Problem Statement
Given an array of unsorted numbers and a target, remove all occurances of target from it. After removing target in-place return the new length of the array.
In-place: Do not use Extra Space.
Example:
Input: arr: [1, 4, 9, 2, 3, 4, 4, 10, 4], target: 4 Output: 5 Explanation: After removing all occurances of 4 from arr. The new length of the array will be 5 with [1, 9, 2, 3, 10].
Code
Java
C++
Python
class Innoskrit {
public static int removeTarget(int arr[], int target) {
int unique = 0;
for(int duplicate = 0; duplicate < arr.length; duplicate++) {
if(arr[duplicate] != target) {
arr[unique] = arr[duplicate];
unique++;
}
}
return unique;
}
public static void main(String[] args) {
int arr[] = new int[] {1, 4, 9, 2, 3, 4, 4, 10, 4};
int target = 4;
int ans = Innoskrit.removeTarget(arr, target);
System.out.println(ans);
}
}
#include<bits/stdc++.h>
using namespace std;
class Innoskrit {
public:
static int removeTarget(vector<int> &arr, int target) {
int unique = 0;
for(int duplicate = 0; duplicate < arr.size(); duplicate++) {
if(arr[duplicate] != target) {
arr[unique] = arr[duplicate];
unique++;
}
}
return unique;
}
};
int main() {
vector<int> arr = {1, 4, 9, 2, 3, 4, 4, 10, 4};
int target = 4;
int ans = Innoskrit::removeTarget(arr, target);
cout << ans;
return 0;
}
class Innoskrit:
@staticmethod
def remove_target(arr, target):
unique = 0
for duplicate in range(0, len(arr)):
if arr[duplicate] != target:
arr[unique] = arr[duplicate]
unique += 1
return unique
if __name__ == '__main__':
arr = [1, 4, 9, 2, 3, 4, 4, 10, 4]
target = 4
ans = Innoskrit.remove_target(arr, target)
print(ans)
Output:
5
Time and Space Complexity
Time Complexity: O(N)
Space Complexity: O(1)