k-th Archives - Huahua's Tech Road

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)

C++

// 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)

C++

// 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;

}

};

You are given a 2D matrix of size m x n, consisting of non-negative integers. You are also given an integer k.

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 k^th 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

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)

C++

// 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];

}

};

Posts tagged as “k-th”

花花酱 LeetCode 2099. Find Subsequence of Length K With the Largest Sum

Solution 1: Full sorting + Hashtable

C++

Solution 2: k-th element

C++

花花酱 LeetCode 1738. Find Kth Largest XOR Coordinate Value

Solution: DP

C++