Algorithm
Problem Name: 378. Kth Smallest Element in a Sorted Matrix
Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix.
Note that it is the kth smallest element in the sorted order, not the kth distinct element.
You must find a solution with a memory complexity better than O(n2).
Example 1:
Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8 Output: 13 Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13
Example 2:
Input: matrix = [[-5]], k = 1 Output: -5
Constraints:
n == matrix.length == matrix[i].length1 <= n <= 300-109 <= matrix[i][j] <= 109- All the rows and columns of
matrixare guaranteed to be sorted in non-decreasing order. 1 <= k <= n2
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int kthSmallest(int** matrix, int matrixRowSize, int matrixColSize, int k) {
int kth;
int i, j, l, r, c, *last_row;
last_row = malloc(matrixColSize * sizeof(int));
//assert(last_row);
memset(last_row, -1, matrixColSize * sizeof(int));
r = c = 0;
kth = matrix[r][c];
last_row[c] = r;
while (-- k) {
r = matrixRowSize - 1; c = matrixColSize - 1;
kth = matrix[r][c]; // the biggest value
l = 0; // line was previous tested
i = 1; // line to be tested
for (j = 0; j < matrixColSize && i != 0; j ++) {
// for every column, look at the next row of the one was already taken
i = last_row[j] + 1;
if (i < matrixRowSize && i != l && kth > matrix[i][j]) {
kth = matrix[i][j];
r = i; c = j;
}
l = i;
}
last_row[c] = r; // mark this one being taken
}
free(last_row);
return kth;
}
Input
#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
priority_queue < int, vector<int>, greater < int>>pq;
for(auto x: matrix)
for(auto y: x) pq.push(y);
while(--k) pq.pop();
return pq.top();
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int kthSmallest(int[][] matrix, int k) {
int numOfRows = matrix.length;
PriorityQueue < int[]> pq = new PriorityQueue<>((o1, o2) -> o1[2] - o2[2]);
for (int j = 0; j < numOfRows; j++) {
pq.add(new int[]{0, j, matrix[0][j]});
}
for (int i = 0; i < k - 1; i++) {
int[] removed = pq.poll();
if (removed[0] == numOfRows - 1) {
continue;
}
pq.add(new int[]{removed[0] + 1, removed[1], matrix[removed[0] + 1][removed[1]]});
}
return pq.peek()[2];
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
/**
* @param {number[][]} matrix
* @param {number} k
* @return {number}
*/
const kthSmallest = function(matrix, k) {
let lo = matrix[0][0],
hi = matrix[matrix.length - 1][matrix[0].length - 1] + 1; //[lo, hi)
while (lo < hi) {
let mid = Math.floor(lo + (hi - lo) / 2);
let count = 0,
j = matrix[0].length - 1;
for (let i = 0; i < matrix.length; i++) {
while (j >= 0 && matrix[i][j] > mid) j--;
count += j + 1;
}
if (count < k) lo = mid + 1;
else hi = mid;
}
return lo;
};
console.log(kthSmallest([[-5]], 1));
console.log(kthSmallest([[1, 2], [1, 3]], 4));
console.log(kthSmallest([[1, 5, 9], [10, 11, 13], [12, 13, 15]], 8));
console.log(kthSmallest([[1, 2], [1, 3]], 2));
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def kthSmallest(self, matrix, k):
return sorted(itertools.chain(*matrix))[k - 1]
Input
Output