**Given an array and a value, how to implement a function to remove all instances of that value in place and return the new length? The order of elements can be changed. It doesn't matter what you leave beyond the new length.**

__Question:__
For example, if the input array is {4, 3, 2, 1, 2, 3, 6},
the result array after removing value 2 contains numbers {4, 3, 1, 3, 6}, and
the new length of the remaining array is 5.

**The most straightforward solution for this problem is to scan the whole array from the beginning to end. When a target number is scanned, remove it and move all number behind it backward. The overall complexity is O(**

__Analysis:__*n*

^{2}), since we have to move O(

*n*) numbers when a target value is removed.

We can notice that it is not required to keep the order for
the remaining numbers, and it does not care what numbers left beyond the new
length. Therefore, we can move all target numbers to be removed to the end of the
array.

Two pointers are defined to solve this problem: The first
pointer (denoted as p1) moves
forward until it reaches a number equal to the target value, which is initialized
to the beginning of array. The other (denoted as p2)
moves backward until it reaches a number not equal to the target value, which
is initialized to the end of array. Two numbers pointed by p1 and p2
are swapped. We repeat the moving and swapping operations until all target
numbers are scanned.

The sample code is shown as below:

unsigned int Remove(int*
numbers, unsigned int
length, int n)

{

if(numbers
== NULL || length < 1)

return
0;

int* p1 =
numbers;

int* p2 =
numbers + length - 1;

while(p1
< p2)

{

while(*p1
!= n && (p1 - numbers) < length)

++p1;

while(*p2
== n && (p2 - numbers) > 0)

--p2;

if(p1
< p2)

{

*p1 = *p2;

*p2 = n;

}

}

return p1 -
numbers;

}

Because p1 points
to the first target number in the array after scanning and swap, all elements
at the left side of p2 are the
remaining numbers. The new length can be calculated by the difference between
the beginning of array and p1.

Since it is only required to scan the whole array once, and
it costs O(1) time to swap a target value to the end of array, the overall time
complexity is O(

*n*) for an array with*n*numbers.The discussion about this problem is included in my book <Coding Interviews: Questions, Analysis & Solutions>, with some revisions. You may find the details of this book on Amazon.com, or Apress.

The author Harry He owns all the rights of this post. If you are going
to use part of or the whole of this ariticle in your blog
or webpages, please add a reference to http://codercareer.blogspot.com/. If you are going to use it in your books, please contact him via zhedahht@gmail.com . Thanks.

I don't get why we can't use a simple solution with a pointer, like this one:

ReplyDeletepublic int removeNumber(int[] A, int n) {

if (A == null || A.length == 0) return;

int i = 0;

for (int j=0; j<A.length; j++)

if (A[j] != n) A[i++] = A[j];

return i; // The new dimension of the array

}

Complexity is still O(n). What am I missing?

Your solution works. However, its time complexity is O(n) if the length of array is n. The time complexity of the solution in post above is O(k), if there are k target numbers in the array. Usually k is less than n, so the solution illustrated in the post is more efficient than yours.

DeleteHarry, I think you're wrong and your solution is O(n). Since the statements ++p1 and ++p2 run for a total of n times.

DeleteI mean ++p1 and --p2.

DeleteThank you anon for your comments. You are right because the time efficiency is O(n) indeed. However, the number of data moves is O(k) in my solution, but it has O(n) data moves in the first reply by "E.".

DeleteThe algorithm given by "E", could also achieve O(k) data moves by only moves at i>0, and his solution keeps the orignial order of the array, I guess should be better

DeleteHello Harry,

DeleteI am shocked, shocked, that there is such article exist!! But I really think you did a great job highlighting some of the key No. 32 - Remove Numbers in Array in the entire space.

Pretty new to all this, just switched to AWS from 1and1. They used to track the number of visitors and page views and other stats. How can I do this with AWS, just to see if I'm getting 5 vs 500 visitors per day? Any help/advice is appreciated.

Build a new AMI by first rotating up an example from a trusted AMI then add packages and components required AWS Training USA . Be cautious of placing sensitive data on an AMI. For example, access credentials of yours should be added to an instance after spinup with a database, position an external volume that operates your MySQL data after spinup as well.

Appreciate your effort for making such useful blogs and helping the community.

Thank you,

Kevin

Harry, why the exchange? If it does not matter what is beyond the new end, why not just copy the value from the end and leave it at that. You save one memory write operation per match.

ReplyDeleteSo the if could contain only: *p1 = *p2--;

何老师，假如输入是 A[2,2,2],length=3，n=2.你的这个函数还是原样输出数组A，这符合题意么？

ReplyDelete我想Harry是对的，因为他的函数输出的是0, 如果是你的case的话，所以是一个从numbers起始，长度为0的array

Delete你的这个思路在leetcode在线测试平台上运行有误

ReplyDelete你看这个双指针思路如何：

ReplyDeleteint removeElement(int A[], int length, int elem) {

int cur = 0;

for(int i =0; i< length; i++)

{

if(A[i] == elem)

continue;

A[cur]=A[i];

cur++;

}

return cur;

}

我觉得你这个做法是最机智的

Delete这是我的做法，用一前一后双指针不断替换。

ReplyDeletepublic int removeNumberInArray(int[] num, int target){

int i = 0, j = num.length-1;

while(i < j){

if(num[i]==target){

int temp = num[j];

num[j] = num[i];

num[i] = temp;

j--;

}else{

i++;

}

}

if(num.length==0){return 0;}

return num[i]==target?i:i+1;

}

Why we cannot just count how many times target value appears in the array and return array.length - count? There is no requirement to provide cleaned array, just its resulting length, so why bother?

ReplyDeleteThe bets use of collecting secondary data is when you have complex data and unable to find any solution.

ReplyDeleteHi Harry,

ReplyDeleteIn total awe…. So much respect and gratitude to you folks for pulling off such amazing blogs without missing any points on the No. 32 - Remove Numbers in Array . Kudos!

I joined the Amazon affiliate site and received my access key id and my secret key.

There was a problem downlaoding and saving the secret key so I created another one.

I am submitting this information to Amazon to complete my affilaite site but No. 32 - Remove Numbers in Array is accepting the information.

I would appreciate any information as to why this may be ocurring.

I look forward to see your next updates.

Kind Regards,

Irene Hynes

Greetings Mate,

ReplyDeleteThree cheers to you ! Hooray!!! I feel like I hit the jackpot on Remove Numbers in Array

I created a new AWS account but then realised a better thing to do would be to change the email address of an existing account to the new email address. I closed the new account but can't change the email address of the existing account to the new email as it now says "The new e-mail address you have provided is already in use. AWS Training USA Please use a different e-mail address."

Very useful article, if I run into challenges along the way, I will share them here.

,Merci

Radhey