# algorithm to remove element from an array

This can be achieved in many ways, some of them are discussed here
– standard way when existing order is important: traverse through the elements, once element_to_be_deleted is found, shift remaining elements one position towards left. The complexity of this algorithm is O(n)
– when order of array is not important: take two pointers; one at the beginning of array and another at the end of the array. Increment the starting pointer till the element_to_be_deleted is found and decrement the ending pointer till element is not equals to element_to_be_deleted and then swap elements at start_pointer and end_pointer. The complexity of this algorithm is O(n)
– brute force algorithm. The complexity of this algorithm is O(n*n)

standard way when existing order is important @ O(n)

```	public int removeNumber(int[] a, int n) {

if (a == null || a.length == 0)
return -1;
int i = 0;
for (int j=0; j<a.length; j++)
if (a[j] != n)
a[i++] = a[j];

System.out.println("New size of the array is "+i);
return i; // The new dimension of the array
}
```

when order of array is not important @ O(n)

```	public void removeElementO_n(int[] a, int element) {
int startPtr = 0;
int endPtr = a.length-1;
//{1,4,2,4,3,5,4,4};
while(startPtr < endPtr) {
while(endPtr > 0 && a[endPtr] == element) {
a[endPtr]=-1;
endPtr--;
}

while(startPtr < a.length && a[startPtr] != element) {
startPtr++;
}

if(startPtr < endPtr) {
a[startPtr] = a[endPtr];
a[endPtr] = -1;
startPtr++;
endPtr--;
}
}
}
```

brute force algorithm @ O(n*n)

```public void removeElementO_n2(int[] a, int element) {
int ct=0;
int ln = a.length;
//{1,4,2,4,3,5,4,4};
for (int i = 0; i < a.length; i++) {
if(a[i] == element) {
ct++;
for (int j = i; j < a.length; j++) {
if((j+1)<a.length) {
a[j] = a[j+1];
}
}
a[ln-ct] = -1;
}
}
}
```

### One Response to algorithm to remove element from an array

1. Sainik Kumar Singhal says:

Your first method will actually remove the last element from array.

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