How to approach this two dimensional array problem? - algorithm

Question

Problem (THE QUESTION)
Strokes to paint
Alex wants to paint a picture. In one stroke, Alex can only paint the same colored cells which are joined via some edge.
Given the painting as a 2-dimensional array of letters indicating colors, determine the minimum number of strokes to completely paint the picture.
Example: The canvas height, h = 3 and width, w = 5 is to be painted with picture=["aabba", "aabba", "aaacb"]. The diagram below shows the 5 strokes needed to paint the canvas. It takes two strokes each for colors a and b, and one for c
a a b b a
a a b b a
a a a c b
Function Description Complete the function strokesRequired in the editor below. The function must return an integer, the minimum number of strokes required to paint the canvas.
strokesRequired has the following parameter(s): picture[picture[0],...picture[h-1]] an array of strings where each string represents one row of the picture to be painted
Constraints
1 <= h <= 10^5
1<= w <= 10^5
1 <= h*w <= 10^5
len(pictureffl) = w (where 0 <= i < h)
picture[i][j] <- (a, b, c) (where 0 <= i < h and 0 <= j < w)
Hello.. so i attended one company interview and they asked me this problem and iam not getting any ideas please help

Answer 1

class Paint:
def __init__(self, row, col, arr):
self.ROW = row
self.COL = col
self.arr = arr
def visit(self, i, j, visited):
ele = self.arr[i][j]
for k in range(i,self.ROW):
for l in range(j, self.COL):
if self.arr[k][l]==ele:
visited[k][l]=True
v=l
if l>0 and self.arr[k][l-1]==ele and not visited[k][l-1]:
self.visit(k, l-1, visited)
if k>0 and self.arr[k-1][l]==ele and not visited[k-1][l]:
self.visit(k-1, l, visited)
elif l>=v:
break
# 2D matrix
def count_cells(self):
# Make an array to mark visited cells.
# Initially all cells are unvisited
visited = [[False for j in range(self.COL)]for i in range(self.ROW)]
# Initialize count as 0 and travese
count = 0
for i in range(self.ROW):
for j in range(self.COL):
# If a cell value false then not visited yet
# then visit
if visited[i][j] == False:
# Visit all cells in the array
self.visit(i, j, visited)
print(visited)
count += 1
return count
arr = ["aabba", "aabba", "aaacb"]
row = len(arr)
col = len(arr[0])
p = Paint(row, col, arr)
print (p.count_cells())

Answer 2

function visit(picture, i, j, visitedBoxes) {
const currentElem = picture[i][j];
if (picture[i][j] === currentElem) {
visitedBoxes[i][j] = true;
// go in four directions
// south
if (i + 1 < picture.length && picture[i+1][j] === currentElem && visitedBoxes[i+1][j] === false) {
visit(picture, i+1, j, visitedBoxes);
}
// west
if (j+ 1 < picture[i].length && picture[i][j+1] === currentElem && visitedBoxes[i][j+1] === false) {
visit(picture, i, j+1, visitedBoxes);
}
// north
if (i > 0 && picture[i-1][j] === currentElem && visitedBoxes[i-1][j] === false) {
visit(picture, i-1, j, visitedBoxes);
}
// west
if (j > 0 && picture[i, j-1] === currentElem && visitedBoxes[i, j-1] === false) {
visit(picture, i, j-1, visitedBoxes);
}
}
}
function countStrokes(picture) {
const visitedBoxes = [];
for (let i = 0; i < picture.length; i++) {
visitedBoxes[i] = [];
for(let j = 0; j < picture[i].length; j++) {
visitedBoxes[i][j] = false;
}
}
let srokesCount = 0;
for (let i = 0; i < picture.length; i++) {
for (let j = 0; j < picture[i].length; j++) {
if (!visitedBoxes[i][j]) {
visit(picture, i, j, visitedBoxes);
srokesCount++;
}
}
}
console.log('Strokes Count', srokesCount);
}
countStrokes(['aaaba', 'ababa', 'aacba']);
This will output 5.
Also you can use
function printVisited(visitedBoxes) {
for (let i = 0; i < visitedBoxes.length; i++) {
let str = ''
for(let j = 0; j < visitedBoxes[i].length; j++) {
str += visitedBoxes[i][j] ? '1 ': '0 ';
}
console.log(str);
}
console.log('-------------');
}
to print after each loop.
Output
1 1 1 0 0
1 0 1 0 0
1 1 0 0 0
-------------
1 1 1 1 0
1 0 1 1 0
1 1 0 1 0
-------------
1 1 1 1 1
1 0 1 1 1
1 1 0 1 1
-------------
1 1 1 1 1
1 1 1 1 1
1 1 0 1 1
-------------
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
-------------
Strokes Count 5