Problem Statement
Given an integer array nums
and an integer val
, remove all occurrences of val
in nums
in-place. The order of the elements may be changed. Then return the number of elements in nums
which are not equal to val
.
Consider the number of elements in nums
which are not equal to val
be k
, to get accepted, you need to do the following things:
- Change the array
nums
such that the firstk
elements ofnums
contain the elements which are not equal toval
. The remaining elements ofnums
are not important as well as the size ofnums
. - Return
k
.
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 implementation assert 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.
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).
Constraints:
0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100
The goal is to remove all occurrences of val
from the array nums
in-place and return the count of remaining elements. The algorithm should ensure that the first k
elements of nums
contain only elements not equal to val
. The order of elements can be changed, and elements beyond k
are not relevant.
-
Use Two Pointers:
- A
writeIndex
pointer to track where the next non-val
element should be placed. - A
readIndex
pointer to iterate over the array.
- A
-
Traverse the Array (
readIndex
):- If
nums[readIndex]
is not equal toval
, move it tonums[writeIndex]
and incrementwriteIndex
.
- If
-
Final Step:
- Return
writeIndex
, which represents the count of elements that are not equal toval
.
- Return
- Time Complexity: (O(n)) (We iterate through the array once)
- Space Complexity: (O(1)) (No extra space is used apart from variables)
function removeElement(nums, val) {
let writeIndex = 0; // Pointer to place the next valid element
for (let readIndex = 0; readIndex < nums.length; readIndex++) {
if (nums[readIndex] !== val) {
nums[writeIndex] = nums[readIndex];
writeIndex++;
}
}
return writeIndex;
}
let nums1 = [3, 2, 2, 3];
let val1 = 3;
let k1 = removeElement(nums1, val1);
console.log(k1, nums1.slice(0, k1)); // Output: 2, [2, 2]
let nums2 = [0, 1, 2, 2, 3, 0, 4, 2];
let val2 = 2;
let k2 = removeElement(nums2, val2);
console.log(k2, nums2.slice(0, k2)); // Output: 5, [0, 1, 3, 0, 4]
- Initialize
writeIndex = 0
(starting position for non-val
elements). - Loop through the array (
readIndex
from 0 tonums.length - 1
):- If
nums[readIndex] !== val
, copynums[readIndex]
tonums[writeIndex]
and incrementwriteIndex
.
- If
- Return
writeIndex
, which represents the new length of the filtered array.