Suppose we have a n x n binary matrix. We can perform an operation on it like, at one step we select two adjacent rows and swap them. We have to count number of minimum swaps required, so that all nodes above the major diagonal of the matrix is 0. If there is no such solution, then return -1.
So, if the input is like
0 | 1 | 0 |
0 | 1 | 1 |
1 | 0 | 0 |
then the output will be 2 because −
To solve this, we will follow these steps:
n := row count of matrix
m := make an array of size n and fill with n
for i in range 0 to n - 1, do
for j in range n-1 to 0, decrease by 1, do
if matrix[i, j] is same as 1, then
m[i] := n-j-1
come out from the loop
t := 0, ans := 0
for i in range 0 to n - 1, do
t := t + 1
flag := False
for j in range i to n - 1, do
if m[j] >= n-t, then
ans := ans + j-i
flag := True
come out from loop
if flag is false, then
return -1
update m[from index i+1 to j] by m[from index i to j-1]
return ans
Let us see the following implementation to get better understanding −
Example
def solve(matrix): n = len(matrix) m = [n] * n for i in range(n): for j in range(n-1,-1,-1): if matrix[i][j] == 1: m[i] = n-j-1 break t,ans = 0,0 for i in range(n): t += 1 flag = False for j in range(i,n): if m[j] >= n-t: ans += j-i flag = True break if not flag: return -1 m[i+1:j+1] = m[i:j] return ans matrix = [[0,1,0],[0,1,1],[1,0,0]] print(solve(matrix))
Input
[[0,1,0],[0,1,1],[1,0,0]]
Output
2