Spiral Matrix - Amazon Top Interview Questions
Problem Statement :
Given a 2-d array matrix, return elements in spiral order starting from matrix[0][0].
Constraints
n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[6, 9, 8],
[1, 8, 0],
[5, 1, 2],
[8, 0, 3],
[1, 6, 4],
[8, 8, 10]
]
Output
[6, 9, 8, 0, 2, 3, 4, 10, 8, 8, 1, 8, 5, 1, 8, 1, 0, 6]
Solution :
Solution in C++ :
vector<int> solve(vector<vector<int>>& matrix) {
vector<int> a;
if (matrix.size() == 0) {
return a;
}
int top, down, left, right;
int direction = 0;
top = 0;
down = matrix.size() - 1;
left = 0;
right = matrix[0].size() - 1;
while (left <= right && top <= down) {
if (direction == 0) {
for (int i = left; i <= right; i++) {
a.push_back(matrix[top][i]);
}
top++;
}
else if (direction == 1) {
for (int i = top; i <= down; i++) {
a.push_back(matrix[i][right]);
}
right--;
}
else if (direction == 2) {
for (int i = right; i >= left; i--) {
a.push_back(matrix[down][i]);
}
down--;
}
else if (direction == 3) {
for (int i = down; i >= top; i--) {
a.push_back(matrix[i][left]);
}
left++;
}
direction = (direction + 1) % 4;
}
return a;
}
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[][] matrix) {
if (matrix.length == 0) {
return new int[0];
}
int[] nums = new int[matrix.length * matrix[0].length];
int startRow = 0, startCol = 0, endRow = matrix.length - 1, endCol = matrix[0].length - 1,
k = 0;
while (startRow <= endRow && startCol <= endCol) {
for (int i = startCol; i <= endCol; i++) {
nums[k] = matrix[startRow][i];
k++;
}
startRow++;
for (int i = startRow; i <= endRow; i++) {
nums[k] = matrix[i][endCol];
k++;
}
endCol--;
if (startRow <= endRow) {
for (int i = endCol; i >= startCol; i--) {
nums[k] = matrix[endRow][i];
k++;
}
endRow--;
}
if (startCol <= endCol) {
for (int i = endRow; i >= startRow; i--) {
nums[k] = matrix[i][startCol];
k++;
}
startCol++;
}
}
return nums;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
mat = []
if len(matrix) == 0:
return mat
a, b = len(matrix) - 1, len(matrix[0]) - 1
c, d = 1, 0
i, j = 0, 0
while len(mat) < (len(matrix) * len(matrix[0])):
# loop for top left to right
while j <= b:
mat.append(matrix[i][j])
j += 1
j -= 1
i += 1
# loop for rightmost top to bottom
while i <= a:
mat.append(matrix[i][j])
i += 1
i -= 1
j -= 1
# To check if no of row is odd we might traverse this in top left to right loop
if i == (len(matrix) - 1) / 2:
break
# loop for bottom right to left
while j >= d:
mat.append(matrix[i][j])
j -= 1
j += 1
i -= 1
# To check if no of column is odd we might traverse this in rightmost top to bottom loop
if j == (len(matrix[0]) - 1) / 2:
break
# loop for leftmost bottom to top
while i >= c:
mat.append(matrix[i][j])
i -= 1
i += 1
j += 1
a -= 1
b -= 1
c += 1
d += 1
return mat
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