Leetcode Array101 Remove Element in GO

Accepted Solutions Runtime Distribution

Accepted Solutions Memory Distribution

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.

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

func removeElement(nums []int, val int) int {
    length := len(nums) - 1
    i := 0
    for i <= length {
        if nums[i] == val {
            for j := i; j < length; j++ {
                nums[j] = nums[j+1]
            }
            nums[length] = 0 // It does not matter.
            length--
        } else {
            i++
        }
    }
    return length + 1
}

• First I thought I need to use for loop, So I implemented i value as for loop with range.
• Later I realized that When value will be shifted, i+1 value would be in nums(0).
• As a result, If Clause cannot check this value.
• So I changed to i<= length.
• Because length has a number which represents numbers in array.
• Also length value can be used as the last array location.
• That means I just need to put 0 value on it after shift.
• After this, the length decrement.
• Last step is returning about length +1. because the length shows array length.
  https://go.dev/play/p/z89pv5eCU1v

sample 0 ms submission

func removeElement(nums []int, val int) int {
    left, right := 0, len(nums)

    for left < right {
        if nums[left] == val {
            right--
            nums[left], nums[right] = nums[right], nums[left]
        } else {
            left++
        }
    }

    return left
}

• In this code, Two pointers are used. Left, Right
• First Left and right point each end side.
• If there is not value to remove, len(nums) equals the numbers.
• but if clause finds the value to remove, right moves to left(-1)
• After that they switch both values, left and right.
• Right array has a value to be ignored or removed.
• If the value is not found from left array, left moves to right(+1)
• When left pointer reaches to right, For loop ends with a value of numbers.
• There is no need to modificaion on return.

sample 2000 KB submission

func removeElement(nums []int, val int) int {
    hits := 0

    for _, v := range nums {
        if v != val {
            nums[hits] = v
            hits++
        }
    }

    return hits
}

• Looks very clean code.
• For loop outputs a value in array.
• hits variable is used as index and acts as counter.
• They are sorted by subtracting the number to be excluded in order.
• Very simple