Return any such subsequence as an integer array of length k.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [2,1,3,3], k = 2
Output: [3,3]
Explanation:
The subsequence has the largest sum of 3 + 3 = 6.

Example 2:

Input: nums = [-1,-2,3,4], k = 3
Output: [-1,3,4]
Explanation: 
The subsequence has the largest sum of -1 + 3 + 4 = 6.

Example 3:

Input: nums = [3,4,3,3], k = 2
Output: [3,4]
Explanation:
The subsequence has the largest sum of 3 + 4 = 7. 
Another possible subsequence is [4, 3].

Constraints:

• 1 <= nums.length <= 1000
• -105 <= nums[i] <= 105
• 1 <= k <= nums.length

Solution 1: Full sorting + Hashtable

Make a copy of the original array, sort it in whole and put k largest elements in a hashtable.

Scan the original array and append the number to the answer array if it’s in the hashtable.

Time complexity: O(nlogn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<int> maxSubsequence(vector<int>& nums, int k) {    
    vector<int> ans;
    vector<int> s(nums);
    sort(rbegin(s), rend(s));
    unordered_map<int, int> m;
    for (int i = 0; i < k; ++i) ++m[s[i]];
    for (int x : nums) 
      if (--m[x] >= 0) 
        ans.push_back(x);
    return ans;
  }
};

Solution 2: k-th element

Use quick select / partial sort to find the k-th largest element of the array. Any number greater than that must be in the sequence. The tricky part is: what about the pivot itself? We couldn’t include ’em all blindly. Need to figure out how many copies of the pivot is there in the k largest and only include as many as of that in the final answer.

Time complexity: O(n+k)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<int> maxSubsequence(vector<int>& nums, int k) {    
    vector<int> ans;
    vector<int> s(nums);
    nth_element(begin(s), begin(s) + k - 1, end(s), greater<int>());
    const int pivot = s[k - 1];
    // How many pivots in the largest k elements.
    int left = count(begin(s), begin(s) + k, pivot);    
    for (size_t i = 0; i != nums.size(); ++i)
      if (nums[i] > pivot || (nums[i] == pivot && --left >= 0)) 
        ans.push_back(nums[i]);
    return ans;
  }
};

The post 花花酱 LeetCode 2099. Find Subsequence of Length K With the Largest Sum appeared first on Huahua's Tech Road.

The value of coordinate (a, b) of the matrix is the XOR of all matrix[i][j] where 0 <= i <= a < m and 0 <= j <= b < n (0-indexed).

Find the kth largest value (1-indexed) of all the coordinates of matrix.

Example 1:

Input: matrix = [[5,2],[1,6]], k = 1
Output: 7
Explanation: The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.

Example 2:

Input: matrix = [[5,2],[1,6]], k = 2
Output: 5
Explanation: The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.

Example 3:

Input: matrix = [[5,2],[1,6]], k = 3
Output: 4
Explanation: The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.

Example 4:

Input: matrix = [[5,2],[1,6]], k = 4
Output: 0
Explanation: The value of coordinate (1,1) is 5 XOR 2 XOR 1 XOR 6 = 0, which is the 4th largest value.

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 1000
• 0 <= matrix[i][j] <= 106
• 1 <= k <= m * n

Solution: DP

Similar to 花花酱 LeetCode 304. Range Sum Query 2D – Immutable

xor[i][j] = matrix[i][j] ^ xor[i – 1][j – 1] ^ xor[i – 1][j] ^ xor[i][j- 1]

Time complexity: O(mn)
Space complexity: O(mn)

// Author: Huahua
class Solution {
public:
  int kthLargestValue(vector<vector<int>>& matrix, int k) {    
    const int m = matrix.size(), n = matrix[0].size();
    vector<int> v;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        v.push_back(matrix[i][j] 
                      ^= (i ? matrix[i - 1][j] : 0) 
                       ^ (j ? matrix[i][j - 1] : 0) 
                       ^ (i * j ? matrix[i - 1][j - 1] : 0));
    nth_element(begin(v), begin(v) + k - 1, end(v), greater<int>());    
    return v[k - 1];
  }
};

The post 花花酱 LeetCode 1738. Find Kth Largest XOR Coordinate Value appeared first on Huahua's Tech Road.