Suppose we have one m x n matrix. and another value k. Here the value of coordinate (a, b) of the matrix is the XOR of all matrix[i, j] where i in range (0 to a) and j in range (0 to b). We have to find the kth largest value (1-indexed) of all the coordinates of matrix.
So, if the input is like
5 | 2 |
1 | 6 |
And k = 1, then the output will be 7 because the value of coordinate (0,1) is 5 XOR 2 = 7, and this is the largest one
To solve this, we will follow these steps −
- m := row count, n := column count
- for i in range 0 to m - 1, do
- for j in range 0 to n - 1, do
- if j is non-zero, then
- matrix[i, j] := matrix[i, j] XOR matrix[i, j-1]
- if j is non-zero, then
- for j in range 0 to n - 1, do
- seen := a new map
- count := 0
- for i in range 0 to n - 1, do
- for j in range 0 to m - 1, do
- if j is non-zero, then
- matrix[j, i] := matrix[j, i] XOR matrix[j-1, i]
- seen[matrix[j, i]] = (1+seen[matrix[j, i]] if possible, otherwise 1)
- count := count + 1
- if count > k, then
- min_value := minimum of seen
- seen[min_value] := seen[min_value] - 1
- if seen[min_value] is 0, then
- delete min_value-th element from seen
- if j is non-zero, then
- for j in range 0 to m - 1, do
- return minimum of seen
Example
Let us see the following implementation to get better understanding −
def solve(matrix, k): m, n = len(matrix), len(matrix[0]) for i in range(m): for j in range(n): if j: matrix[i][j] ^= matrix[i][j-1] seen = {} count = 0 for i in range(n): for j in range(m): if j: matrix[j][i] ^= matrix[j-1][i] seen[matrix[j][i]] = seen.get(matrix[j][i], 0) + 1 count += 1 if count > k: min_value = min(seen) seen[min_value] -= 1 if not seen[min_value]: seen.pop(min_value) return min(seen) matrix = [[5,2],[1,6]] k = 1 print(solve(matrix, k))
Input
[[5,2],[1,6]], 1
Output
7