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 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.
Problem statement taken from: https://leetcode.com/problems/remove-element
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
Explanation
Brute force approach
The brute force approach that first appears is to create a new array and copy all the elements to this new array except val.
Then copy over this new array to the original array. But since the problem statement already mentions we have to do this in place, we cannot create a new array.
The time complexity of the above approach is O(N), but the space complexity is O(N) too.
Using two pointers
We can reduce the space complexity and modify the array in place using two pointers.
Let's check the algorithm.
- if nums.size() == 0
- return 0
- set i, j = 0
- loop for i = 0; i < nums.size() - 1; i++
- if nums[i] != val
- nums[j++] = nums[i] // assign nums[i] to nums[j] and then increment j.
- if nums[i] != val
- nums[j++] = nums[i] // assign nums[i] to nums[j] and then increment j.
- return j
C++ solution
class Solution {
public:
int removeElement(vector<int>& nums, int val) {
if(nums.size() == 0){
return 0;
}
int i, j = 0;
for(i = 0; i < nums.size() - 1; i++){
if(nums[i] != val){
nums[j++] = nums[i];
}
}
if(nums[i] != val){
nums[j++] = nums[i];
}
return j;
}
};
Golang solution
func removeElement(nums []int, val int) int {
if len(nums) == 0 {
return 0
}
i, j := 0, 0
for ; i < len(nums) - 1; i++ {
if nums[i] != val {
nums[j] = nums[i]
j++
}
}
if nums[i] != val {
nums[j] = nums[i]
j++
}
return j
}
Javascript solution
var removeElement = function(nums, val) {
if( nums.length == 0 ){
return 0;
}
let i = 0, j = 0;
for(; i < nums.length - 1; i++){
if( nums[i] != val ){
nums[j++] = nums[i];
}
}
if( nums[i] != val ){
nums[j++] = nums[i];
}
return j;
};
Let's dry-run our algorithm to see how the solution works.
Input: nums = [3, 2, 2, 3], val = 3
Step 1: if nums.size() == 0
4 == 0
false
Step 2: set i, j = 0, 0
Step 3: loop for i = 0; i < nums.size() - 1
i < 3
0 < 3
true
nums[i] != val
nums[0] != 3
3 != 3
false
i++
i = 1
Step 4: loop for i < nums.size() - 1
i < 3
1 < 3
true
nums[i] != val
nums[1] != 3
2 != 3
true
nums[j++] = nums[i]
nums[j] = nums[1]
nums[0] = 2
j++
j = 1
i++
i = 2
nums = [2, 2, 2, 3]
Step 4: loop for i < nums.size() - 1
i < 3
2 < 3
true
nums[i] != val
nums[1] != 3
2 != 3
true
nums[j++] = nums[i]
nums[j] = nums[1]
nums[1] = 2
j++
j = 2
i++
i = 3
nums = [2, 2, 2, 3]
Step 4: loop for i < nums.size() - 1
i < 3
3 < 3
false
Step 5: if nums[i] != val
nums[3] != 3
3 != 3
false
Step 6: return j
So we return the answer as 2.
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