I have just written a binary search on c++, using arrays, but it isn't working for all of my tests.
#include <iostream>
using namespace std;
int bSearch(int arr[], int item);
int main() {
int testArr[] = {1, 3, 5, 7, 9, 10, 12, 13, 15, 16, 18, 20};
int result = bSearch(testArr, 18);
cout << "The result of binary search is " << result << endl;
return 0;
}
int bSearch(int arr[], int item) {
int start = 0;
int middle(0), guess(0);
int finish = sizeof(arr);
while(start <= finish) {
middle = (start + finish) / 2;
guess = arr[middle];
if(guess == item)
return middle;
else if(guess > item)
finish = middle - 1;
else
start = middle + 1;
}
return -1;
}
Can you explain me, why is it so?
In bSearch, the parameter arr is not an array, but a pointer to an int. There is no information on if it points to an array of ints or the number of elements may be part of such an array.so sizeof(arr) will be the size of a pointer (typically 4 or 8).
You'll need to pass number of elements the array holds to bSearch, or use one of the standard containers that track the size (std::vector or std::array).
Related
I'm implementing Binary search. But I don't want to pass the size of an array as an argument in the binarySearch function. I'm trying to find array size in function. I use sizeof operator, but the output of the binary search is wrong. when I try to pass the size of array as an argument then the output is fine. My question is why & what is the problem for calculating array size in function? is it possible to calculate array size in function?
Here is my approach:
code
#include <bits/stdc++.h>
using namespace std;
void binarySearch(int arr[], int value)
{
int c = sizeof(arr) / sizeof(arr[0]);
//cout << c;
int low = 0, high = c - 1, notFound = 1;
while (low <= high)
{
int mid = (low + high) / 2;
if (arr[mid] == value)
{
cout << "Found at index " << mid << endl;
notFound = 0;
break;
}
if (arr[mid] < value)
{
low = mid + 1;
}
else
{
high = mid - 1;
}
}
if (notFound == 1)
{
cout << "Not Found" << endl;
}
}
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 9, 56, 88};
int x = 3;
binarySearch(arr, x);
//cout << p << endl;
}
output:
tempCodeRunnerFile.cpp:5:22: warning: 'sizeof' on array function parameter 'arr' will return size of 'int*' [-Wsizeof-array-argument]
int c = sizeof(arr) / sizeof(arr[0]);
^
tempCodeRunnerFile.cpp:3:23: note: declared here
void binarySearch(int arr[], int value)
~~~~^~~~~
Not Found
My expected Output:
Found at index 2
My question is why & what is the problem for calculating array size in function
The problem is that arr is a pointer to an element of the array. The size of the pointer has nothing to do with the size of the array, which is why your attempted sizeof(arr) cannot work.
The warning message also explains this quite well.
is it possible to calculate array size in function?
Given only a pointer to an element, it is generally1 not possible to determine the size of the array. That is why you must pass the size as an argument.
Another approach which I prefer is to encapsulate the pointer and the length in a class. There is a standard class template for this purpose: std::span.
void binarySearch(std::span<int> arr, int value)
{
int c = arr.size();
...
// call as in your example:
int arr[] = {1, 2, 3, 4, 5, 6, 9, 56, 88};
int x = 3;
binarySearch(arr, x);
1 There are techniques that allow this, but they have draw backs.
One technique is to choose one element value as a "terminator" aka "sentinel" which signifies the end of the array. That way, you can do a linear search for the terminator to determine the length. This is commonly used in C with strings, where the null terminator character terminates the string. This technique isn't an option if you cannot dedicate any of the values as a terminator. The other obvious drawback is the linear complexity of the size calculation which is asymptotically more expensive than the binary search that you're implementing
There is a further possibility that you could use, but calling it with too many different array sized might cause code bloat.
template<std::size_t N>
void binarySearch(int arr[N], int value)
And then use N instead of c.
I wanted to write a function which upon being called deletes an element from an array given that the parameters passed in the deleteArray function were the array, its length and the value of the element to be deleted.
Tried breaking out of the for loop while transversing through the array if the element was found and then tried using i's value in another for loop to replace the current elements with their next element.
like array[j] = array[j + 1]
Here is the code:
#include <iostream>
using namespace std;
void deleteElement(int[], int, int);
int main() {
int array1[] = { 1, 4, 3, 5, 6 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, length, 4);
cout << "\nIn main function\n";
for (int i = 0; i < length; i++) {
cout << array1[i];
}
return 0;
}
void deleteElement(int array2[], int length, int element) {
int i = 0;
for (int i; i < length; i++) {
if (array2[i] == element) {
for (int j = i; j < length; j++) {
array2[j] = array2[j + 1];
}
break;
}
}
if (i == (length - 1)) {
cout << ("Element doesn't exist\n");
}
cout << "Testing OP in deleteElement\n";
for (int i = 0; i < length; i++) {
cout << array2[i];
}
}
Expected:
Testing OP in deleteElement
14356
In main function
1356
Actual:
Testing OP in deleteElement
14356
In main function
14356
The problem is rather silly:
At the beginning of deleteElement(), you define i with int i = 0;, but you redefine another variable i as a local index in each for loop. The for loop introduces a new scope, so the int i definition in the first clause of the for loop defines a new i, that shadows the variable with the same name defined in an outer scope.
for (int i; i < length; i++) {
And you do not initialize this new i variable.
There are 2 consequences:
undefined behavior in the first loop as i is uninitialized. The comparison i < length might fail right away.
the test if (i == (length - 1)) { tests the outer i variable, not the one that for iterated on. Furthermore, the test should be if (i == length) {
There are other issues:
the nested for loop iterates once too many times: when j == length - 1, accessing array[j + 1] has undefined behavior.
you do not update length, so the last element of the array is duplicated. You must pass length by reference so it is updated in the caller's scope.
Here is a corrected version:
#include <iostream>
using namespace std;
void deleteElement(int array2[], int& length, int element);
int main() {
int array1[] = { 1, 4, 3, 5, 6 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, &length, 4);
cout << "\nIn main function\n";
for (int i = 0; i < length; i++) {
cout << array1[i] << " ";
}
return 0;
}
void deleteElement(int array2[], int& length, int element) {
int i;
for (i = 0; i < length; i++) {
if (array2[i] == element)
break;
}
if (i == length) {
cout << "Element doesn't exist\n";
} else {
length -= 1;
for (; i < length; i++) {
array2[i] = array2[i + 1];
}
}
cout << "Testing OP in deleteElement\n";
for (i = 0; i < length; i++) {
cout << array2[i] << " ";
}
}
If you use the algorithm function std::remove, you can accomplish this in one or two lines of code without writing any loops whatsoever.
#include <algorithm>
#include <iostream>
void deleteElement(int array2[], int& length, int element)
{
int *ptr = std::remove(array2, array2 + length, element);
length = std::distance(array2, ptr);
}
int main()
{
int array1[] = { 1, 4, 3, 5, 6 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, length, 4);
for (int i = 0; i < length; ++i)
std::cout << array1[i];
}
Output:
1356
Note that we could have written the deleteElement function in a single line:
void deleteElement(int array2[], int& length, int element)
{
length = std::distance(array2, std::remove(array2, array2 + length, element));
}
Basically, std::remove moves the removed element to the end of the sequence, and returns a pointer to the beginning of the removed elements.
Thus to get the distance from the beginning of the array to where the removed elements are located, usage of std::distance is done to give us our new length.
To remove only the first found element, std::find can be used, and then std::copy over the elements, essentially wiping out the item:
void deleteElement(int array2[], int& length, int element)
{
int *ptr = std::find(array2, array2 + length, element);
if ( ptr != array2 + length )
{
std::copy(ptr+1,array2 + length, ptr);
--length;
}
}
int main()
{
int array1[] = { 1, 4, 3, 5, 4, 6, 9 };
int length = sizeof(array1) / sizeof(array1[0]); //For length of array
deleteElement(array1, length, 4);
for (int i = 0; i < length; ++i)
std::cout << array1[i];
}
Output:
135469
There is no need for multiple loops in deleteElement. Additionally, your removal will fail to remove all elements (e.g. 4 in your example) if your array contains more than one 4, e.g.
int array1[] = { 1, 4, 3, 4, 5 };
You can simplify your deleteElement function and handle removing multiple occurrences of element simply by keeping a count of the number of times the element is found and by using your counter as a flag to control removal, e.g.:
void deleteElement(int array2[], int& length, int element)
{
int found = 0; /* flag indicating no. element found */
for (int i = 0; i < length; i++) { /* iterate over each element */
if (array2[i] == element) { /* check if matches current */
found += 1; /* increment number found */
continue; /* get next element */
}
if (found) /* if matching element found */
array2[i-found] = array2[i]; /* overwrite elements to end */
}
length -= found; /* update length based on no. found & removed */
}
Updating your example main() to show both pre-delete and post-delete, you could do something like the following:
int main (void) {
int array1[] = { 1, 4, 3, 4, 5 };
int length = sizeof array1 / sizeof *array1; //For length of array
cout << "\nBefore Delete\n";
for (int i = 0; i < length; i++)
cout << " " << array1[i];
cout << '\n';
deleteElement(array1, length, 4);
cout << "\nAfter Delete\n";
for (int i = 0; i < length; i++)
cout << " " << array1[i];
cout << '\n';
}
Example Use/Output
Which in the case where you array contains 1, 4, 3, 4, 5 would result in:
$ ./bin/array_del_elem
Before Delete
1 4 3 4 5
After Delete
1 3 5
While you are using an array of type int (of which there are many in both legacy and current code), for new code you should make use of the containers library (e.g. array or vector, etc...) which provide built in member functions to .erase() elements without you having to reinvent the wheel.
Look things over and let me know if you have further questions.
This is because the length of the array is never updated after deleting. Logically the length should decrease by 1 if the element was deleted.
To fix this, either
Pass the length by reference and decrease it by 1 if the element is actually deleted. OR
Return from the deleteElement some value which indicates that the element was deleted. And based of that, decrease the value of length in the main function.
Recalculating the array length will not help because the element is not actually deleted in memory. So the memory allocated to he array remains same.
Other issues:
The first for loop in deleteElement should run till j < length - 1.
The for loop creates a local variable i, which shadows the i variable in outer scope, so the outer i is never updated and always remains = 0
I have some code that rotates a number array to the left but instead, I need it to rotate it to the right. There is other code online that rotates array to the right but that code lets you only rotate numbers in the middle of the array.
I have tried decrementing the loops differently & and changing where its initialized but doesn't seem to rotate the correct way.
Expected output: if array is this {1, 2, 3, 4, 5, 6, 7}. Then it should look like: {7, 1, 2, 3, 4, 5, 6}
Current output: {2, 3, 4, 5, 6, 7, 1}
#include <iostream>
using namespace std;
/*Function to left Rotate arr[] of size n by 1*/
void leftRotatebyOne(int arr[], int n);
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
int i;
for (i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n-1; i++)
arr[i] = arr[i+1];
arr[i] = temp;
}
/* utility function to print an array */
void printArray(int arr[], int size)
{
int i;
for(i = 0; i < size; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
printArray(arr, 7);
leftRotate(arr, 1, 7);
cout << "___" << endl;
printArray(arr, 7);
getchar();
return 0;
}
leftRotateByOne is the key function here. The others can stay the same. Have a look at what it is doing, preferably with a pen and paper to keep track of the operations:
Keeps a copy of the first element.
Moves all elements to the "left" (that is, to the element with index
one less), being careful not to overwrite anything you need later.
Puts the first element in the last place.
So you need to do the opposite:
Keep a copy of the last element.
Moves all elements to the "right" (that is, to the element with index
one more), being careful not to overwrite anything you need later.
Puts the last element in the first place.
For example:
void rightRotatebyOne(int arr[], int n)
{
int i, last;
last = arr[n-1];
for (i = n-1; i > 0; i--)
arr[i] = arr[i-1];
arr[0] = last;
}
I am using sublime text 2 and I am trying to program bubble sort, and every time I run the code below it gives me an error on bubbleSort(num[5], terms); the error is
ERROR: no matching function for call to 'bubbleSort'.
Can anyone tell me why this is happening.
The code is:
#include <iostream>
using namespace std;
void bubbleSort(int arr[], int term) {
for(int i = 0; i < term; ++i) {
for(int index = 0; index < term-i-1; ++index) {
if(arr[index] < arr[index + 1]) {
int swap;
swap = arr[index];
arr[index] = arr[index + 1];
arr[index + 1] = swap;
}
}
}
for(int counter = 0; counter < term; counter++) {
cout << arr[counter] << endl;
}
}
int main() {
cout << "Hi in this program I will do bubble sort" << endl;
cout << "The numbers are 2, 9, 5, 10, 6"<< endl;
int num[5] = {2, 9, 5, 10, 6};
int terms = sizeof (num) / sizeof (num[0]);
bubbleSort(num[5], terms);
//answer = [2, 5, 6, 9, 2, 10]
}
Although a good question should be accompanied by a complete example of the compiler error, this answer should be fine.
void bubbleSort(int arr[], int term) {
// ...
}
It's a function which accepts an array of integers as first parameter and an integer as second one.
When you try to invoke it with:
int num[5] = {2, 9, 5, 10, 6};
// ...
bubbleSort(num[5], terms);
You're passing num[5] which is not an array of integer, but it should be an element of the array num, then an integer itself.
In short you're calling the function passing
bubbleSort(INT, INT);
and not, as requested by the function
bubbleSort(ARRAY_INT, INT);
That' why the compiler doesn't find a function which names bubbleSort and accept two integers.
Additional note
It's a little be out of context, but I want to suggest you to improve your C++ base skills, because the expression:
num[5]
It's perfectly wrong in your code, because it tries to access to the 6-th element in the array (which is composed by only 5 elements), that'll produce an out-of-bound behaviour.
Thanks for helping me now I know that I am supposed to use
bubbleSort(num, terms);
Pseudo Code:
int arr[ 5 ] = { 4, 1, 3, 2, 6 }, x;
x = find(3).arr ;
x would then return 2.
The syntax you have there for your function doesn't make sense (why would the return value have a member called arr?).
To find the index, use std::distance and std::find from the <algorithm> header.
int x = std::distance(arr, std::find(arr, arr + 5, 3));
Or you can make it into a more generic function:
template <typename Iter>
size_t index_of(Iter first, Iter last, typename const std::iterator_traits<Iter>::value_type& x)
{
size_t i = 0;
while (first != last && *first != x)
++first, ++i;
return i;
}
Here, I'm returning the length of the sequence if the value is not found (which is consistent with the way the STL algorithms return the last iterator). Depending on your taste, you may wish to use some other form of failure reporting.
In your case, you would use it like so:
size_t x = index_of(arr, arr + 5, 3);
Here is a very simple way to do it by hand. You could also use the <algorithm>, as Peter suggests.
#include <iostream>
int find(int arr[], int len, int seek)
{
for (int i = 0; i < len; ++i)
{
if (arr[i] == seek) return i;
}
return -1;
}
int main()
{
int arr[ 5 ] = { 4, 1, 3, 2, 6 };
int x = find(arr,5,3);
std::cout << x << std::endl;
}
The fancy answer:
Use std::vector and search with std::find
The simple answer
Use a for loop
If the array is unsorted, you will need to use linear search.
#include <vector>
#include <algorithm>
int main()
{
int arr[5] = {4, 1, 3, 2, 6};
int x = -1;
std::vector<int> testVector(arr, arr + sizeof(arr) / sizeof(int) );
std::vector<int>::iterator it = std::find(testVector.begin(), testVector.end(), 3);
if (it != testVector.end())
{
x = it - testVector.begin();
}
return 0;
}
Or you can just build a vector in a normal way, without creating it from an array of ints and then use the same solution as shown in my example.
int arr[5] = {4, 1, 3, 2, 6};
vector<int> vec;
int i =0;
int no_to_be_found;
cin >> no_to_be_found;
while(i != 4)
{
vec.push_back(arr[i]);
i++;
}
cout << find(vec.begin(),vec.end(),no_to_be_found) - vec.begin();
We here use simply linear search. At first initialize the index equal to -1 . Then search the array , if found the assign the index value in index variable and break. Otherwise, index = -1.
int find(int arr[], int n, int key)
{
int index = -1;
for(int i=0; i<n; i++)
{
if(arr[i]==key)
{
index=i;
break;
}
}
return index;
}
int main()
{
int arr[ 5 ] = { 4, 1, 3, 2, 6 };
int n = sizeof(arr)/sizeof(arr[0]);
int x = find(arr ,n, 3);
cout<<x<<endl;
return 0;
}
You could use the STL algorithm library's find function provided
#include <iostream>
#include <algorithm>
using std::iostream;
using std::find;
int main() {
int length = 10;
int arr[length] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int* found_pos = find(arr, arr + length, 5);
if(found_pos != (arr + length)) {
// found
cout << "Found: " << *found_pos << endl;
}
else {
// not found
cout << "Not Found." << endl;
}
return 0;
}
There is a find(...) function to find an element in an array which returns an iterator to that element. If the element is not found, the iterator point to the end of array.
In case the element is found, we can simply calculate the distance of the iterator from the beginning of the array to get the index of that element.
#include <iterator>
using namespace std;
int arr[ 5 ] = { 4, 1, 3, 2, 6 }
auto it = arr.find(begin(arr), end(arr), 3)
if(it != end(arr))
cerr << "Found at index: " << (it-begin(arr)) << endl;
else
cerr << "Not found\n";