Number of paths in a grid (backtracking) in JavaScript
$begingroup$
I was asked to solve this problem today for one of my interviews. I
would like to ask for feedback and a code review since I was rusty on the backtracking problem.
You’re testing a new driverless car that is located at the Southwest
(bottom-left) corner of an n×n grid. The car is supposed to get to the
opposite, Northeast (top-right), corner of the grid. Given n, the size
of the grid’s axes, write a function numOfPathsToDest that returns the
number of the possible paths the driverless car can take.
For convenience, let’s represent every square in the grid as a pair
(i,j). The first coordinate in the pair denotes the east-to-west axis,
and the second coordinate denotes the south-to-north axis. The initial
state of the car is (0,0), and the destination is (n-1,n-1).
The car must abide by the following two rules: it cannot cross the
diagonal border. In other words, in every step the position (i,j)
needs to maintain i >= j. See the illustration above for n = 5. In
every step, it may go one square North (up), or one square East
(right), but not both. E.g. if the car is at (3,1), it may go to (3,2)
or (4,1).
Explain the correctness of your function, and analyze its time and
space complexities.
Example:
input: n = 4
output: 5 # since there are five possibilities:
# “EEENNN”, “EENENN”, “ENEENN”, “ENENEN”, “EENNEN”,
# where the 'E' character stands for moving one step
# East, and the 'N' character stands for moving one step
# North (so, for instance, the path sequence “EEENNN”
# stands for the following steps that the car took:
# East, East, East, North, Nor
Time Complexity: the function still calculates every square south-east to the diagonal, leaving the time complexity to be $O(n^2)$.
Space Complexity: the space complexity is reduced to $O(n)$ since we are memoizing only the last two rows.
function numOfPathsToDest(n) {
//num_paths(i,j)=num_paths(i,j-1)+num_paths(i-1,j)
let num_paths = ;
for (let i =0; i <n; i++){
num_paths[i] = [1];
}
for (let i = 1; i < n; i++){
for (let j = 0; j <= i; j++){
let top = num_paths[i-1][j] || 0;
let left = num_paths[i][j-1] || 0;
num_paths[i][j] = left + top;
}
}
return num_paths[n-1][n-1];
}
javascript interview-questions
edited 18 mins ago
Jamal♦
30.3k11117227
asked 1 hour ago
NinjaGNinjaG
787528
$endgroup$
add a comment |
$begingroup$
I was asked to solve this problem today for one of my interviews. I
would like to ask for feedback and a code review since I was rusty on the backtracking problem.
You’re testing a new driverless car that is located at the Southwest
(bottom-left) corner of an n×n grid. The car is supposed to get to the
opposite, Northeast (top-right), corner of the grid. Given n, the size
of the grid’s axes, write a function numOfPathsToDest that returns the
number of the possible paths the driverless car can take.
For convenience, let’s represent every square in the grid as a pair
(i,j). The first coordinate in the pair denotes the east-to-west axis,
and the second coordinate denotes the south-to-north axis. The initial
state of the car is (0,0), and the destination is (n-1,n-1).
The car must abide by the following two rules: it cannot cross the
diagonal border. In other words, in every step the position (i,j)
needs to maintain i >= j. See the illustration above for n = 5. In
every step, it may go one square North (up), or one square East
(right), but not both. E.g. if the car is at (3,1), it may go to (3,2)
or (4,1).
Explain the correctness of your function, and analyze its time and
space complexities.
Example:
input: n = 4
output: 5 # since there are five possibilities:
# “EEENNN”, “EENENN”, “ENEENN”, “ENENEN”, “EENNEN”,
# where the 'E' character stands for moving one step
# East, and the 'N' character stands for moving one step
# North (so, for instance, the path sequence “EEENNN”
# stands for the following steps that the car took:
# East, East, East, North, Nor
Time Complexity: the function still calculates every square south-east to the diagonal, leaving the time complexity to be $O(n^2)$.
Space Complexity: the space complexity is reduced to $O(n)$ since we are memoizing only the last two rows.
function numOfPathsToDest(n) {
//num_paths(i,j)=num_paths(i,j-1)+num_paths(i-1,j)
let num_paths = ;
for (let i =0; i <n; i++){
num_paths[i] = [1];
}
for (let i = 1; i < n; i++){
for (let j = 0; j <= i; j++){
let top = num_paths[i-1][j] || 0;
let left = num_paths[i][j-1] || 0;
num_paths[i][j] = left + top;
}
}
return num_paths[n-1][n-1];
}
javascript interview-questions
edited 18 mins ago
Jamal♦
30.3k11117227
asked 1 hour ago
NinjaGNinjaG
787528
$endgroup$
add a comment |
$begingroup$
I was asked to solve this problem today for one of my interviews. I
would like to ask for feedback and a code review since I was rusty on the backtracking problem.
You’re testing a new driverless car that is located at the Southwest
(bottom-left) corner of an n×n grid. The car is supposed to get to the
opposite, Northeast (top-right), corner of the grid. Given n, the size
of the grid’s axes, write a function numOfPathsToDest that returns the
number of the possible paths the driverless car can take.
For convenience, let’s represent every square in the grid as a pair
(i,j). The first coordinate in the pair denotes the east-to-west axis,
and the second coordinate denotes the south-to-north axis. The initial
state of the car is (0,0), and the destination is (n-1,n-1).
The car must abide by the following two rules: it cannot cross the
diagonal border. In other words, in every step the position (i,j)
needs to maintain i >= j. See the illustration above for n = 5. In
every step, it may go one square North (up), or one square East
(right), but not both. E.g. if the car is at (3,1), it may go to (3,2)
or (4,1).
Explain the correctness of your function, and analyze its time and
space complexities.
Example:
input: n = 4
output: 5 # since there are five possibilities:
# “EEENNN”, “EENENN”, “ENEENN”, “ENENEN”, “EENNEN”,
# where the 'E' character stands for moving one step
# East, and the 'N' character stands for moving one step
# North (so, for instance, the path sequence “EEENNN”
# stands for the following steps that the car took:
# East, East, East, North, Nor
Time Complexity: the function still calculates every square south-east to the diagonal, leaving the time complexity to be $O(n^2)$.
Space Complexity: the space complexity is reduced to $O(n)$ since we are memoizing only the last two rows.
function numOfPathsToDest(n) {
//num_paths(i,j)=num_paths(i,j-1)+num_paths(i-1,j)
let num_paths = ;
for (let i =0; i <n; i++){
num_paths[i] = [1];
}
for (let i = 1; i < n; i++){
for (let j = 0; j <= i; j++){
let top = num_paths[i-1][j] || 0;
let left = num_paths[i][j-1] || 0;
num_paths[i][j] = left + top;
}
}
return num_paths[n-1][n-1];
}
javascript interview-questions
edited 18 mins ago
Jamal♦
30.3k11117227
asked 1 hour ago
NinjaGNinjaG
787528
$endgroup$
I was asked to solve this problem today for one of my interviews. I
would like to ask for feedback and a code review since I was rusty on the backtracking problem.
You’re testing a new driverless car that is located at the Southwest
(bottom-left) corner of an n×n grid. The car is supposed to get to the
opposite, Northeast (top-right), corner of the grid. Given n, the size
of the grid’s axes, write a function numOfPathsToDest that returns the
number of the possible paths the driverless car can take.
For convenience, let’s represent every square in the grid as a pair
(i,j). The first coordinate in the pair denotes the east-to-west axis,
and the second coordinate denotes the south-to-north axis. The initial
state of the car is (0,0), and the destination is (n-1,n-1).
The car must abide by the following two rules: it cannot cross the
diagonal border. In other words, in every step the position (i,j)
needs to maintain i >= j. See the illustration above for n = 5. In
every step, it may go one square North (up), or one square East
(right), but not both. E.g. if the car is at (3,1), it may go to (3,2)
or (4,1).
Explain the correctness of your function, and analyze its time and
space complexities.
Example:
input: n = 4
output: 5 # since there are five possibilities:
# “EEENNN”, “EENENN”, “ENEENN”, “ENENEN”, “EENNEN”,
# where the 'E' character stands for moving one step
# East, and the 'N' character stands for moving one step
# North (so, for instance, the path sequence “EEENNN”
# stands for the following steps that the car took:
# East, East, East, North, Nor
Time Complexity: the function still calculates every square south-east to the diagonal, leaving the time complexity to be $O(n^2)$.
Space Complexity: the space complexity is reduced to $O(n)$ since we are memoizing only the last two rows.
function numOfPathsToDest(n) {
//num_paths(i,j)=num_paths(i,j-1)+num_paths(i-1,j)
let num_paths = ;
for (let i =0; i <n; i++){
num_paths[i] = [1];
}
for (let i = 1; i < n; i++){
for (let j = 0; j <= i; j++){
let top = num_paths[i-1][j] || 0;
let left = num_paths[i][j-1] || 0;
num_paths[i][j] = left + top;
}
}
return num_paths[n-1][n-1];
}
javascript interview-questions
javascript interview-questions
edited 18 mins ago
Jamal♦
30.3k11117227
asked 1 hour ago
NinjaGNinjaG
787528
edited 18 mins ago
Jamal♦
30.3k11117227
asked 1 hour ago
NinjaGNinjaG
787528
edited 18 mins ago
Jamal♦
30.3k11117227
edited 18 mins ago
Jamal♦
30.3k11117227
edited 18 mins ago
Jamal♦
30.3k11117227
30.3k11117227
asked 1 hour ago
NinjaGNinjaG
787528
asked 1 hour ago
NinjaGNinjaG
787528
asked 1 hour ago
NinjaGNinjaG
787528
787528
add a comment |
add a comment |
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