Program to count number of operations required to all cells into same color in Python
Suppose we have a two-dimensional matrix M. Now in each cell contains a value that represents its color, and adjacent cells (top, bottom, left, right) with the same color are to be grouped together. Now, consider an operation where we set all cells in one group to some color. Then finally find the minimum number of operations required so that every cell has the same color. And when the color is transformed, it cannot be set again.
So, if the input is like
Then the output will be 2, as We can fill the group with 2 as color into 1 and then fill 3 with 1.
To solve this, we will follow these steps−
if matrix is empty, then
Define a function dfs() . This will take i, j, matrix, val
n := row count of matrix, m := col count of matrix
if i < 0 or i > n - 1 or j < 0 or j > m - 1, then
if matrix[i, j] is same as -1, then
if matrix[i, j] is same as val, then
matrix[i, j] := -1
dfs(i, j + 1, matrix, val)
dfs(i + 1, j, matrix, val)
dfs(i, j - 1, matrix, val)
dfs(i - 1, j, matrix, val)
otherwise,
From the main method, do the following−
n := row count of matrix, m := col count of matrix
d := empty map
for i in range 0 to n-1, do
for j in range 0 to m-1, do
val := matrix[i, j]
if val is not same as -1, then
d[val] := d[val] + 1
dfs(i, j, matrix, val)
sort dictionary elements of f based on their values
safe := last element of l
res := 0
for each key value pairs k and v of d, do
return res
Let us see the following implementation to get better understanding −
Example
Live Demo
from collections import defaultdict
class Solution:
def solve(self, matrix):
if not matrix:
return 0
def dfs(i, j, matrix, val):
n, m = len(matrix), len(matrix[0])
if i < 0 or i > n - 1 or j < 0 or j > m - 1:
return
if matrix[i][j] == -1:
return
if matrix[i][j] == val:
matrix[i][j] = -1
dfs(i, j + 1, matrix, val)
dfs(i + 1, j, matrix, val)
dfs(i, j - 1, matrix, val)
dfs(i - 1, j, matrix, val)
else:
return
n, m = len(matrix), len(matrix[0])
d = defaultdict(int)
for i in range(n):
for j in range(m):
val = matrix[i][j]
if val != -1:
d[val] += 1
dfs(i, j, matrix, val)
l = sorted(d,key=lambda x: d[x])
safe = l[-1]
res = 0
for k, v in d.items():
if k != safe:
res += v
return res
ob = Solution()
matrix = [
[2, 2, 2, 2],
[1, 1, 1, 1],
[2, 3, 2, 1]
]
print(ob.solve(matrix))
Input
matrix = [[2, 2, 2, 2],[1, 1, 1, 1],[2, 3, 2, 1]]
Output
2
Published on 05-Oct-2020 17:21:08
