Given a sorted array nums, remove the duplicates in-place such that each element appear only once 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.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Solution:

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int removeDuplicates(vector<int>& nums) {
    const int n = nums.size();
    if (n <= 1) return n;        
    int ans = 0;
    int i = 0;
    while (i < n) {
      nums[ans++] = nums[i];
      int j = i + 1;
      while (j < n && nums[j] == nums[j - 1]) ++j;
      i = j;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 26. Remove Duplicates from Sorted Array appeared first on Huahua's Tech Road.

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

Solution

Time complexity: O(n!)

Space complexity: O(n + k)

// Author: Huahua
// Running time: 20 ms
class Solution {
public:
  vector<vector<int>> permuteUnique(vector<int>& nums) {
    sort(begin(nums), end(nums));
    vector<vector<int>> ans;
    vector<int> used(nums.size());
    vector<int> cur;
    dfs(nums, cur, used, ans);
    return ans;
  }
private:
  void dfs(const vector<int>& nums, vector<int>& cur, vector<int>& used, vector<vector<int>>& ans) {
    if (cur.size() == nums.size()) {
      ans.push_back(cur);
      return;
    }
    for (int i = 0; i < nums.size(); ++i) {
      if (used[i]) continue;
      // Same number can be only used once at each depth.
      if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;
      used[i] = 1;
      cur.push_back(nums[i]);
      dfs(nums, cur, used, ans);
      cur.pop_back();
      used[i] = 0;
    }
  }
};

Related Problems

• 花花酱 LeetCode 784. Letter Case Permutation

The post 花花酱 LeetCode 47. Permutations II appeared first on Huahua's Tech Road.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

Example:

<b>Input:</b>
[4,3,2,7,8,2,3,1]

<b>Output:</b>
[2,3]

class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
        vector<int> ans;
        for(auto num : nums) {
            int index = abs(num)-1;
            if (nums[index] < 0)
                ans.push_back(index+1);
            nums[index]*=-1;
        }
        return ans;
    }
};

The post 花花酱 LeetCode 422. Find All Duplicates in an Array appeared first on Huahua's Tech Road.