Algorithm for deleting multiple array elements and shifting array - c++

Question

I have an array let's say with 5 items, if element[i] is less than 3 need to move element[i+1] in place of element[i].
int array[5] = {4, 2, 3, 5, 1};
int number = 3;
for (int i = 0; i < number; i++)
{
if (array[i] > number)
{
for (int j = 0; j < i - 1; j++)
{
array[j] = array[j + 1];
}
number = number - 1;
}
}
expected result is array = {2, 3, 1, anyNumber, anyNumber};

Answer 1

A O(n) working code for the above problem.. But as others pointed out in the comments.. You end up with an array that is using less space then allocated to it..
#include<stdio.h>
int main()
{
int arr[] = {4, 2, 3, 5, 1};
int* temp1 = arr;
int* temp2 = arr;
int i, n1 = 5, n2 = 5;
for(i = 0; i < n1; i++)
{
if(*temp2 >= 3)
{
*temp1 = *temp2;
temp1++;
temp2++;
}
else
{
n2--; //the number of elements left in the array is denoted by n2
temp2++;
}
}
}

Answer 2

Nested loops give you O(n2) complexity, and non-obvious code.
Better use std::remove_if:
int array[5] = {4, 2, 3, 5, 1};
int number = 3;
remove_if( begin( array ), end( array ), [=]( int x ) { return x>number; } );
Disclaimer: code untouched by compiler's hands.

Answer 3

Try this code. You should not decrease number at each step. Also, the second loop should start at i and stop at the end of array:
int array[5] = {4, 2, 3, 5, 1};
int number = 3;
for (int i = 0; i < number; i++)
{
if (array[i] > number)
{
for (int j = i; j < 5; j++)
{
array[j] = array[j + 1];
}
}
}

Answer 4

Here's a more compact and idiomatic (that's how I view it anyway) way to remove items from an array:
#include <iostream>
#include <algorithm>
#include <iterator>
int main()
{
int array[] = {4, 2, 3, 5, 1};
int* begin = array;
int* end = begin + sizeof(array)/sizeof(array[0]);
int number = 3;
end = std::remove_if(begin, end, [&number](int v) {return v > number;});
std::copy(begin, end, std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
return 0;
}
Just for comparison, here's a version using std::vector:
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
int main()
{
std::vector<int> array = {4, 2, 3, 5, 1};
int number = 3;
auto end = std::remove_if(array.begin(), array.end(), [&number](int v) {return v > number;});
std::copy(array.begin(), end, std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
return 0;
}

Answer 5

As an alternative, if you want to keep your items, but denote what will be at some later time, "removed", the algorithm that can be used is stable_partition:
#include <algorithm>
#include <iostream>
#include <iterator>
#include <functional>
int main()
{
int vValues[] = {4,2,3,5,1};
// partition the values on left and right. The left side will have values
// <= 3, and on right >3. The return value is the partition point.
int *p = std::stable_partition(vValues, vValues + 5,
std::bind2nd(std::less_equal<int>(), 3));
// display information
std::cout << "Partition is located at vValues[" << std::distance(vValues, p) << "]\n";
std::copy(vValues, vValues + 5, std::ostream_iterator<int>(std::cout, " "));
}
Output:
Partition is located at vValues[3]
2 3 1 4 5
You will see that 2,3,1 are on the left of partition p, and 4,5 are on the right of the partition p. So the "removed" items start at where p points to. The std::partition ensures the elements are still in their relative order when done.

Answer 6

I created my own example, hope this helps people as a reference:
// Removing an element from the array. Thus, shifting to the left.
void remove(){
int a[] = {1, 2, 3, 4, 5, 6};
int size = sizeof(a)/sizeof(int); // gives the size
for(int i = 0; i < size; i++){
cout << "Value: " << a[i] << endl;
}
int index = 2; // desired index to be removed
for(int i = 0; i < size; i++){
if(i == index){
for(int j = i; j < size; j++){
a[j] = a[j+1];
}
}
}
size--; // decrease the size of the array
cout << "\nTesting output: " << endl;
for(int i = 0; i < size; i++){
cout << "Value: " << a[i] << endl;
}
}
int main(){
remove();
return 0;
}