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Find Non-Duplicate Number Instances (easy)

## Problem Statement

Given an array of sorted numbers, move all non-duplicate number instances at the beginning of the array in-place. The relative order of the elements should be kept the same and you should **not use any extra space** so that the solution has constant space complexity i.e., O(1).

Move all the unique number instances at the beginning of the array and after moving return the length of the subarray that has no duplicate in it.

**Example 1**:

```
Input: [2, 3, 3, 3, 6, 9, 9]
Output: 4
Explanation: The first four elements after moving element will be [2, 3, 6, 9].
```

**Example 2**:

```
Input: [2, 2, 2, 11]
Output: 2
Explanation: The first two elements after moving elements will be [2, 11].
```

**Constraints:**

- 1 <= nums.length <= 3 * 10<sup>4</sup>
`-100 <= nums[i] <= 100`

`nums`

is sorted in non-decreasing order.

## Try it yourself

Try solving this question here:

Python3

Python3

. . .

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