Remove Element from Array LeetCode solution in Java

Remove Element from Array LeetCode solution in Java | Programming blog

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.

See full problem description on LeetCode :-

Deleting items from an array

Lets see solutions :-

Solution 1 :-

class Solution {
    public int removeElement(int[] nums, int val) {
    int temp = 0;
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] != val) {
            nums[temp] = nums[i];
            temp++;
        }
    }
    return temp;
    }
}

Input and Output :-

Input
nums = [3,2,2,3], val = 3
Output
2, nums = [2,2]
___________________
Input
nums = [0,1,2,2,3,0,4,2], val = 2
Output
5, nums = [0,1,4,0,3]

Explanation :-

We have to remove given value from array.

Initialize any temporary variable with 0.
Loop through given array nums.
In if condition check given val is not matching with current array value.
If val is not match with current array element, then assign nums[i] array element to current array element

nums [3, 2, 2, 3], val = 3
i = 0, temp = 0.
0 is less than array length 4, so it goes in loop.

nums[0] != 3 | 3 != 3 | false. it does not go into if loop.

i = 1, temp = 0.

nums[1] != 3 | 2 != 3 | true. it goes into if loop
nums[temp] = nums[i] | nums[0] = nums[1] | assign 1st element to 0th element into nums | nums [2, 2, 2, 3]
temp++ | 1

i = 2, temp = 1, numa = [2, 2, 2, 3]

nums[2] != 3 | 2 != 3 | true. it goes into if loop
nums[temp] = nums[i] | nums[1] = nums[2] | assign 2st element to 1th element into nums | nums [2, 2, 2, 3]
temp++ | 2

i = 3, temp = 2, numa = [2, 2, 2, 3]

nums[3] != 3 | 3 != 3 | false. it does not goes into if loop.

i = 4, temp = 2, numa = [2, 2, 2, 3]
Now i become 4, so i does goes into fro loop. and we got answer.

It doesn't matter what you leave beyond the returned length. For example if you return 2 with nums = [2,2,3,3] or nums = [2,2,0,0], your answer will be accepted.
(From leetcode).

Solution 2 :-

public int removeElement(int[] nums, int val) {
    int i = 0;
    int length = nums.length;
    while (i < length) {
        if (nums[i] == val) {

            // assign last array element to current array position 
            nums[i] = nums[length - 1];
            // reduce array size by one
            length--;
        } else {
            i++;
        }
    }
    return length;
}

In above solution we simply assign last array element to current array index.

Explanation :-

nums[0, 1, 2, 2, 3, 0, 4, 2], val = 2, length = 8, i = 0

First two time nums[i] == val is false so it goes to else condition and increment i value.

nums[0, 1, 2, 2, 3, 0, 4, 2], val = 2, length = 8, i = 2

nums[i] == val | 2 == 2 becomes true and goes in if condition.

nums[i] = nums[length - 1] | nums[2] = nums[7]. So nums array last element is 2 and we are copy 2 to current index 2. and decrement length by 1.

nums[0, 1, 2, 2, 3, 0, 4, 2], val = 2, length = 7, i = 2

nums[i] == val | 2 == 2 becomes true and goes in if condition.

nums[i] = nums[length - 1] | nums[2] = nums[6]. So nums array last element is 4 and we are copy 4 to current index 2. and decrement length by 1.

nums[0, 1, 4, 2, 3, 0, 4, 2], val = 2, length = 6, i = 2
nums[0, 1, 4, 2, 3, 0, 4, 2], val = 2, length = 6, i = 3

nums[i] == val | 2 == 2 becomes true and goes in if condition.

nums[i] = nums[length - 1] | nums[3] = nums[5]. So nums array last element is 0 and we are copy 0 to current index 3. and decrement length by 1.

nums[0, 1, 4, 0, 3, 0, 4, 2], val = 2, length = 5, i = 3
nums[0, 1, 4, 0, 3, 0, 4, 2], val = 2, length = 5, i = 4
nums[0, 1, 4, 0, 3, 0, 4, 2], val = 2, length = 5, i = 5
And now, i < length | 5 < 5 becoms false and return length that is our answer.

Happy Coding.

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