LeetCode | Remove Element | Amazon Interview Questions | Geek Hacker
Problem Statement
Given an integer array nums
and an integer val
, remove all occurrences of val
in nums
in-place. The relative order of the elements may be changed.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums
. More formally, if there are k
elements after removing the duplicates, then the first k
elements of nums
should hold the final result. It does not matter what you leave beyond the first k
elements.
Return k
after placing the final result in the first k
slots of nums
.
Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.
Custom Judge
The judge will test your solution with the following code:
int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
// It is sorted with no values equaling val.int k = removeElement(nums, val); // Calls your implementationassert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
assert nums[i] == expectedNums[i];
}
If all assertions pass, then your solution will be accepted.
Constraints
0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100
Examples
Example 1
Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).
Example 2
Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).
Analysis
In order to remove all occurrences of val
in nums
in-place, we use a counter variable to keep track of array elements other than val
. Then, we scan the array for elements not equal to val
.
Algorithm
- Declare a counter variable
count
, to count the number of elements except val. - Scan the array left to right.
- If the current element is not equal to
val
, add it tonums
in the place ofcount
.
Python 3 Code
def removeElement(self, nums, val):
count = 0
for i in range(len(nums)):
if nums[i] != val:
nums[count] = nums[i]
count += 1
return count
Complexity
Time Complexity
O(n) because we scanned the array once.
Space Complexity
O(1) because we haven’t used any data structure as per the problem requirements.
For more LeetCode problems’ solutions, visit my GitHub repo.
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