Look at this code, In my VS code or some online compilers it gives favorable output, but when I'm submitting this on HackerRank or this online compiler I'm getting wrong output but only when I provide input as: 1 1 1 100...and I'm not able to spot the Error?..I'm providing question for reference.
/*There will be two arrays of integers. Determine all integers that satisfy
the following two conditions:
The elements of the first array are all factors of the integer being
considered
The integer being considered is a factor of all elements of the second array
These numbers are referred to as being between the two arrays. Determine how
many such numbers exist.
*/
#include <iostream>
int main()
{
int count1, count2;
int num1[20], num2[20];
std::cin >> count1 >> count2;
for (int i = 0; i < count1; i++)
{
std::cin >> num1[i];
}
for (int i = 0; i < count2; i++)
{
std::cin >> num2[i];
}
int occurence_firstarray = 0, occurence_secondarray = 0, totalvalid_occurence = 0;
for (int i = num1[count1 - 1]; i < num2[1]; i++)
{
occurence_firstarray = 0;
occurence_secondarray = 0;
for (int j = 0; j < count1; j++)
{
if (i % num1[j] == 0)
{
occurence_firstarray++;
}
}
if (occurence_firstarray == count1)
{
for (int p = 0; p < count2; p++)
{
if (num2[p] % i == 0)
{
occurence_secondarray++;
}
}
}
if (occurence_secondarray == count2)
{
totalvalid_occurence++;
}
}
std::cout << totalvalid_occurence;
return (0);
}
Considering your inputs: 1 1 1 100,
What are you doing in this code is You are pointing on index beyond your second array's size limit i < num2[1], this is why you are getting wrong outputs.
Do some changes as, i <= num2[0];
I am trying to create a vector that holds a user summited value worth of my class object. Unfortunately, I am running into an error that states my vector's subscript is out of range. My main guess is there might be something wrong with the way I am analyzing each element of the vector or possibly the way I am creating the vector. Any suggestions would be gladly appreciated!
...
unsigned int population = 3;
vector<Platypus> pop;
for(unsigned int i = 0; i < population; i++) {
pop.push_back(Platypus());
}
for (unsigned int i = 0; i < length; i++) {
for (unsigned int j = 0; j < population; j++) {
if (i%(int)fabs(4 - eatAmount) == 0) {
pop.at(j).eat();
}
if(pop.at(j).getAge() >= 24 &&
pop.at(j).getGender()=='F' &&
pop.at(j).getAlive()==1 &&
pop.at(j).getMutant()!=0){
pop.push_back(pop.at(j).hatch());
}
if (i > 30) {
pop.at(j).age_me(1);
}
if (j % 3 == 0 && j!=0) {
pop.at(j).fight(pop.at(j - 1));
}
pop.at(j).print(cout);
}
}
for (int i = pop.size(); i > 0; i--) {
if (pop[i].getAlive() == 0) {
pop.erase(pop.begin()+i);
}
}
}
This is you culprit:
for (int i = pop.size(); i > 0; i--) {
if (pop[i].getAlive() == 0) { // HERE
pop.erase(pop.begin()+i);
}
}
On the first loop iteration, i is initialised to the size of the vector. To turn this into a valid index, you need to subtract 1, or otherwise you go out of bounds.
I'm trying to read characters (numbers) from a string, and then passing them as integers in a vector. Currently something like this:
int p;
std::string line = "1111222111"
std::vector<int> vec;
for(int i = 0; i < str.size(); i++)
{
if(line.find('0', static_cast<std::size_t>(i)) != line.npos)
p = 0;
else if(line.find('1', static_cast<std::size_t>(i)) != line.npos)
p = 1;
else if(line.find('2', static_cast<std::size_t>(i)) != line.npos)
p = 2;
vec.push_back(p);
}
When doing this, the vector ends only with 0's. Keep in mind that this is not the actual code, just a vague representation of the part where I have my problem. I think I must be doing something wrong in the line.find stuff, but I don't really know any other way to use it (not that I actually know how to really use it...). Thanks in advance guys!
You can make use of the isdigit function to determine if the character is a digit. If so, you can make use of the fact that ASCII orders the digits adjacent.
You can thus solve the problem with the following code:
for(int i = 0; i < str.size(); i++) {
int c = str[i]-'0';
if(c >= 0 && c <= 9) {
vec.push_back(c);
}
}
#include <vector>
#include <string>
int main(int argc, char *argv[])
{
std::vector<int> ans;
const std::string str = "1234";
for (int i = 0; i < str.size(); ++i) {
if ('0' <= str[i] && str[i] <= '9')
ans.push_back(str[i] - '0');
}
return 0;
}
I think you should instead of this :
for(int i = 0; i < str.size(); i++)
have this:
for(int i = 0; i < line.size(); i++)
str is not defined in your example. I think you meant line.
This question already has answers here:
Deleting duplicates in the array
(2 answers)
Closed 8 years ago.
I need to write a program where I have to print the numbers which occur twice or more than twice in the array. To make things simpler, I am working on a sorted array.
Here is my code.
#include <stdio.h>
int main()
{
int n,i;
int a[10]={2,2,2,4,6,6,9,10,10,11};
printf("%d\n",a[10]);
for(i=0;i<10;++i)
{
if(a[i]==a[i+1] && a[i]!=a[i-1])
printf("%d ",a[i]);
else
continue;
}
return 0;
}
The code seems to work fine but I do not like my code because at some point, the loop compares the value of a[0] with a[-1] and a[9] with a[10] and both of these, a[-1] and a[10], are garbage values. I am sure there are better ways to do it but I am unable to think of any.
Also, I need to extend the above program to count the frequency of duplicate numbers.
Help is appreciated. Thanks!
First, you can't access a[10] in your printf line, this is outside your array.
This should work fine.
#include <stdio.h>
int main()
{
int n,i;
int a[10]={2,2,2,4,6,6,9,10,10,11};
for(i=0; i < 9; ++i)
{
if(a[i] == a[i+1])
if(i > 0 && a[i] == a[i-1])
continue;
else
printf("%d ",a[i]);
}
return 0;
}
Edit: Or you can use the shorter yet harder to read:
for(i=0; i < 9; ++i)
{
if(a[i] == a[i+1] && (!i || a[i] != a[i-1]))
printf("%d ",a[i]);
}
See code below for the solution, which will print only the duplicate numbers from the array and how many times they occur.
I added the int c which is used for your count. It is initially set to 1, and increased by 1 for each duplicate number.
#include <stdio.h>
using namespace std;
int main() {
int n,i,c;
int a[10]={2,2,2,4,6,6,9,10,10,11};
c = 1;
for(i=0; i < 9; ++i)
{
if (a[i] == a[i+1]) {
n = a[i];
c += 1;
} else {
if (c > 1) {
printf("Number: %d, Occurences: %d \n",n,c);
c = 1;
}
}
}
return 0;
}
Since the array is already sorted, it'd be easier to do something like this, with another loop inside the for loop.
int duplicateCount = 1; // Must be at least one instance of a value in the array
for(int i = 0; i < 9; ++i) {
int j = i;
while(j != 9 && a[j] == a[j + 1]) {
++j;
++duplicateCount;
}
i = j; // j is now at end of list of duplicates, so put i at the end as well,
// to be incremented to the next unique value at end of for loop iteration
printf("%d: %d\n", a[j], duplicateCount);
duplicateCount = 1; // Reset for next values
}
If you want to count the frequency of all the numbers, you can easily turn duplicateCount into an array of values to count the frequency of each unique value. For a better solution, you could use another data structure, such as a map.
You are really going to need to have two indexes that walk through your array.
Here's a start:
int i = 0;
int j;
while (i < 10) {
for (j = i+1; j < 10; ++j) if (a[i] != a[j]) break;
if (j !+ i+i) printf("%d\n", a[i]);
i = j;
}
You could use a functor, overload operator() and return a set (unique values). Assuming your array is sorted it's just a matter of comparing the previous one with the next and inserting it in the set if they equal. If they are not sorted then you have to go through the whole array for every entry. Below are two examples with output as an explanation.
#include <set>
using namespace std;
Unsorted array:
struct UnsortedArrayDuplicateEntries {
set<int> operator()(int* array, int size) {
set<int> duplicates;
for (int i = 0; i < size - 1; i++)
for (int j = i + 1;j < size; j++)
if (array[i] == array[j])
duplicates.insert(array[j]);
return duplicates;
}
};
Sorted array:
struct SortedArrayDuplicateEntries {
set<int> operator()(int* array, int size) {
set<int> duplicates;
for (int i = 0; i < size - 1; i++)
if (array[i] == array[i+1])
duplicates.insert(array[j]);
return duplicates;
}
};
Test sorted:
SortedArrayDuplicateEntries d;
int sorted[10]={2,2,2,4,6,6,9,10,10,11};
set<int> resultSorted = d(sorted,10);
for (int i : resultSorted) cout << i << endl;
Output sorted:
2
6
10
Test unsorted:
UnsortedArrayDuplicateEntries d;
int unsorted[10]={2,6,4,2,10,2,9,6,10,11};
set<int> resultUnsorted = d(unsorted,10);
for (int i : resultUnsorted) cout << i << endl;
Output unsorted:
2
6
10
I hope it helps!
#include <stdio.h>
int main(){
int i, a[10]={2,2,2,4,6,6,9,10,10,11};
int size = 10;
for(i=0;i<size;++i){
int tmp = a[i];
int c = 1;
while(++i < size && tmp == a[i])
++c;
if(c > 1)
printf("%d times %d\n", tmp, c);
--i;
}
return 0;
}
I need to get the unique value from 2 int arrays
Duplicate is allowed
There is just one unique value
like :
int arr1[3]={1,2,3};
int arr2[3]={2,2,3};
and the value i want to get is :
int unique[]={1}
how can i do this?
im already confused in my 'for' and 'if'
this was not homework
i know how to merge 2 arrays and del duplicate values
but i alse need to know which array have the unique value
plz help me :)
and here is some code i did
int arr1[3]={1,2,3}
int arr2[3]={2,2,3}
int arrunique[1];
bool unique = true;
for (int i=0;i!=3;i++)
{
for (int j=0;j!=3;j++)
{
if(arr1[i]==arr2[j])
{
unique=false;
continue;
}
else
{
unique=true;
}
if(unique)
{
arrunique[0]=arr1[i]
break;
}
}
cout << arrunique[0];
Assuming:
You have two arrays of different length,
The arrays are sorted
The arrays can have duplicate values in them
You want to get the list of values that only appear in one of the arrays
including their duplicates if present
You can do (untested):
// Assuming arr1[], arr2[], and lengths as arr1_length
int i = 0,j = 0, k = 0;
int unique[arr1_length + arr2_length];
while(i < arr1_length && j < arr2_length) {
if(arr1[i] == arr2[j]) {
// skip all occurrences of this number in both lists
int temp = arr1[i];
while(i < arr1_length && arr1[i] == temp) i++;
while(j < arr2_length && arr2[j] == temp) j++;
} else if(arr1[i] > arr2[j]) {
// the lower number only occurs in arr2
unique[k++] = arr2[j++];
} else if(arr2[j] > arr1[i]) {
// the lower number only occurs in arr1
unique[k++] = arr1[i++];
}
}
while(i < arr1_length) {
// if there are numbers still to read in arr1, they're all unique
unique[k++] = arr1[i++];
}
while(j < arr2_length) {
// if there are numbers still to read in arr2, they're all unique
unique[k++] = arr2[j++];
}
Some alternatives:
If you don't want the duplicates in the unique array, then you can skip all occurrences of this number in the relevant list when you assign to the unique array.
If you want to record the position instead of the values, then maintain two arrays of "unique positions" (one for each input array) and assign the value of i or j to the corresponding array as appropriate.
If there's only one unique value, change the assignments into the unique array to return.
Depending on your needs, you might also want to look at set_symmetric_difference() function of the standard library. However, its treatment of duplicate values makes its use a bit tricky, to say the least.
#include <stdio.h>
#include <stdlib.h>
int cmp ( const void *a , const void *b )
{
return *(int *)a - *(int *)b;
}
int main()
{
int arr1[5] = {5,4,6,3,1};
int arr2[3] = {5, 8, 9};
int unique[8];
qsort(arr1,5,sizeof(arr1[0]),cmp);
printf("\n");
qsort(arr2,3,sizeof(arr2[0]),cmp);
//printf("%d", arr1[0]);
int i = 0;
int k = 0;
int j = -1;
while (i < 5 && k < 3)
{
if(arr1[i] < arr2[k])
{
unique[++j] = arr1[i];
i++;
}
else if (arr1[i] > arr2[k])
{
unique[++j] = arr2[k];
k++;
}
else
{
i++;
k++;
}
}
//int len = j;
int t = 0;
if(i == 5)
{
for(t = k; t < 3; t++)
unique[++j] = arr2[t];
}
else
for(t = i; t < 5; t++)
unique[++j] = arr2[t];
for(i = 0; i <= j; i++)
printf("%d ", unique[i]);
return 0;
}
This is my codes,though there is a good answer .
I didn't realize the idea that know which array have the unique value.
I also think that the right answer that you chose didn't , either.
Here is my version of the algorithm for finding identical elements in sorted arrays on Python in C ++, it works in a similar way
def unique_array(array0 : (int), array1 : (int)) -> (int):
index0, index1, buffer = 0, 0, []
while index0 != len(array0) and index1 != len(array1):
if array0[index0] < array1[index1]:
buffer.append(array0[index0])
index0 += 1
elif array0[index0] > array1[index1]:
buffer.append(array1[index1])
index1 += 1
else:
index0 += 1; index1 += 1
buffer.extend(array0[index0 : len(array0)])
buffer.extend(array1[index1 : len(array1)])
return buffer