assigned it a value:
cin >> given[n];
is causing the assert failure because you're trying to change the first (second, third etc) character in a string with a length of zero. To fix the assert problem (but not the program, which will always return 0%), just initialise the string:
string given = "ZZZZZZZZZZ";
To fix the rest of the stuff (btw this isn't the only way):
Change:
string inputAnswers(string given); //for both prototype and function.
to:
void inputAnswers(string& given); //pass by reference instead of pass by value.
//also get rid of "return given;"
Change:
int n = 0; //the n here is different to the one in the next line
for (int n = 0; n < 10; n++); //this n's scope begins and ends here thanks to the semicolon
{//the code here is executed once, this isn't in the loop!
if (given[n] != correctAnswers[n]) //we're using the first n here, which is 0.
numRight -= 1;
}
to:
for (int n = 0; n < 10; n++) //only one n variable and no semicolon
{// now this is in the loop and will execute 10 times.
if (given[n] != correctAnswers[n])
numRight -= 1;
}
Don't bother with this line:
int numRight = 0; //Set at 0 and then never changed.
and change:
numCorrect(correctAnswers, given);
to:
int numRight = numCorrect(correctAnswers, given); //declared when necessary and assigned the correct value
You either want to reserve enough space in your vector to hold 10 characters, or use push_back to populate the vector. Indexing a vector with [] won't grow the vector for you.
EDIT:
Ignore the first part about reserve. That doesn't stop the debug assertion. You will want to change this
cin >> given[n];
To something like this:
char input;
cin >> input;
given.push_back(input);
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 8 years ago.
Improve this question
I was asked this question in an interview:
Given an array with the input string, display the output as shown below
Input
INDIA
Output
INDA
****
*
I iterated through the array and stored each character as a key in std::map with value as number of occurrence. Later I iterate the map and print the asteriks and reduce the value in the map for each character.
Initially, I was asked not to use any library. I gave a solution which needed lot of iterations. For every character, iterate the complete array till the index to find previous occurrences and so on.
Is there any better way, e.g. better complexity, such as faster operation, by which this can be achieved?
Essentially what you are asking is how to implement map without using the STL code, as using some kind of data structure which replicates the basic functionality of map is pretty much the most reasonable way of solving this problem.
There are a number of ways of doing this. If your keys (here the possible characters) come from a very large set where most elements of the set don't appear (such as the full Unicode character set), you would probably want to use either a tree or a hash table. Both of these data structures are very important with lots of variations and different ways of implementing them. There is lots of information and example code about the two structures around.
As #PeterG said in a comment, if the only characters you are going to see are from a set of 256 8-bit chars (eg ASCII or similar), or some other limited collection like the upper-case alphabet you should just use an array of 256 ints and store a count for each char in that.
here is another one:
You can see it working HERE
#include <stdio.h>
int main()
{
int i,j=0,f=1;
char input[50]={'I','N','D','I','A','N','A','N'};
char letters[256]={0};
int counter[256]={0};
for(i=0;i<50;i++)
{
if(input[i])
counter[input[i]]++;
if(counter[input[i]]==1)
{
putchar(input[i]);
letters[j]=input[i];
j++;
}
}
putchar('\n');
while(f)
{
f=0;
for(i=0;i<j;i++)
if(counter[letters[i]])
{
putchar('*');
counter[letters[i]]--;
f=1;
}
else
{
putchar(' ');
}
putchar('\n');
}
return 0;
}
If the alphabet under consideration is fixed, it can be done in two passes:
Create an integer array A with the size of the alphabet, initialized with all zeros.
Create a boolean array B with size of the input, initialize with all false.
Iterate the input; increase for every character the corresponding content of A.
Iterate the input; output a character if its value it B is false and set its value in B to true. Finally, output a carriage return.
Reset B.
Iterate input as in 4., but print a star if if the character's count in A is positive, then decrease this count; print a space otherwise.
Output a carriage return; loop to 5 as long as there are any stars in the output generated.
This is interesting. You shouldnt use a stl::map because that is not a hashmap. An stl map is a binary tree. An unordered_map is actually a hash map. In this case we dont need either. We can use a simple array for char counts.
void printAstr(std::string str){
int array[256] ;// assumining it is an ascii string
memset(array, 0, sizeof(array));
int astrCount = 0;
for(int i = 0; i < str.length()-1; i++){
array[(int) str[i]]++;
if(array[(int) str[i]] > 1) astrCount++;
}
std::cout << str << std::endl;
for(int i = 0; i < str.length()-1;i++) std::cout << "* ";
std::cout << std::endl;
while(astrCount != 0){
for(int i= 0; i< str.length() - 1;i++){
if(array[(int) str[i]] > 1){
std::cout << "* ";
array[(int) str[i]]--;
astrCount--;
}else{
std::cout << " ";
}
}
std::cout << std::endl;
}
}
pretty simple just add all values to the array, then print them out the number of times you seem them.
EDIT: sorry just made some logic changes. This works now.
The following code works correctly. I am assuming that you can't use std::string and take note that this doesn't take overflowing into account since I didn't use dynamic containers. This also assumes that the characters can be represented with a char.
#include <iostream>
int main()
{
char input[100];
unsigned int input_length = 0;
char letters[100];
unsigned int num_of_letters = 0;
std::cin >> input;
while (input[input_length] != '\0')
{
input_length += 1;
}
//This array acts like a hash map.
unsigned int occurrences[256] = {0};
unsigned int max_occurrences = 1;
for (int i = 0; i < input_length; ++i)
{
if ((occurrences[static_cast<unsigned char>(input[i])] += 1) == 1)
{
std::cout<< " " << (letters[num_of_letters] = input[i]) << " ";
num_of_letters += 1;
}
if (occurrences[static_cast<unsigned char>(input[i])] > max_occurrences)
{
max_occurrences = occurrences[static_cast<unsigned char>(input[i])];
}
}
std::cout << std::endl;
for (int row = 1; row <= max_occurrences; ++row)
{
for (int i = 0; i < num_of_letters; ++i)
{
if (occurrences[static_cast<unsigned char>(letters[i])] >= row)
{
std::cout << " * ";
}
else
{
std::cout << " ";
}
}
std::cout << std::endl;
}
return 0;
}
The question is marked as c++ but It seems to me that the answers not are all quite C++'ish, but could be quite difficult to achieve a good C++ code with a weird requirement like "not to use any library". In my approach I've used some cool C++11 features like in-class initialization or nullptr, here is the live demo and below the code:
struct letter_count
{
char letter = '\0';
int count = 0;
};
int add(letter_count *begin, letter_count *end, char letter)
{
while (begin != end)
{
if (begin->letter == letter)
{
return ++begin->count;
}
else if (begin->letter == '\0')
{
std::cout << letter; // Print the first appearance of each char
++begin->letter = letter;
return ++begin->count;
}
++begin;
}
return 0;
}
int max (int a, int b)
{
return a > b ? a : b;
}
letter_count *buffer = nullptr;
auto testString = "supergalifragilisticoespialidoso";
int len = 0, index = 0, greater = 0;
while (testString[index++])
++len;
buffer = new letter_count[len];
for (index = 0; index < len; ++index)
greater = max(add(buffer, buffer + len, testString[index]), greater);
std::cout << '\n';
for (int count = 0; count < greater; ++count)
{
for (index = 0; buffer[index].letter && index < len; ++index)
std::cout << (count < buffer[index].count ? '*' : ' ');
std::cout << '\n';
}
delete [] buffer;
Since "no libraries are allowed" (except for <iostream>?) I've avoided the use of std::pair<char, int> (which could have been the letter_count struct) and we have to code many utilities (such as max and strlen); the output of the program avobe is:
supergaliftcod
**************
* ******* *
* *** *
* *
*
*
My general solution would be to traverse the word and replace repeated characters with an unused nonsense character. A simple example is below, where I used an exclamation point (!) for the nonsense character (the input could be more robust, some character that is not easily typed, disallowing the nonsense character in the answer, error checking, etc). After traversal, the final step would be removing the nonsense character. The problem is keeping track of the asterisks while retaining the original positions they imply. For that I used a temp string to save the letters and a process string to create the final output string and the asterisks.
#include <iostream>
#include <string>
using namespace std;
int
main ()
{
string input = "";
string tempstring = "";
string process = "";
string output = "";
bool test = false;
cout << "Enter your word below: " << endl;
cin >> input;
for (unsigned int i = 0; i < input.length (); i++)
{ //for the traversed letter, traverse through subsequent letters
for (unsigned int z = i + 1; z < input.length (); z++)
{
//avoid analyzing nonsense characters
if (input[i] != '!')
{
if (input[i] == input[z])
{ //matched letter; replace with nonsense character
input[z] = '!';
test = true; //for string management later
}
}
}
if (test)
{
tempstring += input[i];
input[i] = '*';
test = false; //reset bool for subsequent loops
}
}
//remove garbage symbols and save to a processing string
for (unsigned int i = 0; i < input.size (); i++)
if (input[i] != '!')
process += input[i];
//create the modified output string
unsigned int temp = 0;
for (unsigned int i = 0; i < process.size (); i++)
if (process[i] == '*')
{ //replace asterisks with letters stored in tempstring
output += tempstring[temp];
temp++;
}
else
output += process[i];
//output word with no repeated letters
cout << output << endl;
//output asterisks equal to output.length
for (unsigned int a = 0; a < output.length (); a++)
cout << "*";
cout << endl;
//output asterisks for the letter instances removed
for (unsigned int i = 0; i < process.size (); i++)
if (process[i] != '*')
process[i] = ' ';
cout << process << endl << endl;
}
Sample output I received by running the code:
Enter your word below:
INDIA
INDA
****
*
Enter your word below:
abcdefgabchijklmnop
abcdefghijklmnop
****************
***
It is possible just using simple array to keep count of values.
#include<iostream>
#include<string>
using namespace std;
int main(){
string s;
char arr[10000];
cin>>s;
int count1[256]={0},count2[256]={0};
for(int i=0;i<s.size();++i){
count1[s[i]]++;
count2[s[i]]++;
}
long max=-1;
int j=0;
for(int i=0;i<s.size();++i){
if(count1[s[i]]==count2[s[i]]){ //check if not printing duplicate
cout<<s[i];
arr[j++]=s[i];
}
if(count2[s[i]]>max)
max=count2[s[i]];
--count1[s[i]];
}
cout<<endl;
for(int i =1; i<=max;++i){
for(int k=0;k<j;++k){
if(count2[arr[k]]){
cout<<"*";
count2[arr[k]]--;
}
else
cout<<" ";
}
cout<<endl;
}
}
I have this homework question:
Write and test a program that read in n integers (max value for n is 20), each integer has
a value between 0 and 100 inclusive. You program should then print out the unique values
among the input numbers and the count of these values.
Sample input:
Enter a the number of integers = 8
Enter 8 integers: 5 6 7 6 6 17 17 35
Sample output:
Number 5: 1
Number 6: 3
Number 7: 1
Number 17: 2
Number 35: 1
This is what I did:
#include<iostream>
using namespace std;
int main(){
int a[20], n;
cout<< "Please enter the number of integers= ";
cin>> n;
cout<<"Please enter"<< n<<" integers: ";
for (int i=0; i<n; i++)
cin >> a[i];
for (int k=0; k< n; k++){
int sum=0;
for (int i=0; i< n; i++){
if (a[i]==a[k])
sum= sum+1;
}
cout<< "Number "<< a[k]<<" : "<< sum<< endl;
}
}
Consider that when you iterate through your list, you're checking all values with both i and k. So essentially, if you had a list of 1 1 2 2, then the first one will count itself, and the 1 at a[1]. The second 1 will count the first 1 and itself, giving you your repeated output.
A way to simplify this would be to make use of a hash_map, or some similar structure (I'm not as familiar with C++) that maps a key to a value and doesn't allow repeats. This would allow you to record the unique numbers as keys, and increment them with only one pass through the list. The advantage to using the hashMap is that you make your program linear (although I don't think that's really a concern at this stage).
The simplest way to solve your problem, however would be to use a Bin sort technique. The underlying idea here is that your number range is simply 0 to 100, meaning you could create bins for 0 to 100 and increment each one. Again, this is Java code, and doesn't have any actual input for a.
// Count is the key, it uses indexes from 0 to 100, with null values of
// 0 after initialized. Simply iterate the loop, and use the value of
// a[k] to increment the corresponding count in the count array.
// Finally, print the results
int[] a = new int[20];
int[] count = new int [101];
for (int k = 0; k < a.length; k++){
count[a[k]]++;
for (int i = 0; i < count.length; i++){
if (count[i] > 0)
System.out.println(i + ": " + count[i]);
}
Add another bool b[20] ,initialize it with true. Then every time you detect a[k] is a dupe, you set b[k] = false. Only print a[k] if b[k] == true
for (int k = 0; k < n; k++) {
if (!b[k]) {
continue;
}
int sum = 0;
for (int i = 0; i < n; i++) {
if (a[i] == a[k]) {
sum = sum + 1;
b[i] = false;
}
}
cout << "Number " << a[k] << " : " << sum << endl;
}
You have to keep a running count of the items you processed in a separate array, and before running your inner loop to count the items, check if he item you're trying to count isn't in your second array already.
Before you print the result, check if you already printed it for this number
Here's my new attempt.
A quick fix (while not the most professional) would be to create another loop checking for repeats right before printing out.
I took your current big loop and turned it into an even bigger monster.
I also tested it out and it works for me. =D
for (int k=0; k< n; k++){
int sum=0;
for (int i=0; i< n; i++)
{
if (a[i]==a[k])
sum= sum+1;
}
bool repeat = false;
for(int i = 0; i < k; i++)
{
if(a[k] == a[i])
{
repeat = true;
}
}
if(!repeat)
cout<< "Number "<< a[k]<<" : "<< sum<< endl;
}
An alternate implementation (and more memory-hungry with your current limit of 20 input values) would be to create an array of 100 "count" values. Increment the appropriate item for each input value, then iterate through the count array outputting non-zero values.
Apparently that description wasn't good enough... perhaps some code would help (NOTE:This code is untested, but should be enough for you to understand the concept):
#include<iostream>
using namespace std;
int main(){
int a[101], n, v;
cout<< "Please enter the number of integers= ";
cin>> n;
cout<<"Please enter"<< n<<" integers: ";
for (int i=0; i<n; i++)
{
cin >> v;
a[v] ++;
}
for (int k=0; k< 100; k++){
if (a[k] > 0)
{
cout<< "Number "<< k + 1 <<" : "<< a[k] << endl;
}
}
}
}
getting familiar with Standart Template Library is the key to writing good programs in my humble opinion, since this is a homework, you do the controlling for 0 and 100 ;-))
#include <iostream>
#include <map>
using std::cin;
using std::map;
using std::cout;
using std::endl;
int main()
{
int limit = 20;
int cnt=0;
int n;
map<int, int> counters;
while( cnt++ < limit )
{
cin >> n;
++counters[n];
}
for(map<int, int>::iterator it = counters.begin();
it!=counters.end(); ++ it)
cout << it->first << " " << it->second << endl;
return 0;
}