Remove Element from Unsorted Array - Easy

-

 Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}


Solution:

The requirement here is to remove all occurrences of an element 'val' from the array. 

We can solve this problem using many approaches but the one that I am going to talk about here is two-pointer approach.  

The fast pointer or the forward looking pointer here is 'i' and the slow pointer is 'j'. Whenever we encounter the value 'val' that is to be removed we note it's index by pointing the slow pointer 'j' to it's index and then move all the elements backwards. We repeat this step till all the occurrence of element 'val' is removed.


Time Complexity: O(n)

Space Complexity: O(1)

Comments

Post a Comment

Popular posts from this blog

Longest Common Prefix

-
Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string  "" . Example 1: Input: ["flower","flow","flight"] Output: "fl" Example 2: Input: ["dog","racecar","car"] Output: "" Explanation: There is no common prefix among the input strings. Note: All given inputs are in lowercase letters  a-z . Solution: The requirement here is to find longest common prefix amongst an array of strings.  There are many approaches to this problem but the one that I am going to discuss here uses Trie to store the array of strings and then scan the trie to get the longest common prefix. To know more about Trie, you can visit here . Algorithm: 1) Store each string from the array of strings in Trie. 2) Get the deepest Trie Node such that all the nodes from the root of the trie to that node has only one children. Space Complexity: O(n) ...
Read more

Roman To Integer

-
Roman numerals are represented by seven different symbols:  I ,  V ,  X ,  L ,  C ,  D  and  M . Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, two is written as  II  in Roman numeral, just two one's added together. Twelve is written as,  XII , which is simply  X  +  II . The number twenty seven is written as  XXVII , which is  XX  +  V  +  II . Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not  IIII . Instead, the number four is written as  IV . Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as  IX . There are six instances where subtraction is used: I  can be placed before  V  (5) and  X  (10) to ...
Read more

Two Sum

-
Given an array of integers, return   indices   of the two numbers such that they add up to a specific target. You may assume that each input would have  exactly  one solution, and you may not use the  same  element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[ 0 ] + nums[ 1 ] = 2 + 7 = 9, return [ 0 , 1 ]. Solution: Since for each element 'x' present in the array, we will have to check whether a number 'y' exists in the array such that sum of two numbers is target and return indices of the number.  We can also say that, for any element 'x' present in the array , find a number 'target-x' which is also present in the array. One approach to the problem is to scan the complete array for finding number 'target-x' for each element 'x'.  Another approach is to use dictionaries. For each element 'x' in the array, check whether the dictionary contains element 'target-x'. So here, first we will scan the...
Read more