Kth Smallest Element in a Sorted Matrix – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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
C++ Kth Smallest Element in a Sorted Matrix LeetCode Solution
class Solution { // 20 ms, faster than 98.92%
public:
int m, n;
int kthSmallest(vector<vector<int>>& matrix, int k) {
m = matrix.size(), n = matrix[0].size(); // For general, the matrix need not be a square
int left = matrix[0][0], right = matrix[m-1][n-1], ans = -1;
while (left <= right) {
int mid = (left + right) >> 1;
if (countLessOrEqual(matrix, mid) >= k) {
ans = mid;
right = mid – 1; // try to looking for a smaller value in the left side
} else left = mid + 1; // try to looking for a bigger value in the right side
}
return ans;
}
int countLessOrEqual(vector<vector<int>>& matrix, int x) {
int cnt = 0, c = n – 1; // start with the rightmost column
for (int r = 0; r < m; ++r) {
while (c >= 0 && matrix[r][c] > x) –c; // decrease column until matrix[r][c] <= x
cnt += (c + 1);
}
return cnt;
}
};
Java Kth Smallest Element in a Sorted Matrix LeetCode Solution
class Solution { // 0 ms, faster than 100%
int m, n;
public int kthSmallest(int[][] matrix, int k) {
m = matrix.length; n = matrix[0].length; // For general, the matrix need not be a square
int left = matrix[0][0], right = matrix[m-1][n-1], ans = -1;
while (left <= right) {
int mid = (left + right) >> 1;
if (countLessOrEqual(matrix, mid) >= k) {
ans = mid;
right = mid – 1; // try to looking for a smaller value in the left side
} else left = mid + 1; // try to looking for a bigger value in the right side
}
return ans;
}
int countLessOrEqual(int[][] matrix, int x) {
int cnt = 0, c = n – 1; // start with the rightmost column
for (int r = 0; r < m; ++r) {
while (c >= 0 && matrix[r][c] > x) –c; // decrease column until matrix[r][c] <= x
cnt += (c + 1);
}
return cnt;
}
}
Python 3 Kth Smallest Element in a Sorted Matrix LeetCode Solution
class Solution { // 0 ms, faster than 100%
int m, n;
public int kthSmallest(int[][] matrix, int k) {
m = matrix.length; n = matrix[0].length; // For general, the matrix need not be a square
int left = matrix[0][0], right = matrix[m-1][n-1], ans = -1;
while (left <= right) {
int mid = (left + right) >> 1;
if (countLessOrEqual(matrix, mid) >= k) {
ans = mid;
right = mid – 1; // try to looking for a smaller value in the left side
} else left = mid + 1; // try to looking for a bigger value in the right side
}
return ans;
}
int countLessOrEqual(int[][] matrix, int x) {
int cnt = 0, c = n – 1; // start with the rightmost column
for (int r = 0; r < m; ++r) {
while (c >= 0 && matrix[r][c] > x) –c; // decrease column until matrix[r][c] <= x
cnt += (c + 1);
}
return cnt;
}
}
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