Question
Read the Question here https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/
Answer
Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.size() == 0) return 0;
int i = 0;
for (int j = 1; j < nums.size(); j++) {
if (nums[i] != nums[j]) {
i++;
nums[i] = nums[j];
}
}
return i + 1;
}
};
This algorithm will work only on sorted array .
The function take a vector of integer (as a reference) and return an integer length of modifies vector after removing the duplicate integer. The function start with a if condition which check if the vector is empty and if it is true it will return 0 , if it is false and there is elements in vector it will continue. The algorithm uses two pointer i and j which itreate over the vector and and check for duplicate integers. The pointer i starts with the first element and the for loop where j starts with second element. Inside the for loop, if statement check if index i element is not equal to index j element, and if this is the case, this mean the index j element is not duplicate of index i element , so the element j is assigned to i+1. Then the pointer i is increment with 1. This process continue till the end of the vector.
At the end of the function, the pointer i+1 is returned as the length of the modified vector after removing duplicates.
Its time complextity is O(n) as we are using only 1 for loop so the time complexity in depend on the size of the vector.
This is how I debug, you can go through it to see what is happening in the code.
vector -> 0 0 1 1 1 2 2 3 3 4
i=0 -> ↑
j=1 -> ↑
if (0!=0) ❌
----------------------------------------
Vector -> 0 0 1 1 1 2 2 3 3 4
i=0 -> ↑
j=2 -> ↑
if (0!=1) ✅
Vector -> 0 0 1 1 1 2 2 3 3 4
i=1 -> ↑
nums[i] = nums[j]
nums[1] = nums[2]
0 = 1
Vector modified (0 is replaced with 1)
↓
Vector -> 0 1 1 1 1 2 2 3 3 4
----------------------------------------
Vector -> 0 1 1 1 1 2 2 3 3 4
i=1 -> ↑
j=3 -> ↑
if (1!=1)❌
----------------------------------------
Vector -> 0 1 1 1 1 2 2 3 3 4
i=1 -> ↑
j=4 -> ↑
if (1!=1) ❌
----------------------------------------
Vector -> 0 1 1 1 1 2 2 3 3 4
i=1 -> ↑
j=5 -> ↑
if (1!=2)✅
Vector -> 0 1 1 1 1 2 2 3 3 4
i=2 -> ↑
nums[i] = nums[j]
nums[2] = nums[5]
1 = 2
Vector modified (1 is replaced with 2)
↓
Vector -> 0 1 2 1 1 2 2 3 3 4
----------------------------------------
Vector -> 0 1 2 1 1 2 2 3 3 4
i=2 -> ↑
j=6 -> ↑
if (2!=2) ❌
----------------------------------------
Vector -> 0 1 2 1 1 2 2 3 3 4
i=2 -> ↑
j=7 -> ↑
if (2!=3)✅
Vector -> 0 1 2 1 1 2 2 3 3 4
i=3 -> ↑
nums[i] = nums[j]
nums[3] = nums[7]
1 = 3
Vector modified (1 is replaced with 3)
↓
Vector -> 0 1 2 3 1 2 2 3 3 4
----------------------------------------
Vector -> 0 1 2 3 1 2 2 3 3 4
i=3 -> ↑
j=8 -> ↑
if (3!=3) ❌
----------------------------------------
Vector -> 0 1 2 3 1 2 2 3 3 4
i=3 -> ↑
j=9 -> ↑
if (3!=4)✅
Vector -> 0 1 2 3 1 2 2 3 3 4
i=4 -> ↑
nums[i] = nums[j]
nums[4] = nums[9]
1 = 4
Vector modified (1 is replaced with 4)
↓
Vector -> 0 1 2 3 4 2 2 3 3 4
----------------------------------------
For Loop Stop
----------------------------------------
Answer
Vector -> 0 1 2 3 4 2 2 3 3 4
return i+1 = 5
Top comments (0)