Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

Example:

matrix = [ [ 1, 5, 9], [10, 11, 13], [12, 13, 15] ],

k = 8,

return 13.

**Java Solution**

This problem is similar to Search a 2D Matrix II. The start point of such a sorted matrix is left-bottom corner.

public int kthSmallest(int[][] matrix, int k) { int m=matrix.length; int lower = matrix[0][0]; int upper = matrix[m-1][m-1]; while(lower<upper){ int mid = lower + ((upper-lower)>>1); int count = count(matrix, mid); if(count<k){ lower=mid+1; }else{ upper=mid; } } return upper; } private int count(int[][] matrix, int target){ int m=matrix.length; int i=m-1; int j=0; int count = 0; while(i>=0&&j<m){ if(matrix[i][j]<=target){ count += i+1; j++; }else{ i--; } } return count; } |

Simpler solution

public int kthSmallest1(int[][] matrix, int k) {

k = k-1;

int n=matrix[0].length;

int row = k/n;

int col = k%n;

return matrix[row][col];

}

because we are calculating the mid using the formula

lower + ((upper-lower) / 2). Hence it will definetely will fall within the range

Another solution.

Memory consumption: n

Complexity: < O(k * (n + 1))

int kthSmallest(int[][] matrix, int k) {

int rows[] = new int[matrix[0].length];

int cur = matrix[0][0];

while (true) {

Integer next = null;

for (int col = 0; col 0; ) {

int row = rows[col];

if (row < matrix.length) {

int value = matrix[row][col];

if (value != cur) {

if (next == null) {

next = value;

} else {

next = Math.min(next, value);

}

}

if (value == cur) {

if (--k == 0) {

return cur;

}

++rows[col];

continue;

}

}

++col;

}

`cur = next;`

}

}

Sorry.. But how do you guarantee the mid(calculated by lower + (upper – lower)/2) do exist in the matrix.. What if it doesn’t?