I need a program that sorts an array of integers without using conditional statements. Numbers are in the range from 0 to 100 and don't repeat.
#include <iostream>
using namespace std;
int main() {
int arr[] = { 34, 12, 24, 65, 63, 22 };
int arraySize = (sizeof(arr) / sizeof(*arr));
unsigned char buf[101] = { 0 };
for (int k = 0; k < arraySize; k++) {
buf[arr[k]]++;
}
unsigned char i = 0;
for (int k = 0; k <= 100; k++) {
arr[i] = k;
i += buf[k];
}
for (int a : arr) {
cout << a << endl;
}
system("pause");
return 0;
}
This program works but I get the error after closing of the command prompt:
Run-Time Check Failure #2 - Stack around the variable 'arr' was corrupted.
Is there a way to fix it?
The problem is that your code writes past the end of the array. It happens after you have encountered the last element in the counted sequence, but before the array buf has been exhausted, i.e.
for (int k = 0; k <= 100; k++) {
arr[i] = k;
i += buf[k];
}
When you add the highest element, which is 65, to the result, i reaches 6, so assigning a[i] becomes illegal. See what's going on by adding an extra element to your array, setting it to -1, and watching what happens to it (it gets set to 100; demo 1).
You can fix it by adding an early exit condition to stop as soon as you filled the array back, i.e.
for (int k = 0; i < arraySize && k <= 100; k++) {
arr[i] = k;
i += buf[k];
}
Now the -1 past the end of "active" part of our array remains -1 (demo).
The logic of the second loop is wrong. You have six numbers in arr, no doubles, which means that a total of six elements in buf will be set to 1.
That means that after a while, the value of i will be 6, which you then use as an index into arr, but index 6 is the seventh element in an array, leading you to write out of bounds.
Related
I have been doing this problem for 2 days now, and I still can't figure out how to do this properly.
In this program, I have to input the number of sticks available (let's say 5). Then, the user will be asked to input the lengths of each stick (space-separated integer). Let's say the lengths of each stick respectively are [4, 4, 3, 3, 4]. Now, I have to determine if there are pairs (2 sticks of same length). In this case, we have 2 (4,4 and 3,3). Since there are 2 pairs, we can create a canvas (a canvas has a total of 2 pairs of sticks as the frame). Now, I don't know exactly how to determine how many "pairs" there are in an array. I would like to ask for your help and guidance. Just note that I am a beginner. I might not understand complex processes. So, if there is a simple (or something that a beginner can understand) way to do it, it would be great. It's just that I don't want to put something in my code that I don't fully comprehend. Thank you!
Attached here is the link to the problem itself.
https://codeforces.com/problemset/problem/127/B
Here is my code (without the process that determines the number of pairs)
#include<iostream>
#include<cmath>
#define MAX 100
int lookForPairs(int numberOfSticks);
int main(void){
int numberOfSticks = 0, maxNumOfFrames = 0;
std::cin >> numberOfSticks;
maxNumOfFrames = lookForPairs(numberOfSticks);
std::cout << maxNumOfFrames << std::endl;
return 0;
}
int lookForPairs(int numberOfSticks){
int lengths[MAX], pairs = 0, count = 0, canvas = 0;
for(int i=0; i<numberOfSticks; i++){
std::cin >> lengths[i];
}
pairs = floor(count/2);
canvas = floor(pairs/2);
return count;
}
I tried doing it like this, but it was flawed. It wouldn't work when there were 3 or more integers of the same number (for ex. [4, 4, 3, 4, 2] or [5. 5. 5. 5. 6]). On the first array, the count would be 6 when it should only be 3 since there are only three 4s.
for(int i=0; i<numberOfSticks; i++){
for (int j=0; j<numberOfSticks; j++){
if (lengths[i] == lengths[j] && i!=j)
count++;
}
}
Instead of storing all the lengths and then comparing them, count how many there are of each length directly.
These values are known to be positive and at most 100, so you can use an int[100] array for this as well:
int counts[MAX] = {}; // Initialize array to all zeros.
for(int i = 0; i < numberOfSticks; i++) {
int length = 0;
std::cin >> length;
counts[length-1] += 1; // Adjust for zero-based indexing.
}
Then count them:
int pairs = 0;
for(int i = 0; i < MAX; i++) {
pairs += counts[i] / 2;
}
and then you have the answer:
return pairs;
Just an extension to molbdnilo's answer: You can even count all pairs in one single iteration:
for(int i = 0; i < numberOfSticks; ++i)
{
if(std::cin >> length) // catch invalid input!
{
pairs += flags[length] == 1; // add a pair if there is already a stick
flags[length] ^= 1; // toggle between 0 and 1...
}
else
{
// some appropriate error handling
}
}
Note that I skipped subtracting 1 from the length – which requires the array being one larger in length (but now it can be of smallest type available, i.e. char), while index 0 just serves as an unused sentinel. This variant would even allow to use bitmaps for storing the flags, though questionable if, with a maximum length that small, all this bit fiddling would be worth it…
You can count the number of occurrences using a map. It seems that you are not allowed to use a standard map. Since the size of a stick is limited to 100, according to the link you provided, you can use an array, m of 101 items (stick's minimum size is 1, maximum size is 100). The element index is the size of the stick. The element value is the number of sticks. That is, m[a[i]] is the number of sticks of size a[i]. Demo.
#define MAX 100
int n = 7;
int a[MAX] = { 1,2,3,4,1,2,3 };
int m[MAX + 1]; // maps stick len to number of sticks
void count()
{
for (int i = 0; i < n; ++i)
m[a[i]]++;
}
int main()
{
count();
for (int i = 1; i < MAX + 1; ++i)
if (m[i])
std::cout << i << "->" << m[i] << std::endl;
}
Your inner loop is counting forward from the very beginning each time, making you overcount the items in your array. Count forward from i , not zero.
for(int i=0; i<numberOfSticks; i++)
{
for (int j=i; j<numberOfSticks; j++) { // count forward from i (not zero)
if (lengths[i] == lengths[j] && i!=j)
{ // enclosing your blocks in curly braces , even if only one line, is easier to read
count++; // you'll want to store this value somewhere along with the 'length'. perhaps a map?
}
}
}
I'm trying to find a way to iterating while comparing the element with the next element to find what is the biggest element in the array. But, the output i want keep repeating as much as the loop run.
int main(){
int array[4];
for ( int i = 0; i < 4; i++){
cin >> array[i];
}
for (int i:array){
for (int j = 1; j < 4; j++){
if (i < array[j]){
break;
}
if (i > array[j] ){
cout << i;
}
}
}
}
You can use the following program to find the biggest element in the array. Note that there is no need to use two for loops as you did in your code snippet.
#include <iostream>
int main()
{
int array[4] = {1,10, 13, 2};
int arraySize = sizeof(array)/sizeof(int);//note that you can also use std::size() with C++17
int startingValue = array[0];
for(int i = 1; i < arraySize; ++i)//start from 1 instead of 0 since we already have array[0]
{
if(array[i] > startingValue)
{
startingValue = array[i];
}
}
//print out the biggest value
std::cout<<"the biggest element in the array is: "<<startingValue<<std::endl;
return 0;
}
Your program is reapeating output because you have the cout inside the if which is satisfied multiple times(depending upon how big the array is and what elements it contains). For example, if you try your example on the array int array[] = {23,2,13,6,52,9,3,78}; then the output of your program will be 2323231313652525293787878 . So the output is reapeating more than 2 times. You can instead use the version i gave that uses only 1 for loop and prints the correct biggest element only once.
Note that you can also use std::size with C++17 while sizeof(array)/sizeof(int) works with all C++ versions.
I am trying to implement a small program that iterates over a 2d vector backwards and adds the value of the element.
If that element have already been added then I want to overwrite the value of the vector with 99.
So for example if number of climbs is four then add the of the program points should have the value 5 and the vector should look like this at the end
{1, 1 ,1},
{99, 1, 1},
{99(should start here), 99, 99}
But I keep getting a segmentation fault and I don't know whether I am iterating over the vector backwards incorrectly.
This is my full code
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<vector<int>> vect
{ {3}, //finish here
{1, 1 ,1},
{2, 1, 1},
{1, 1, 1} //start here
};
int points = 0;
for (int i = 0; i < vect.size(); i++)
{
for (int j = 0; j < vect[i].size(); j++)
{
cout << vect[i][j] << " ";
}
cout << endl;
}
int visited = 99;
int number_of_climbs = 4;
for(int i = 4; i >= 0; i--)
for (int j = 0; j < number_of_climbs; j++)
{
if(vect[i][j] != 99)
{
points += vect[i][j];
vect[i][j] = 99;;
continue;
}
}
return 0;
}
Looping over things backwards always trips me up too, especially when I accidentally declare my loop using an unsigned int. Your backwards loop is close to correct, but can be simplified into:
for (int i = vect.size() - 1; i >= 0; --i)
{
for (int j = vect[i].size() - 1; j >= 0; --j)
{
}
}
You need to start at size() - 1 because the only indices available are [0, size) (that is EXCLUSIVE of size), so in a vector of 4 elements, that's 0, 1, 2 and 3 (and you need to start at 3), which is likely the cause of your segmentation fault since you start your initial loop at 4 so it's immediately out-of-bounds.
Here in the inner loop you can see we account for the vectors being different sizes by just using the size of each vector.
It's also possible to use reverse iterators, but iterators can be confusing to people:
for (auto outer = vect.rbegin(); outer != vect.rend(); ++outer)
{
for (auto inner = outer->rbegin(); inner != (*outer).rend(); ++inner)
{
}
}
Note I've shown both ways to access the iterator, using it-> or (*it). Within the inner loop, you will simply use *inner to get the actual value held by the vector, so if you simply had std::cout << *inner << " "; that would print the 2d vector on one line in reverse order.
Even though we're looping in reverse, we still use ++
I have Array A[9]= {1,2,3,4,5,6,7,8,9} and I need to delete the numbers which are not dividing by 2. The code I tried to do:
int main()
{
int n;
ifstream fd(Cdf);
fd>>n; // read how many numbers are in the file.
int A[n];
for(int i = 0; i < n; i++)
{
fd >> A[i]; //read the numbers from file
}
for(int i = 0; i < n; i ++) // moving the numbers.
{
if(A[i] % 2 !=0)
{
for(int j = i; j < n; j++)
{
A[i] = A[i+1];
}
}
}
fd.close();
return 0;
}
But I get numbers like 224466888. what I need to do to get 2,4,6,8?
I need to delete numbers in the same array.
First you should use std::vector for dynamic size arrays.
Second, for removing numbers that are even in a vector, you can do :
std::vector<int> inf = {12,0,5,6,8};
auto func = [](int i){return i % 2 != 0;};
inf.erase(std::remove_if(inf.begin(),inf.end(),func), inf.end());
EDIT :
Ok, so you can still do this without std::vectors, but it will be uglier :
#include <algorithm>
int res[] = {2,5,9,8,6,7};
int size = 6;
auto func = [](int i){return i % 2 != 0;};
int new_size = std::remove_if(res,res + size, func) - res;
All the data you want is in [0, new_size[ range, the other part of your array is now garbage.
Your removal loop is indexing with the wrong variable:
for(int j = i; j < n; j++)
{
A[i] = A[i+1];
}
You're using i, which doesn't change in the loop.
Change it to j. You also need to subtract one from the upper limit, as you'd step outside of the array otherwise when accessing A[j + 1].
for(int j = i; j < n - 1; j++)
{
A[j] = A[j + 1];
}
An array can't be used for your purpose. It is allocated on stack and its size can't be changed dynamically (you can't change the size of an array in general, not only when it is allocated on stack).
You could allocate a second array and keep reallocating it with realloc everytime you add a new element but that's not the good way to do it. You are working with C++ so just use a std::vector<int> and your problems will be solved:
std::vector<int> evenArray;
evenArray.reserve(sizeof(A)/sizeof(A[0])/2);
if (number is even) {
evenArray.pushBack(number);
}
Mind that vector stores elements contiguously so this is legal:
int *evenA = &evenArray[0];
For your inner for loop you should be referencing j, not i.
for(int j = i; j < n - 1; j++)
{
A[j] = A[j+1];
}
Otherwise, what's the point of creating j?
Of course, this also means if you read the whole array back you will display all the characters that were shifted (which will just be equal to the last number). So, you should probably keep track of the new length of the array and just iterate to that instead of the end of the array.
EDIT:
In the inner for loop you need to loop to n - 1 otherwise when you have A[j + 1] it will go off the end of the array when you to change it, which may or may not give you a runtime error.
I haven't done any programming classes for a few years, so please forgive any beginner mistakes/methods of doing something. I'd love suggestions for the future. With the code below, I'm trying to check the values of two arrays (sorted already) and put them into a combined array. My solution, however inefficient/sloppy, is to use a for loop to compare the contents of each array's index at j, then assign the lower value to index i of the combinedArray and the higher value to index i+1. I increment i by 2 to avoid overwriting the previous loop's indexes.
int sortedArray1 [5] = {11, 33, 55, 77, 99};
int sortedArray2 [5] = {22, 44, 66, 88, 00};
combinedSize = 10;
int *combinedArray;
combinedArray = new int[combinedSize];
for(int i = 0; i <= combinedSize; i+=2)
{
for(int j = 0; j <= 5; j++)
{
if(sortedArray1[j] < sortedArray2[j])
{
combinedArray[i] = sortedArray1[j];
combinedArray[i+1] = sortedArray2[j];
}
else if(sortedArray1[j] > sortedArray2[j])
{
combinedArray[i] = sortedArray2[j];
combinedArray[i+1] = sortedArray1[j];
}
else if(sortedArray1[j] = sortedArray2[j])
{
combinedArray[i] = sortedArray1[j];
combinedArray[i+1] = sortedArray2[j];
}
}
}
for(int i = 0; i < combinedSize; i++)
{
cout << combinedArray[i];
cout << " ";
}
And my result is this
Sorted Array 1 contents: 11 33 55 77 99
Sorted Array 2 contents: 0 22 44 66 88
5 77 5 77 5 77 5 77 5 77 Press any key to continue . . .
In my inexperienced mind, the implementation of the sorting looks good, so I'm not sure why I'm getting this bad output. Advice would be fantastic.
what about this:
int i=0,j=0,k=0;
while(i<5 && j<5)
{
if(sortedArray1[i] < sortedArray2[j])
{
combinedArray[k]=sortedArray1[i];
i++;
}
else
{
combinedArray[k]=sortedArray2[j];
j++;
}
k++;
}
while(i<5)
{
combinedArray[k]=sortedArray1[i];
i++;k++;
}
while(j<5)
{
combinedArray[k]=sortedArray2[j];
j++; k++;
}
Firstly, there are some immediate problems with how you use C++:
You use = instead of == for equality check (hence causing undesired value assignments and the if-condition to return true when it shouldn't);
Your outer loops upper boundary is defined as i <= 10, while the correct boundary check would be i < 10;
You have a memory leak at the end of the function because you fail to de-allocate memory. You need a delete [] combinedArray at the end.
Secondly, your outer loop iterates through all values of the destination array, and in each step uses an inner loop to iterate through all values of the source arrays. That is not what you want. What you want is one loop counting from j=0 to j<5 and iterating through the source arrays. The positions in the destination array are then determined as 2*j and 2*j+1, and there is no need for an inner loop.
Thirdly, as explained in the comment, a correct implementation of sorted-list merge needs two independent counters j1 and j2. However, your current input is hardwired into the code, and if you replace 00 with 100, your current algorithm (after the corrections above are made) will actually work for the given input.
Finally, but less importantly, I wonder why your destination array is allocated on the heap using new. As long as you are dealing with small arrays, you may allocate it on the stack just like the source arrays. If, however, you allocate it on the heap, better use a std::unique_ptr<>, possibly combined with std::array<>. You'll get de-allocation for free then without having to think of putting a delete [] statement at the end of the function.
Before even looking at the implementation, check the algorithm and write it down with pen and paper. The first thing that pops is that you are assuming that the first two elements in the result will come one from each source array. That need not be the case, consider two arrays where all elements in one are smaller than all elements in the other and the expected result:
int a[] = { 1, 2, 3 };
int b[] = { 4, 5, 6 };
If you want the result sorted, then the first three elements come all from the first array. With that in mind think on what you really know about the data. In particular, both arrays are sorted, which means that the first elements will be smaller than the rest of the elements in the respective array. The implication of this is that the smaller element is the smaller of the heads. By putting that element into the result you have reduced the problem to a smaller set. You have a' = { 2, 3 }, b = { 4, 5, 6 } and res = { 1 } and a new problem that is finding the second element of res knowing that a' and b are sorted.
Figure out in paper what you need to do, then it should be straight forward to map that to code.
So, I modified your code to make it work. Actually it would be good idea to have two pointer/index for two sorted arrays. So that you can update your corresponding pointer after adding it to your combinedArray. Let me know if you don't understand any part of this code. Thanks.
int sortedArray1 [5] = {11, 33, 55, 77, 99};
int sortedArray2 [5] = {0, 22, 44, 66, 88};
int combinedSize = 10;
int *combinedArray;
combinedArray = new int[combinedSize];
int j = 0;
int k = 0;
for(int i = 0; i < combinedSize; i++)
{
if (j < 5 && k < 5) {
if (sortedArray1[j] < sortedArray2[k]) {
combinedArray[i] = sortedArray1[j];
j++;
} else {
combinedArray[i] = sortedArray2[k];
k++;
}
}
else if (j < 5) {
combinedArray[i] = sortedArray1[j];
j++;
}
else {
combinedArray[i] = sortedArray2[k];
k++;
}
}
for(int i = 0; i < combinedSize; i++)
{
cout << combinedArray[i];
cout << " ";
}
cout<<endl;
The else if(sortedArray1[j] = sortedArray2[j]), did you mean else if(sortedArray1[j] == sortedArray2[j])?
The former one will assign the value of sortedArray2[j] to sortedArray1[j] -- and that's the reason that why you get 5 77 5 77...
But where's the 5 come from? There's no 5 in either sortedArray, yet I find for(int j = 0; j <= 5; j++) must be something wrong. The highest index of a size N array is N-1 rather than N in C/C++(but not in Basic).. so use j<5 as the condition, or you may fall into some situation which is hard to explain or predict..
After all, there's problem in your algorithm itself, every time the outer loop loops, it will at last compare the last elements in the two arrays, which makes the output to repeat two numbers.
So you need to correct your algorithm too, see Merge Sort.
Slightly different approach, which is IMHO a bit cleaner:
//A is the first array, m its length
//B is the second array, n its length
printSortedAndMerged(int A[], int m, int B[], int n){
int c[n+m];
int i=0, j=0;
for(int k=0; k < n+m; k++){
if(i < m && j < n){
if(A[i] < B[j]){
c[k] = A[i];
i++;
}
else{
c[k] = B[j];
j++;
}
continue; //jump to next iteration
}
if(i < m){ // && ~(j < n)
//we already completely traversed B[]
c[k] = A[i];
i++;
continue;
}
if(j < n){ // %% ~(i < m)
//we already completely traversed A[]
c[k] = B[j];
j++;
continue;
}
//we should never reach this
cout << "Wow, something wrong happened!" << endl;
}//for
for(int i=0; i<n+m; i++){
cout << c[i] << endl;
}
}
Hope it helps.