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What it teaches: Keep an element only if it differs from the last kept one: read/write pointers on sorted data.
Practise it on judges as “Remove Duplicates from Sorted Array”.
nums is sorted. Remove duplicates in place so each value appears once, keeping order. Return the number of unique values k; the first k elements of nums must hold them.
Example 1
Input: nums = [1, 1, 2] Output: 2
nums starts with [1, 2].
Example 2
Input: nums = [0, 0, 1, 1, 1, 2, 2, 3, 3, 4] Output: 5
nums starts with [0, 1, 2, 3, 4].
Constraints
Pattern clues in the wording
These clues point to Two Pointers: Read and Write: A fast pointer reads every element and a slow pointer marks where the next kept element should be written.
Stuck? Take one hint at a time
class Solution {
public int removeDuplicates(int[] nums) {
int write = 1;
return write;
}
}Write your solution locally or in your editor for now. The in-browser runner (Java first, then Python, C++ and more) will run these tests right here.
Test cases
| # | Input | Expected |
|---|---|---|
| 1 | nums = [1,1,2] | 2 |
| 2 | nums = [0,0,1,1,1,2,2,3,3,4] | 5 |
+ 1 hidden test the code runner will check
From slow to fast
write = 1. For each read from 1, if nums[read] != nums[write − 1], copy it to nums[write++].
class Solution {
public int removeDuplicates(int[] nums) {
int write = 1;
for (int read = 1; read < nums.length; read++) {
if (nums[read] != nums[write - 1]) nums[write++] = nums[read];
}
return write;
}
}Verdict: One pass in place.
Before you submit
Test these inputs
Mistakes people make
nums[read − 1] works here too, but fails in the "at most two copies" follow-up; comparing with the last written value generalises.Interview
What if each value may appear at most twice?