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Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
ans = -1
dx = [1, -1, 0, 0]
dy = [0, 0, 1, -1]
def rec(i, j, path=0):
nonlocal ans, xd, yd
if i < 0 or j < 0 or i >= n or j >= m:
return
if mat[i][j] == 0:
return
if i == xd and j == yd:
ans = max(ans, path)
return
mat[i][j] = 0
for k in range(4):
x = i + dx[k]
y = j + dy[k]
rec(x, y, path + 1)
mat[i][j] = 1
rec(xs, ys)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER FUNC_DEF NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR RETURN IF VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def __longestPathHelper(self, matrix, path, n, m, x, y, xd, yd):
self.visited.add((x, y))
for x_offset, y_offset in self.dirs:
temp_x, temp_y = x + x_offset, y + y_offset
if (
0 <= temp_x < n
and 0 <= temp_y < m
and (temp_x, temp_y) not in self.visited
and matrix[temp_x][temp_y] != 0
):
if temp_x == xd and temp_y == yd:
self.ans = max(self.ans, path + 1)
else:
self.__longestPathHelper(
matrix, path + 1, n, m, temp_x, temp_y, xd, yd
)
self.visited.discard((x, y))
def longestPath(self, matrix, n, m, xs, ys, xd, yd):
if matrix[xs][ys] == 0:
return -1
self.visited, self.ans = set(), -1
self.dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
self.__longestPathHelper(matrix, 0, n, m, xs, ys, xd, yd)
return self.ans | CLASS_DEF FUNC_DEF EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_DEF IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR VAR RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
def solve(mat, n, m, xs, ys, xd, yd, count=0):
if mat[xd][yd] == 0:
return -1
if xs == xd and ys == yd:
return count
if xs < 0 or xs >= n or ys < 0 or ys >= m:
return -1
if mat[xs][ys] in [0, 2]:
return -1
count += 1
mat[xs][ys] = 2
res = max(
solve(mat, n, m, xs + 1, ys, xd, yd, count),
solve(mat, n, m, xs - 1, ys, xd, yd, count),
solve(mat, n, m, xs, ys + 1, xd, yd, count),
solve(mat, n, m, xs, ys - 1, xd, yd, count),
)
mat[xs][ys] = 1
return res
return solve(mat, n, m, xs, ys, xd, yd) | CLASS_DEF FUNC_DEF FUNC_DEF NUMBER IF VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR RETURN NUMBER IF VAR VAR VAR LIST NUMBER NUMBER RETURN NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
def function(i, j, visited, mat, sum1, sum2):
if i < 0 or j < 0 or i >= len(mat) or j >= len(mat[0]) or mat[i][j] == 0:
return
if visited[i][j] == "T":
return
if i == xd and j == yd:
sum2[0] = max(sum2[0], sum1)
return
visited[i][j] = "T"
function(i + 1, j, visited, mat, sum1 + mat[i][j], sum2)
function(i - 1, j, visited, mat, sum1 + mat[i][j], sum2)
function(i, j + 1, visited, mat, sum1 + mat[i][j], sum2)
function(i, j - 1, visited, mat, sum1 + mat[i][j], sum2)
visited[i][j] = "F"
return
sum1 = 0
sum2 = [0]
visited = [["F" for i in range(len(mat[0]))] for j in range(len(mat))]
function(xs, ys, visited, mat, sum1, sum2)
if sum2[0] == 0:
return -1
return sum2[0] | CLASS_DEF FUNC_DEF FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER RETURN IF VAR VAR VAR STRING RETURN IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN ASSIGN VAR VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR STRING RETURN ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR STRING VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER NUMBER RETURN NUMBER RETURN VAR NUMBER |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int):
ans = -1
dir_x = [0, 0, 1, -1]
dir_y = [1, -1, 0, 0]
def isvalid(mat, n, m, xs, ys):
if xs < 0 or ys < 0:
return False
if xs >= n or ys >= m:
return False
if mat[xs][ys] != 1:
return False
return True
def solve(mat, n, m, xs, ys, xd, yd, dist):
nonlocal ans
if xs == xd and ys == yd:
ans = max(ans, dist)
return
for i in range(4):
dist += 1
mat[xs][ys] = -1
xs += dir_x[i]
ys += dir_y[i]
if isvalid(mat, n, m, xs, ys):
solve(mat, n, m, xs, ys, xd, yd, dist)
xs -= dir_x[i]
ys -= dir_y[i]
mat[xs][ys] = 1
dist -= 1
if mat[xs][ys] != 1:
return -1
solve(mat, n, m, xs, ys, xd, yd, 0)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER FUNC_DEF IF VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FOR VAR FUNC_CALL VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER RETURN NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | def backtracking(xs, ys, n, m, xd, yd, arr):
if arr[xs][ys] == 0 or arr[xd][yd] == 0:
return -1
vis = set()
vis.add((xs, ys))
res = []
helper(xs, ys, n, m, xd, yd, arr, vis, res)
if res:
return res[-1] - 1
return -1
def helper(i, j, r, c, xd, yd, arr, vis, res):
if (i, j) == (xd, yd):
if not res:
res.append(len(vis))
else:
res[-1] = max(res[-1], len(tuple(vis)))
for ti, tj in zip([1, -1, 0, 0], [0, 0, 1, -1]):
ni, nj = ti + i, tj + j
if isValid(ni, nj, r, c, arr) and (ni, nj) not in vis:
vis.add((ni, nj))
helper(ni, nj, r, c, xd, yd, arr, vis, res)
vis.remove((ni, nj))
def isValid(i, j, r, c, arr):
if i >= 0 and i < r and j >= 0 and j < c and arr[i][j]:
return True
return False
class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
return backtracking(xs, ys, n, m, xd, yd, mat) | FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR RETURN BIN_OP VAR NUMBER NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR IF VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR LIST NUMBER NUMBER NUMBER NUMBER LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
def check(mat, n, m, i, j, vis):
if i >= 0 and j >= 0 and i < n and j < m and vis[i][j] == 0:
return True
return False
def recur(mat, n, m, xs, ys, xd, yd, vis, rh, ch):
if mat[xs][ys] == 0:
return -1
if xs == xd and ys == yd:
return 0
vis[xs][ys] = 1
res = -1
for i in range(4):
x = xs + rh[i]
y = ys + ch[i]
if check(mat, n, m, x, y, vis):
mxl = recur(mat, n, m, x, y, xd, yd, vis, rh, ch)
if mxl != -1:
res = max(res, mxl + 1)
vis[xs][ys] = 0
return res
vis = [[(0) for i in range(m)] for j in range(n)]
rh = [-1, 0, 0, 1]
ch = [0, -1, 1, 0]
ans = recur(mat, n, m, xs, ys, xd, yd, vis, rh, ch)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
if mat[xs][ys] == 0:
return -1
self.mat = mat
self.n = n
self.m = m
self.ans = 0
self.xd = xd
self.yd = yd
self.move(xs, ys, 1, {(xs, ys): 1})
return self.ans - 1
def move(self, i, j, length, visited):
if i == self.xd and j == self.yd:
self.ans = max(self.ans, length)
return
for xdash, ydash in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
if (
xdash >= 0
and xdash < self.n
and ydash >= 0
and ydash < self.m
and self.mat[xdash][ydash] == 1
and (xdash, ydash) not in visited
):
visited[xdash, ydash] = 1
self.move(xdash, ydash, length + 1, visited)
visited.pop((xdash, ydash))
return | CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER DICT VAR VAR NUMBER RETURN BIN_OP VAR NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FOR VAR VAR LIST BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR RETURN |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
res = [0]
def dfs(r, c, path):
if r == xd and c == yd:
res[0] = max(res[0], path)
return
directions = [[1, 0], [-1, 0], [0, -1], [0, 1]]
visited[r][c] = 1
for dr, dc in directions:
i, j = dr + r, dc + c
if (
i >= 0
and i < n
and j >= 0
and j < m
and mat[i][j] == 1
and visited[i][j] == 0
):
dfs(i, j, path + 1)
visited[r][c] = 0
visited = [[(0) for i in range(m)] for j in range(n)]
dfs(xs, ys, 0)
if res[0] == 0:
return -1
return res[0] | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR LIST NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER NUMBER RETURN NUMBER RETURN VAR NUMBER VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
visited = [[(0) for i in range(m)] for i in range(n)]
ds = ""
ans = [-1]
xMoves = [1, 0, 0, -1]
yMoves = [0, 1, -1, 0]
dir = "DLRU"
def solve(i, j, ds):
if i == xd and j == yd:
ans[0] = max(ans[0], len(ds))
return
for k in range(4):
nI = i + xMoves[k]
nJ = j + yMoves[k]
if (
nI >= 0
and nJ >= 0
and nI < n
and nJ < m
and visited[nI][nJ] != 1
and mat[nI][nJ] == 1
):
visited[nI][nJ] = 1
solve(nI, nJ, ds + dir[k])
visited[nI][nJ] = 0
if mat[xs][ys] == 1:
visited[xs][ys] = 1
solve(xs, ys, ds)
return ans[0] | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR STRING FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR NUMBER VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
ans = -1
def dfs(x, y, visited, dist):
nonlocal ans
if mat[x][y] == 0 or (x, y) in visited:
return
if (x, y) == (xd, yd):
ans = max(dist, ans)
return
visited.add((x, y))
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
nx = x + dx
ny = y + dy
if 0 <= nx < n and 0 <= ny < m:
dfs(nx, ny, visited, dist + 1)
visited.remove((x, y))
dfs(xs, ys, set(), 0)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | def rec(mat, i, j, visited, xd, yd):
if (
i < 0
or j < 0
or i >= len(mat)
or j >= len(mat[0])
or visited[i][j]
or not mat[i][j]
):
return -1
if i == xd and j == yd:
return 0
visited[i][j] = True
x = [0, 0, 1, -1]
y = [1, -1, 0, 0]
ma = -1
for k in range(4):
temp = rec(mat, i + x[k], j + y[k], visited, xd, yd)
if temp != -1:
ma = max(ma, temp + 1)
visited[i][j] = False
return ma
class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
visited = [[(False) for i in range(m)] for j in range(n)]
return rec(mat, xs, ys, visited, xd, yd) | FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
ans = -1
def dfs(x, y, visited, dist):
nonlocal ans
if mat[x][y] == 0 or (x, y) in visited:
return
if (x, y) == (xd, yd):
ans = max(dist, ans)
return
visited.add((x, y))
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
nx = x + dx
ny = y + dy
if 0 <= nx < n and 0 <= ny < m:
dfs(nx, ny, visited, dist + 1)
visited.remove((x, y))
dfs(xs, ys, set(), 0)
return ans
class IntArray:
def __init__(self) -> None:
pass
def Input(self, n):
arr = [int(i) for i in input().strip().split()]
return arr
def Print(self, arr):
for i in arr:
print(i, end=" ")
print()
class IntMatrix:
def __init__(self) -> None:
pass
def Input(self, n, m):
matrix = []
for _ in range(n):
matrix.append([int(i) for i in input().strip().split()])
return matrix
def Print(self, arr):
for i in arr:
for j in i:
print(j, end=" ")
print()
if __name__ == "__main__":
t = int(input())
for _ in range(t):
a = IntArray().Input(2)
b = IntArray().Input(4)
mat = IntMatrix().Input(a[0], a[0])
obj = Solution()
res = obj.longestPath(mat, a[0], a[1], b[0], b[1], b[2], b[3])
print(res) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER RETURN VAR VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR IF VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
path = [(1, 0), (-1, 0), (0, 1), (0, -1)]
visit = [([0] * m) for i in range(n)]
def f(i, j, count):
nonlocal maxi
if i == xd and j == yd:
maxi = max(maxi, count)
return
for di, dj in path:
ni, nj = i + di, j + dj
if (
0 <= ni < n
and 0 <= nj < m
and visit[ni][nj] == 0
and mat[ni][nj] == 1
):
visit[ni][nj] = 1
f(ni, nj, count + 1)
visit[ni][nj] = 0
maxi = -1
visit[xs][ys] = 1
f(xs, ys, 0)
return maxi | CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FOR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
def fun(i, j):
if i == xd and j == yd:
return 0
if i < 0 or i >= n or j < 0 or j >= m or mat[i][j] == 0:
return float("-inf")
mat[i][j] = 0
x = 1 + max(fun(i + 1, j), fun(i - 1, j), fun(i, j + 1), fun(i, j - 1))
mat[i][j] = 1
return x
x = fun(xs, ys)
if x > 0:
return x
return -1 | CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR STRING ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN VAR RETURN NUMBER |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
def haha(arr, i, j, p, q):
if i == p and j == q:
return 0
if i < 0 or j < 0 or i >= n or j >= m or arr[i][j] == 0:
return float("-inf")
ma = arr[i][j]
arr[i][j] = 0
a = haha(arr, i + 1, j, p, q)
b = haha(arr, i, j + 1, p, q)
c = haha(arr, i - 1, j, p, q)
d = haha(arr, i, j - 1, p, q)
arr[i][j] = ma
return 1 + max(a, b, c, d)
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
x = haha(mat, xs, ys, xd, yd)
if x == float("-inf"):
return -1
return x
class IntArray:
def __init__(self) -> None:
pass
def Input(self, n):
arr = [int(i) for i in input().strip().split()]
return arr
def Print(self, arr):
for i in arr:
print(i, end=" ")
print()
class IntMatrix:
def __init__(self) -> None:
pass
def Input(self, n, m):
matrix = []
for _ in range(n):
matrix.append([int(i) for i in input().strip().split()])
return matrix
def Print(self, arr):
for i in arr:
for j in i:
print(j, end=" ")
print()
if __name__ == "__main__":
t = int(input())
for _ in range(t):
a = IntArray().Input(2)
b = IntArray().Input(4)
mat = IntMatrix().Input(a[0], a[0])
obj = Solution()
res = obj.longestPath(mat, a[0], a[1], b[0], b[1], b[2], b[3])
print(res) | CLASS_DEF FUNC_DEF FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR RETURN BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR IF VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, matrix, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
if matrix[xs][ys] == 0 or matrix[xd][yd] == 0:
return -1
visited = [[(False) for _ in range(len(matrix[0]))] for _ in range(len(matrix))]
self.res = 0
r = len(matrix)
c = len(matrix[0])
def dfs(i, j, ans):
if (
i < 0
or j < 0
or i >= r
or j >= c
or matrix[i][j] == 0
or visited[i][j] == True
):
return
if i == xd and j == yd:
self.res = max(self.res, ans)
visited[i][j] = True
dfs(i + 1, j, ans + 1)
dfs(i, j + 1, ans + 1)
dfs(i - 1, j, ans + 1)
dfs(i, j - 1, ans + 1)
visited[i][j] = False
dfs(xs, ys, 0)
return self.res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
ans = -1
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
def fun(i, j, path):
nonlocal ans
mat[i][j] = 2
if i == xd and j == yd:
ans = max(ans, path)
if i - 1 >= 0:
if mat[i - 1][j] == 1:
fun(i - 1, j, path + 1)
if i + 1 < n:
if mat[i + 1][j] == 1:
fun(i + 1, j, path + 1)
if j - 1 >= 0:
if mat[i][j - 1] == 1:
fun(i, j - 1, path + 1)
if j + 1 < m:
if mat[i][j + 1] == 1:
fun(i, j + 1, path + 1)
mat[i][j] = 1
fun(xs, ys, 0)
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER VAR IF VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
def haha(arr, i, j, p, q):
if i == p and j == q:
return 0
if i < 0 or j < 0 or i >= n or j >= m or arr[i][j] == 0:
return float("-inf")
ma = arr[i][j]
arr[i][j] = 0
a = haha(arr, i + 1, j, p, q)
b = haha(arr, i, j + 1, p, q)
c = haha(arr, i - 1, j, p, q)
d = haha(arr, i, j - 1, p, q)
arr[i][j] = ma
return 1 + max(a, b, c, d)
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
x = haha(mat, xs, ys, xd, yd)
if x == float("-inf"):
return -1
return x | CLASS_DEF FUNC_DEF FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR RETURN BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
paths = []
path = []
recStack = set()
recStack.add((xs, ys))
def dfs(r, c):
if mat[r][c] == 0:
return
if [r, c] == [xd, yd]:
paths.append(path.copy())
return
if r - 1 >= 0 and mat[r - 1][c] == 1 and (r - 1, c) not in recStack:
path.append("U")
recStack.add((r - 1, c))
dfs(r - 1, c)
recStack.remove((r - 1, c))
path.pop()
if r + 1 < n and mat[r + 1][c] == 1 and (r + 1, c) not in recStack:
path.append("D")
recStack.add((r + 1, c))
dfs(r + 1, c)
recStack.remove((r + 1, c))
path.pop()
if c - 1 >= 0 and mat[r][c - 1] == 1 and (r, c - 1) not in recStack:
path.append("L")
recStack.add((r, c - 1))
dfs(r, c - 1)
recStack.remove((r, c - 1))
path.pop()
if c + 1 < m and mat[r][c + 1] == 1 and (r, c + 1) not in recStack:
path.append("R")
recStack.add((r, c + 1))
dfs(r, c + 1)
recStack.remove((r, c + 1))
path.pop()
dfs(xs, ys)
mx = -1
for p in paths:
mx = max(mx, len(p))
return mx | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR FUNC_DEF IF VAR VAR VAR NUMBER RETURN IF LIST VAR VAR LIST VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR RETURN IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs: int, ys: int, xd: int, yd: int) -> int:
if not mat:
return mat
if mat[xs][ys] == 0:
return -1
visited = set()
rows, cols = len(mat), len(mat[0])
ans = []
visited = set()
def dfs(i, j, path, visited):
if i == xd and j == yd:
ans.append(path)
return
if (
(i + 1, j) not in visited
and i + 1 >= 0
and i + 1 < rows
and j >= 0
and j < cols
and mat[i + 1][j] == 1
):
visited.add((i, j))
dfs(i + 1, j, path + "D", visited)
visited.remove((i, j))
if (
(i - 1, j) not in visited
and i - 1 >= 0
and i - 1 < rows
and j >= 0
and j < cols
and mat[i - 1][j] == 1
):
visited.add((i, j))
dfs(i - 1, j, path + "U", visited)
visited.remove((i, j))
if (
(i, j - 1) not in visited
and i >= 0
and i < rows
and j - 1 >= 0
and j - 1 < cols
and mat[i][j - 1] == 1
):
visited.add((i, j))
dfs(i, j - 1, path + "L", visited)
visited.remove((i, j))
if (
(i, j + 1) not in visited
and i >= 0
and i < rows
and j + 1 >= 0
and j + 1 < cols
and mat[i][j + 1] == 1
):
visited.add((i, j))
dfs(i, j + 1, path + "R", visited)
visited.remove((i, j))
dfs(xs, ys, "", visited)
if len(ans) == 0:
return -1
maxi = len(ans[0])
for i in range(len(ans)):
maxi = max(maxi, len(ans[i]))
return maxi | CLASS_DEF FUNC_DEF VAR VAR VAR VAR IF VAR RETURN VAR IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR STRING VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR STRING VAR EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR STRING VAR EXPR FUNC_CALL VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR STRING VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def utils(
self, mat, n, m, src_row, src_col, dest_row, dest_col, max_path, cur_path
):
if src_col == dest_col and src_row == dest_row:
max_path[0] = max(max_path[0], path)
return 0
if (
(src_col >= m or src_row >= n)
or (src_row < 0 or src_col < 0)
or mat[src_row][src_col] == 0
):
return 0
mat[src_row][src_col] = 0
self.utils(
mat, n, m, src_row + 1, src_col, dest_row, dest_col, max_path, cur_path + 1
)
self.utils(
mat, n, m, src_row - 1, src_col, dest_row, dest_col, max_path, cur_path + 1
)
self.utils(
mat, n, m, src_row, src_col + 1, dest_row, dest_col, max_path, cur_path + 1
)
self.utils(
mat, n, m, src_row, src_col - 1, dest_row, dest_col, max_path, cur_path + 1
)
mat[src_row][src_col] = 1
return 0
def longestPath(self, mat, n, m, xs, ys, xd, yd):
ans = -1
def dfs(x, y, visited, dist):
nonlocal ans
if mat[x][y] == 0 or (x, y) in visited:
return
if (x, y) == (xd, yd):
ans = max(dist, ans)
return
visited.add((x, y))
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
nx = x + dx
ny = y + dy
if 0 <= nx < n and 0 <= ny < m:
dfs(nx, ny, visited, dist + 1)
visited.remove((x, y))
dfs(xs, ys, set(), 0)
return ans | CLASS_DEF FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def is_safe(self, x, y, r, c, mat):
global vis
if x < 0 or x >= r or y < 0 or y >= c:
return False
return mat[x][y] == 1 and vis[x][y] == False
def bt(self, cur_x, cur_y, xd, yd, r, c, dist, mat):
global ans, vis
if cur_x == xd and cur_y == yd:
ans = max(ans, dist)
return
vis[cur_x][cur_y] = True
x = [0, 0, -1, 1]
y = [-1, 1, 0, 0]
for i in range(len(x)):
new_x = cur_x + x[i]
new_y = cur_y + y[i]
if self.is_safe(new_x, new_y, r, c, mat):
self.bt(new_x, new_y, xd, yd, r, c, dist + 1, mat)
vis[cur_x][cur_y] = False
def longestPath(self, mat, r, c, xs, ys, xd, yd):
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
global ans, vis
ans = 0
vis = [[(False) for j in range(c)] for i in range(r)]
self.bt(xs, ys, xd, yd, r, c, 0, mat)
return ans | CLASS_DEF FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR RETURN NUMBER RETURN VAR VAR VAR NUMBER VAR VAR VAR NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER VAR RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
self.ans = -1
self.dir_delta = [[0, -1], [-1, 0], [0, 1], [1, 0]]
R = n
C = m
def canGoto(r, c):
nonlocal mat, R, C
return 0 <= r < R and 0 <= c < C and mat[r][c] == 1
def backtrack(sr, sc, distance=0):
nonlocal mat, xd, yd
if canGoto(sr, sc):
if sr == xd and sc == yd:
self.ans = max(self.ans, distance)
else:
mat[sr][sc] = 2
for dr, dc in self.dir_delta:
backtrack(sr + dr, sc + dc, distance + 1)
mat[sr][sc] = 1
backtrack(xs, ys)
return self.ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR FUNC_DEF RETURN NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER FUNC_DEF NUMBER IF FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
if mat[xs][ys] == 0:
return -1
vis = [[(False) for _ in range(m)] for _ in range(n)]
mat[xs][ys] = -1
mx = [-1]
self.fn(mat, n, m, xs, ys, xd, yd, 0, vis, mx)
return max(mx)
def fn(self, mat, n, m, xs, ys, xd, yd, path, vis, mx):
if xs == xd and ys == yd:
mx.append(path)
return
for i, j in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
if 0 <= xs + i < n and 0 <= ys + j < m and mat[xs + i][ys + j] == 1:
mat[xs + i][ys + j] = -1
self.fn(mat, n, m, xs + i, ys + j, xd, yd, path + 1, vis, mx)
mat[xs + i][ys + j] = 1 | CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR LIST NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR RETURN FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN FOR VAR VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER IF NUMBER BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR VAR NUMBER |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def isValid(self, mat, x, y, n, m):
if x >= 0 and y >= 0 and x < n and y < m and not vis[x][y] and mat[x][y]:
return True
return False
def solve(self, mat, n, m, x, y, ex, ey, cor, steps):
global maxi
if not self.isValid(mat, x, y, n, m):
return
if x == ex and y == ey:
maxi = max(maxi, steps)
return
vis[x][y] = 1
for i in range(4):
nx = x + cor[i][0]
ny = y + cor[i][1]
self.solve(mat, n, m, nx, ny, ex, ey, cor, steps + 1)
vis[x][y] = 0
def longestPath(self, mat, n, m, sx, sy, ex, ey):
if sx == ex and sy == ey:
return 0
if not mat[sx][sy]:
return -1
global vis
vis = [[(0) for i in range(m)] for j in range(n)]
global maxi
maxi = float("-Inf")
cor = [[1, 0], [-1, 0], [0, 1], [0, -1]]
self.solve(mat, n, m, sx, sy, ex, ey, cor, 0)
if maxi == float("-Inf"):
return -1
return maxi | CLASS_DEF FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF FUNC_CALL VAR VAR VAR VAR VAR VAR RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
def solve(mat, n, m, xs, ys, xd, yd, temp):
if (
xs < 0
or ys < 0
or xs == n
or ys == m
or mat[xs][ys] == 2
or mat[xs][ys] == 0
):
return
nonlocal length
if xs == xd and ys == yd:
if temp > length:
length = temp
return
mat[xs][ys] = 2
solve(mat, n, m, xs - 1, ys, xd, yd, temp + 1)
solve(mat, n, m, xs + 1, ys, xd, yd, temp + 1)
solve(mat, n, m, xs, ys + 1, xd, yd, temp + 1)
solve(mat, n, m, xs, ys - 1, xd, yd, temp + 1)
mat[xs][ys] = 1
length = -1
solve(mat, n, m, xs, ys, xd, yd, 0)
return length | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def __init__(self):
self.ans = -1
def longestPath(self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int):
rows = len(mat)
cols = len(mat[0])
v2 = [[(False) for _ in range(cols)] for _ in range(rows)]
self.path(xs, ys, xd, yd, mat, v2, 0)
return self.ans
def path(self, i, j, end_i, end_j, mat, visited, steps):
if (
i < 0
or j < 0
or i >= len(mat)
or j >= len(mat[0])
or visited[i][j] is True
or mat[i][j] == 0
):
return
if i == end_i and j == end_j:
self.ans = max(self.ans, steps)
return
visited[i][j] = True
self.path(i + 1, j, end_i, end_j, mat, visited, steps + 1),
self.path(i, j + 1, end_i, end_j, mat, visited, steps + 1)
self.path(i - 1, j, end_i, end_j, mat, visited, steps + 1)
self.path(i, j - 1, end_i, end_j, mat, visited, steps + 1)
visited[i][j] = False
return | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
ans = []
visited = [[(0) for _ in range(m)] for _ in range(n)]
def isSafe(x, y):
if (
x >= 0
and y >= 0
and x < n
and y < m
and visited[x][y] != 1
and mat[x][y] == 1
):
return True
return False
def solve(x, y, path):
if x == xd and y == yd:
ans.append(path)
return
visited[x][y] = 1
if isSafe(x + 1, y):
solve(x + 1, y, path + 1)
if isSafe(x, y - 1):
solve(x, y - 1, path + 1)
if isSafe(x, y + 1):
solve(x, y + 1, path + 1)
if isSafe(x - 1, y):
solve(x - 1, y, path + 1)
visited[x][y] = 0
if mat[xs][ys] == 0:
return -1
else:
solve(xs, ys, 0)
if ans:
return max(ans)
else:
return -1 | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER RETURN NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR RETURN FUNC_CALL VAR VAR RETURN NUMBER |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
return (
self.helper(mat, xs, ys, n - 1, m - 1, (xd, yd), 0) if mat[xs][ys] else -1
)
def helper(self, grid, x, y, n, m, end, count):
if (x, y) == end:
return count
matrix = [row.copy() for row in grid]
matrix[x][y] = 0
output = -1
if x > 0 and matrix[x - 1][y]:
output = max(output, self.helper(matrix, x - 1, y, n, m, end, count + 1))
if y > 0 and matrix[x][y - 1]:
output = max(output, self.helper(matrix, x, y - 1, n, m, end, count + 1))
if x < n and matrix[x + 1][y]:
output = max(output, self.helper(matrix, x + 1, y, n, m, end, count + 1))
if y < m and matrix[x][y + 1]:
output = max(output, self.helper(matrix, x, y + 1, n, m, end, count + 1))
return output | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR RETURN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR NUMBER NUMBER VAR FUNC_DEF IF VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
path_lengths = []
visited = set()
pos = xs, ys
dst = xd, yd
length = 0
self.dfs(mat, n, m, pos, dst, visited, length, path_lengths)
if len(path_lengths) == 0:
return -1
return max(path_lengths)
def dfs(self, mat, n, m, pos, dst, visited, length, path_lengths):
if mat[pos[0]][pos[1]] == 0:
return
if pos in visited:
return
if pos == dst:
path_lengths.append(length)
return
visited.add(pos)
if pos[1] < m - 1:
right_pos = pos[0], pos[1] + 1
self.dfs(
mat, n, m, right_pos, dst, visited.copy(), length + 1, path_lengths
)
if pos[0] < n - 1:
down_pos = pos[0] + 1, pos[1]
self.dfs(mat, n, m, down_pos, dst, visited.copy(), length + 1, path_lengths)
if pos[1] > 0:
left_pos = pos[0], pos[1] - 1
self.dfs(mat, n, m, left_pos, dst, visited.copy(), length + 1, path_lengths)
if pos[0] > 0:
up_pos = pos[0] - 1, pos[1]
self.dfs(mat, n, m, up_pos, dst, visited.copy(), length + 1, path_lengths) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR NUMBER VAR NUMBER NUMBER RETURN IF VAR VAR RETURN IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
ans = [-1]
def f(x, y, xd, yd, mat, n, m, vis, ans, path):
if x < 0 or x >= n or y < 0 or y >= m or vis[x][y] or mat[x][y] == 0:
return
if x == xd and y == yd:
ans[0] = max(ans[0], path)
return
vis[x][y] = 1
f(x + 1, y, xd, yd, mat, n, m, vis, ans, path + 1)
f(x - 1, y, xd, yd, mat, n, m, vis, ans, path + 1)
f(x, y + 1, xd, yd, mat, n, m, vis, ans, path + 1)
f(x, y - 1, xd, yd, mat, n, m, vis, ans, path + 1)
vis[x][y] = 0
vis = [([0] * m) for i in range(n)]
f(xs, ys, xd, yd, mat, n, m, vis, ans, 0)
return ans[0] | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST NUMBER FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR NUMBER VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
if mat[xs][ys] == 0 or mat[xd][yd] == 0:
return -1
visited = [([-1] * len(mat[0])) for i in range(len(mat))]
def traverse(i, j, xd, yd):
if (
i < 0
or j < 0
or i >= len(mat)
or j >= len(mat[0])
or mat[i][j] == 0
or visited[i][j] == 1
):
return -999999999
if i == xd and j == yd:
return 0
visited[i][j] = 1
val = -999999999
val = max(val, 1 + traverse(i + 1, j, xd, yd))
val = max(val, 1 + traverse(i - 1, j, xd, yd))
val = max(val, 1 + traverse(i, j + 1, xd, yd))
val = max(val, 1 + traverse(i, j - 1, xd, yd))
visited[i][j] = -1
return val
val = traverse(xs, ys, xd, yd)
if val < 0:
return -1
else:
return val | CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR NUMBER RETURN NUMBER RETURN VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | def util(mat, m, n, xs, ys, xd, yd, visit, path):
if xs == xd and ys == yd:
return path
if xs < 0 or xs >= m or ys < 0 or ys >= n or mat[xs][ys] == 0 or visit[xs][ys]:
return float("inf")
visit[xs][ys] = True
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
dist = 0
for i in range(4):
x, y = xs + dx[i], ys + dy[i]
t = util(mat, m, n, x, y, xd, yd, visit, path + 1)
if t != float("inf") and t > dist:
dist = t
visit[xs][ys] = False
return dist
class Solution:
def longestPath(
self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int
) -> int:
if mat[xd][yd] == 0:
return -1
m, n = len(mat), len(mat[0])
visit = [[(False) for c in range(n)] for r in range(m)]
ans = util(mat, m, n, xs, ys, xd, yd, visit, 0)
if ans == float("inf"):
return -1
else:
return ans | FUNC_DEF IF VAR VAR VAR VAR RETURN VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR VAR RETURN FUNC_CALL VAR STRING ASSIGN VAR VAR VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR STRING VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR VAR |
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1.
Example 1:
Input:
{xs,ys} = {0,0}
{xd,yd} = {1,7}
matrix = 1 1 1 1 1 1 1 1 1 1
1 1 0 1 1 0 1 1 0 1
1 1 1 1 1 1 1 1 1 1
Output: 24
Explanation:
Example 2:
Input:
{xs,ys} = {0,3}
{xd,yd} = {2,2}
matrix = 1 0 0 1 0
0 0 0 1 0
0 1 1 0 0
Output: -1
Explanation:
We can see that it is impossible to
reach the cell (2,2) from (0,3).
Your Task:
You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path.
Expected Time Complexity: O(2^(N*M))
Expected Auxiliary Space: O(N*M)
Constraints:
1 <= N,M <= 10 | class Solution:
def longestPath(self, mat, n, m, xs, ys, xd, yd):
n = len(mat)
m = len(mat[0])
ans = -1
def dfs(i, j, visited, dist):
nonlocal ans
move = [[-1, 0], [1, 0], [0, -1], [0, 1]]
if mat[i][j] == 0 or (i, j) in visited:
return
if (i, j) == (xd, yd):
ans = max(ans, dist)
return
visited.add((i, j))
for x, y in move:
newX = x + i
newY = y + j
if (
newX >= 0
and newX < n
and newY >= 0
and newY < m
and mat[newX][newY] == 1
):
dfs(newX, newY, visited, dist + 1)
visited.remove((i, j))
return ans
ans = dfs(xs, ys, set(), 0)
if ans == None:
return -1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR RETURN IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER IF VAR NONE RETURN NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def l_find_bsearch(self, nums, target):
l = 0
r = len(nums) - 1
while l <= r:
mid = (l + r) // 2
if nums[mid] == target:
r = mid - 1
if nums[mid] < target:
l = mid + 1
if nums[mid] > target:
r = mid - 1
return l
def r_find_bsearch(self, nums, target):
l = 0
r = len(nums) - 1
while l <= r:
mid = (l + r) // 2
if nums[mid] == target:
l = mid + 1
if nums[mid] < target:
l = mid + 1
if nums[mid] > target:
r = mid - 1
return r
def count(self, nums, n, target):
if nums == []:
return 0
l = self.l_find_bsearch(nums, target)
if l >= len(nums) or nums[l] != target:
return 0
r = self.r_find_bsearch(nums, target)
if r < 0 or nums[r] != target:
return 0
return r - l + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF IF VAR LIST RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR VAR RETURN NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
count = 0
i = 0
while i < n:
if arr[i] == x:
count += 1
i += 1
return count | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count1(self, nums, n, x):
count = 0
for num in nums:
if num == x:
count += 1
return count
def count(self, nums, n, x):
def binarySearch(left, right, searchDirection):
result = -1
while left <= right:
mid = (left + right) // 2
val = nums[mid]
if val == x:
result = mid
if searchDirection == "f":
right = mid - 1
else:
left = mid + 1
elif val > x:
right = mid - 1
else:
left = mid + 1
return result
floor = binarySearch(0, n - 1, "f")
ceil = binarySearch(0, n - 1, "c")
if floor != -1 and ceil != -1:
return ceil - floor + 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER RETURN VAR FUNC_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER STRING IF VAR NUMBER VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
l = 0
r = n - 1
left = -1
while l <= r:
mid = (l + r) // 2
if arr[mid] == x:
left = mid
r = mid - 1
elif arr[mid] > x:
r = mid - 1
else:
l = mid + 1
l = 0
r = n - 1
right = -1
while l <= r:
mid = (l + r) // 2
if arr[mid] == x:
right = mid
l = mid + 1
elif arr[mid] > x:
r = mid - 1
else:
l = mid + 1
if right + left == -2:
return 0
else:
return right - left + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER RETURN NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def binary(self, arr, n, x):
s = 0
e = len(arr) - 1
while s <= e:
m = (s + e) // 2
if arr[m] == x:
return m
elif arr[m] < x:
s = m + 1
else:
e = m - 1
return -1
def count(self, arr, n, x):
idx = self.binary(arr, n, x)
if idx == -1:
return 0
k = 1
left = idx - 1
while left >= 0 and arr[left] == x:
k = k + 1
left = left - 1
right = idx + 1
while right < n and arr[right] == x:
k += 1
right += 1
return k | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR RETURN VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
def first(arr, n, x):
low = 0
high = n - 1
s = -1
while low <= high:
mid = low + (high - low) // 2
if arr[mid] < x:
low = mid + 1
elif arr[mid] > x:
high = mid - 1
else:
s = mid
high = mid - 1
return s
def last(arr, n, x):
low = 0
high = n - 1
e = -1
while low <= high:
mid = low + (high - low) // 2
if arr[mid] < x:
low = mid + 1
elif arr[mid] > x:
high = mid - 1
else:
e = mid
low = mid + 1
return e
f = first(arr, n, x)
z = last(arr, n, x)
if z == -1 and f == -1:
return 0
return z - f + 1 | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER RETURN NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
def search(last=False):
res = -1
low, high = 0, n - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] == x:
res = mid
if last:
low = mid + 1
else:
high = mid - 1
elif arr[mid] > x:
high = mid - 1
else:
low = mid + 1
return res
firstOccurence = search()
if firstOccurence == -1:
return 0
lastOccurence = search(True)
return lastOccurence - firstOccurence + 1 | CLASS_DEF FUNC_DEF FUNC_DEF NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR IF VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
first = -1
last = -1
count = 0
for i in range(n):
if arr[i] == x:
first = i
break
if first == -1:
return 0
for i in range(n - 1, -1, -1):
if arr[i] == x:
last = i
break
count = last - first + 1
return count | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
n = len(arr)
first = -1
last = -1
low = 0
high = n - 1
while low <= high:
mid = (low + high) // 2
if (mid == 0 or x > arr[mid - 1]) and arr[mid] == x:
first = mid
break
elif x > arr[mid]:
low = mid + 1
else:
high = mid - 1
low = first if first != -1 else 0
high = n - 1
while low <= high:
mid = (low + high) // 2
if (mid == high or x < arr[mid + 1]) and arr[mid] == x:
last = mid
break
elif x < arr[mid]:
high = mid - 1
else:
low = mid + 1
count = last - first + 1 if first != -1 else 0
return count | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR ASSIGN VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
if n <= 0:
return -1
elif x not in arr:
return 0
else:
c = 0
for i in arr:
if i == x:
c += 1
return c | CLASS_DEF FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
start = 0
end = n - 1
while start <= end:
if arr[start] != x:
start = start + 1
if arr[end] != x:
end = end - 1
if arr[start] == x and arr[end] == x:
break
if start <= end:
return end - start + 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR VAR VAR IF VAR VAR RETURN BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
if x not in arr:
return 0
l = 0
r = len(arr) - 1
while arr[l] != x:
l = l + 1
while arr[r] != x:
r = r - 1
return r - l + 1 | CLASS_DEF FUNC_DEF IF VAR VAR RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
l = 0
r = n - 1
c = 0
last = -1
first = -1
while l <= r:
mid = (l + r) // 2
if arr[mid] == x:
first = mid
l = mid + 1
elif arr[mid] < x:
l = mid + 1
else:
r = mid - 1
l = 0
r = n - 1
while l <= r:
mid = (l + r) // 2
if arr[mid] == x:
r = mid - 1
last = mid
elif arr[mid] < x:
l = mid + 1
else:
r = mid - 1
if first == last and first != -1 and last != -1:
return 1
elif first == -1 and last == -1:
return 0
return first - last + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR NUMBER VAR NUMBER RETURN NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
d = {}
for i in arr:
if i in d:
d[i] += 1
else:
d[i] = 1
c = 0
for i in d:
if i == x:
return d[i]
break
else:
c += 1
if c != 0:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR RETURN VAR VAR VAR NUMBER IF VAR NUMBER RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
left = 0
right = len(arr) - 1
f = 1
w = 1
while left <= right:
mid = (left + right) // 2
if arr[mid] == x:
w = 0
k = mid
while k >= 1 and arr[k] == arr[k - 1]:
f += 1
k -= 1
while mid < len(arr) - 1 and arr[mid] == arr[mid + 1]:
f += 1
mid += 1
break
if arr[mid] == x:
f += 1
break
if arr[mid] < x:
left = mid + 1
if arr[mid] > x:
right = mid - 1
if w:
return 0
return f | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR RETURN NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
i, j = 0, n - 1
while i <= j:
m = (i + j) // 2
if arr[m] >= x:
j = m - 1
else:
i = m + 1
if i < 0 or i >= n or arr[i] != x:
return 0
f = i
i, j = 0, n - 1
while i <= j:
m = (i + j) // 2
if arr[m] <= x:
i = m + 1
else:
j = m - 1
return j - f + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def bin_search(self, flag, l, r, nums, target):
ans = -1
while l <= r:
mid = (l + r) // 2
if nums[mid] < target:
l = mid + 1
elif nums[mid] > target:
r = mid - 1
else:
ans = mid
if flag:
r = mid - 1
else:
l = mid + 1
return ans
def count(self, nums, n, target):
l = 0
r = len(nums) - 1
left = self.bin_search(True, l, r, nums, target)
right = self.bin_search(False, l, r, nums, target)
if left == -1 and right == -1:
return 0
return right - left + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER RETURN NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
s = 0
mid = -1
e = n - 1
while s <= e:
mid = s + (e - s) // 2
if arr[mid] == x:
break
elif arr[mid] > x:
e = mid - 1
else:
s = mid + 1
if mid == -1:
return 0
count = 0
i = mid
while i >= 0 and arr[i] == x:
count += 1
i -= 1
i = mid + 1
while i < n and arr[i] == x:
count += 1
i += 1
return count | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
l, r = 0, n - 1
count = 0
while l <= r:
mid = (l + r) // 2
if arr[mid] == x:
count += 1
i, j = mid, mid
while i - 1 >= 0 and arr[i - 1] == x:
count += 1
i -= 1
while j + 1 < n and arr[j + 1] == x:
count += 1
j += 1
return count
elif arr[mid] > x:
r = mid - 1
else:
l = mid + 1
return count | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR WHILE BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR NUMBER WHILE BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR NUMBER RETURN VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
mydict = dict()
for i in range(n):
if arr[i] not in mydict:
mydict[arr[i]] = 1
else:
mydict[arr[i]] += 1
if x in mydict:
return mydict[x]
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR VAR RETURN VAR VAR RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
left = 0
right = n - 1
while left <= right:
mid = (left + right) // 2
if arr[mid] == x:
result = [mid, mid]
while (
result[0] >= 0
and arr[result[0]] == x
and result[1] < n
and arr[result[1]] == x
):
result[0] -= 1
result[1] += 1
while result[0] >= 0 and arr[result[0]] == x:
result[0] -= 1
while result[1] < n and arr[result[1]] == x:
result[1] += 1
return result[1] - result[0] - 1
elif arr[mid] < x:
left = mid + 1
else:
right = mid - 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR LIST VAR VAR WHILE VAR NUMBER NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR NUMBER NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER VAR VAR NUMBER VAR VAR NUMBER NUMBER WHILE VAR NUMBER VAR VAR VAR NUMBER VAR VAR NUMBER NUMBER RETURN BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
if n == 1:
if x == arr[0]:
return 1
l = 0
r = n - 1
while l < r:
if arr[l] != x:
l = l + 1
if arr[r] != x:
r = r - 1
if arr[r] == x and arr[l] == x:
return r - l + 1
break
return 0 | CLASS_DEF FUNC_DEF IF VAR NUMBER IF VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR VAR VAR RETURN BIN_OP BIN_OP VAR VAR NUMBER RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
res = []
for i in range(n):
if arr[i] == x:
res.append(i)
if len(res) != 0:
return len(res)
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR RETURN NUMBER |
Given a sorted array Arr of size N and a number X, you need to find the number of occurrences of X in Arr.
Example 1:
Input:
N = 7, X = 2
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 4
Explanation: 2 occurs 4 times in the
given array.
Example 2:
Input:
N = 7, X = 4
Arr[] = {1, 1, 2, 2, 2, 2, 3}
Output: 0
Explanation: 4 is not present in the
given array.
Your Task:
You don't need to read input or print anything. Your task is to complete the function count() which takes the array of integers arr, n, and x as parameters and returns an integer denoting the answer.
If x is not present in the array (arr) then return 0.
Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{5}
1 ≤ Arr[i] ≤ 10^{6}
1 ≤ X ≤ 10^{6} | class Solution:
def count(self, arr, n, x):
def rhs(arr, x):
low = 0
high = len(arr) - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] > x:
high = mid - 1
else:
low = mid + 1
return high
def lhs(arr, x):
low = 0
high = len(arr) - 1
while low <= high:
mid = (low + high) // 2
if arr[mid] >= x:
high = mid - 1
else:
low = mid + 1
return low
a = rhs(arr, x)
b = lhs(arr, x)
if arr[a] != x:
return 0
return a - b + 1 | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR RETURN NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | for _ in range(int(input())):
n = input()
if len(n) >= 4:
if n[-4:] == "1000":
print("YES")
else:
print("NO")
else:
print("NO") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | adj = [[0, 1], [2, 1], [3, 1], [4, 1], [0, 1]]
T = int(input())
for _ in range(T):
L = list(input())
M = [int(x) for x in L]
initial = 0
for i in range(len(M)):
a = int(M[i])
initial = adj[initial][a]
if initial == 4:
print("YES")
else:
print("NO") | ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def Helper(index, S, adj, root):
if index == len(S) and root == 4:
return True
if index >= len(S):
return False
ans = False
for node in adj[root]:
if node[1] == S[index]:
ans = ans | Helper(index + 1, S, adj, node[0])
else:
pass
return ans
def function(S, adj):
if len(S) > 3 and S[len(S) - 4 :] == "1000":
return "YES"
return "NO"
def createADJ():
adj = [[] for i in range(5)]
adj[0] = [(0, "0"), (1, "1")]
adj[1] = [(1, "1"), (2, "0")]
adj[2] = [(1, "1"), (3, "0")]
adj[3] = [(4, "0"), (1, "1")]
adj[4] = [(0, "0"), (1, "1")]
return adj
def main():
T = int(input())
for _ in range(T):
adj = createADJ()
S = input()
val = function(S, adj)
print(val)
main() | FUNC_DEF IF VAR FUNC_CALL VAR VAR VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER RETURN VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING RETURN STRING RETURN STRING FUNC_DEF ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER LIST NUMBER STRING NUMBER STRING ASSIGN VAR NUMBER LIST NUMBER STRING NUMBER STRING ASSIGN VAR NUMBER LIST NUMBER STRING NUMBER STRING ASSIGN VAR NUMBER LIST NUMBER STRING NUMBER STRING ASSIGN VAR NUMBER LIST NUMBER STRING NUMBER STRING RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
for i in range(t):
s = str(input())
n = len(s)
if n < 4:
print("NO")
elif s[n - 4] != "0" and s[n - 3] == "0" and s[n - 2] == "0" and s[n - 1] == "0":
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | for _ in range(int(input())):
map = [[0, 1], [2, 1], [3, 1], [4, 1], [0, 1]]
path = input()
curr = 0
for p in path:
curr = map[curr][int(p)]
if curr == 4:
print("YES")
else:
print("NO") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | graph = [
{"0": 0, "1": 1},
{"1": 1, "0": 2},
{"0": 3, "1": 1},
{"1": 1, "0": 4},
{"1": 1, "0": 0},
]
for i in range(0, int(input())):
S = input()
N = len(S)
currentIndex = 0
for i in range(0, N):
currentIndex = graph[currentIndex][S[i]]
if currentIndex == 4:
print("YES")
else:
print("NO") | ASSIGN VAR LIST DICT STRING STRING NUMBER NUMBER DICT STRING STRING NUMBER NUMBER DICT STRING STRING NUMBER NUMBER DICT STRING STRING NUMBER NUMBER DICT STRING STRING NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
for _ in range(t):
n = input()
if len(n) < 4:
print("NO")
elif n[-1] == "0" and n[-2] == "0" and n[-3] == "0" and n[-4] == "1":
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def state(country, roa):
road = int(roa)
if country == "red":
if road == 0:
return "red"
else:
return "blue"
elif country == "blue":
if road == 1:
return "blue"
else:
return "yellow"
elif country == "yellow":
if road == 0:
return "pink"
else:
return "blue"
elif country == "pink":
if road == 0:
return "green"
else:
return "blue"
elif country == "green":
if road == 1:
return "blue"
else:
return "red"
for _ in range(int(input())):
s = input()
c = "red"
for i in range(len(s)):
c = state(c, s[i])
if c == "green":
print("YES")
else:
print("NO") | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR STRING IF VAR NUMBER RETURN STRING RETURN STRING IF VAR STRING IF VAR NUMBER RETURN STRING RETURN STRING IF VAR STRING IF VAR NUMBER RETURN STRING RETURN STRING IF VAR STRING IF VAR NUMBER RETURN STRING RETURN STRING IF VAR STRING IF VAR NUMBER RETURN STRING RETURN STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | from sys import stdin
for _ in range(int(stdin.readline())):
s = stdin.readline().strip()
cond = True
n = len(s)
state = 1
for i in range(n):
if s[i] == "0":
if state != 1 or state != 5:
state += 1
elif state == 5:
state -= 4
elif state != 2:
state = 2
if state == 5:
print("YES")
else:
print("NO") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def path(a):
a = [int(i) for i in a]
color = "red"
for i in a:
if color == "red":
if i == 1:
color = "blue"
elif i == 0:
color = "red"
elif color == "blue":
if i == 1:
color = "blue"
elif i == 0:
color = "yellow"
elif color == "yellow":
if i == 1:
color = "blue"
elif i == 0:
color = "pink"
elif color == "pink":
if i == 1:
color = "blue"
elif i == 0:
color = "green"
elif color == "green":
if i == 1:
color = "blue"
elif i == 0:
color = "red"
return color == "green"
for i in range(int(input())):
if path(input()):
print("YES")
else:
print("NO") | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR STRING FOR VAR VAR IF VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR STRING IF VAR NUMBER ASSIGN VAR STRING IF VAR NUMBER ASSIGN VAR STRING RETURN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR IF FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def next(c, n):
if n == "1":
return "B"
elif c == "R":
return "R"
elif c == "B":
return "Y"
elif c == "Y":
return "P"
elif c == "P":
return "G"
else:
return "R"
for i in range(int(input())):
string = input()
c = "R"
for i in range(len(string)):
n = string[i]
c = next(c, n)
if c == "G":
print("YES")
else:
print("NO") | FUNC_DEF IF VAR STRING RETURN STRING IF VAR STRING RETURN STRING IF VAR STRING RETURN STRING IF VAR STRING RETURN STRING IF VAR STRING RETURN STRING RETURN STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | T = int(input())
l = [[0, 1], [2, 1], [3, 1], [4, 1], [0, 1]]
for _ in range(T):
s = input()
a = 0
for i in s:
x = int(i)
a = l[a][x]
if a == 4:
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | st = dict()
st["R"] = ["R", "B"]
st["B"] = ["Y", "B"]
st["Y"] = ["P", "B"]
st["P"] = ["G", "B"]
st["G"] = ["R", "B"]
tn = int(input())
for _ in range(tn):
si = input()
cs = "R"
for ch in si:
cs = st[cs][int(ch)]
if cs == "G":
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING LIST STRING STRING ASSIGN VAR STRING LIST STRING STRING ASSIGN VAR STRING LIST STRING STRING ASSIGN VAR STRING LIST STRING STRING ASSIGN VAR STRING LIST STRING STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
for _ in range(t):
s = input()
if len(s) >= 4:
x = s[-4] + s[-3] + s[-2] + s[-1]
if x == "1000":
print("YES")
else:
print("NO")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | graph = []
for i in range(0, 5):
l = []
for j in range(0, 2):
l.append(0)
graph.append(l)
graph[0][0] = 0
graph[0][1] = 1
graph[1][0] = 2
graph[1][1] = 1
graph[2][0] = 3
graph[2][1] = 1
graph[3][0] = 4
graph[3][1] = 1
graph[4][0] = 0
graph[4][1] = 1
n = int(input())
while n:
string = input()
pos = 0
for i in string:
pos = graph[pos][int(i)]
if pos == 4:
print("YES")
else:
print("NO")
n -= 1 | ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | for _ in range(int(input())):
s = input()
currnode = "red"
for i in s:
if currnode == "red":
if i == "1":
currnode = "blue"
elif currnode == "blue":
if i == "0":
currnode = "yellow"
elif currnode == "yellow":
if i == "0":
currnode = "pink"
else:
currnode = "blue"
elif currnode == "pink":
if i == "0":
currnode = "green"
else:
currnode = "blue"
elif currnode == "green":
if i == "0":
currnode = "red"
else:
currnode = "blue"
if currnode == "green":
print("YES")
else:
print("NO") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR VAR IF VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | n = int(input())
for i in range(n):
s = input()
j = len(s) - 1
if j > 2:
if s[j - 3] == "1" and s[j - 2] == "0" and s[j - 1] == "0" and s[j] == "0":
print("YES")
else:
print("NO")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def makeAdj():
dic = {}
dic[0, 0] = 0
dic[0, 1] = 1
dic[1, 0] = 2
dic[1, 1] = 1
dic[2, 1] = 1
dic[2, 0] = 3
dic[3, 0] = 4
dic[3, 1] = 1
dic[4, 1] = 1
dic[4, 0] = 0
return dic
adj = makeAdj()
t = int(input())
for i in range(t):
s = str(input())
node = 0
for j in s:
node = adj[node, int(j)]
if node == 4:
print("YES")
else:
print("NO") | FUNC_DEF ASSIGN VAR DICT ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
for i in range(t):
string = input()
place = "R"
for j in range(len(string)):
if place == "R":
if string[j] == "1":
place = "B"
elif place == "B":
if string[j] == "0":
place = "Y"
elif place == "Y":
if string[j] == "1":
place = "B"
elif string[j] == "0":
place = "P"
elif place == "P":
if string[j] == "1":
place = "B"
elif string[j] == "0":
place = "G"
elif place == "G":
if string[j] == "1":
place = "B"
elif string[j] == "0":
place = "R"
if place == "G":
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR VAR STRING ASSIGN VAR STRING IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | for _ in range(int(input())):
a = input()
d = {
"q00": "q0",
"q01": "q1",
"q11": "q1",
"q10": "q2",
"q21": "q1",
"q20": "q3",
"q31": "q1",
"q30": "q4",
"q41": "q1",
"q40": "q0",
}
g = "q0"
for i in a:
g += i
g = d[g]
if g == "q4":
print("YES")
else:
print("NO") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR STRING FOR VAR VAR VAR VAR ASSIGN VAR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | d = {
"red": {"0": "red", "1": "blue"},
"blue": {"0": "yellow", "1": "blue"},
"yellow": {"0": "violet", "1": "blue"},
"violet": {"0": "green", "1": "blue"},
"green": {"0": "red", "1": "blue"},
}
for _ in range(int(input())):
s = input()
c = "red"
f = 0
for i in s:
c = d[c][i]
if c == "green":
print("YES")
else:
print("NO") | ASSIGN VAR DICT STRING STRING STRING STRING STRING DICT STRING STRING STRING STRING DICT STRING STRING STRING STRING DICT STRING STRING STRING STRING DICT STRING STRING STRING STRING DICT STRING STRING STRING STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | for _ in range(int(input())):
s = input()
if len(s) < 4:
print("NO")
elif s[-4:] == "1000":
print("YES")
else:
print("NO") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def main():
t = int(input())
for _ in range(t):
sx = input()
if sx[-4:] == "1000":
print("YES")
else:
print("NO")
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
for i in range(t):
s = input()
l = len(s)
if l >= 4:
if s.endswith("1000"):
print("YES")
else:
print("NO")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER IF FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | def change_of_state(state, cur_no):
if cur_no == 1:
if state == "R":
state = "B"
elif state == "B":
state = "B"
elif state == "Y":
state = "B"
elif state == "P":
state = "B"
elif state == "G":
state = "B"
elif state == "R":
state = "R"
elif state == "B":
state = "Y"
elif state == "Y":
state = "P"
elif state == "P":
state = "G"
elif state == "G":
state = "R"
return state
for _ in range(int(input())):
s = str(input())
state = "R"
for i in range(len(s)):
state = change_of_state(state, int(s[i]))
if state == "G":
print("YES")
else:
print("NO") | FUNC_DEF IF VAR NUMBER IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING IF VAR STRING ASSIGN VAR STRING RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
while t > 0:
t = t - 1
s = input()
a = [int(i) for i in s]
if len(a) < 4:
print("NO")
continue
sample = ["r0r", "r1b", "b1b", "b0y", "y1b", "y0p", "p0g", "p1b", "g1b", "g0b"]
curr_state = "r"
next_state = ""
k = 0
z = 0
for i in a:
z += 1
tran_val = i
for j in sample:
if curr_state == j[0] and tran_val == int(j[1]):
next_state = j[2]
break
curr_state = next_state
if curr_state == "g" and z == len(a):
print("YES")
k = 1
break
if k == 0:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER ASSIGN VAR VAR FOR VAR VAR IF VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR IF VAR STRING VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | path = {
"R0": "R",
"R1": "B",
"B1": "B",
"B0": "Y",
"Y1": "B",
"Y0": "P",
"P1": "B",
"P0": "G",
"G1": "B",
"G0": "R",
}
T = int(input())
for _ in range(T):
s = input()
current = "R"
for c in s:
current = path[current + c]
print("YES" if current == "G" else "NO") | ASSIGN VAR DICT STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR STRING STRING STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | T = int(input())
R = "R"
G = "G"
B = "B"
Y = "Y"
P = "P"
Z = "0"
O = "1"
A = {}
A[R] = {}
A[R][Z] = R
A[R][O] = B
A[B] = {}
A[B][Z] = Y
A[B][O] = B
A[Y] = {}
A[Y][Z] = P
A[Y][O] = B
A[P] = {}
A[P][Z] = G
A[P][O] = B
A[G] = {}
A[G][Z] = R
A[G][O] = B
for t in range(T):
s = input()
c = R
for i in s:
c = A[c][i]
if c == G:
print("YES")
else:
print("NO") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR DICT ASSIGN VAR VAR DICT ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR DICT ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR DICT ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR DICT ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR DICT ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
There are $5$ cities in the country.
The map of the country is given below.
The tour starts from the red city.
Each road is associated with a character.
Initially, there is an empty string.
Every time a road has been travelled the character associated gets appended to the string.
At the green city either the string can be printed or the tour can be continued.
In the problem, you are given a string tell whether it is possible to print the string while following the rules of the country?
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains a single line of input, a string $ s $. The string consists only of $0's$ and $1's$.
-----Output:-----
For each testcase, output "YES" or "NO" depending on the input.
-----Constraints-----
-
$1 \leq T \leq 10000$
-
$1 \leq length of each string \leq 10000$
-
$ 1 \leq Summation length \leq 10^5$
-----Sample Input:-----
1
100
-----Sample Output:-----
NO
-----EXPLANATION:----- | t = int(input())
i = 0
while i < t:
s = input()
y = s[-4:]
if y == "1000":
print("YES")
else:
print("NO")
i = i + 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR NUMBER |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**6)
class Solution:
def largestSumCycle(self, N, Edge):
def dfs(v, path={}, sums=0):
nonlocal ans, visited, Edge
if v in visited or Edge[v] == -1:
return
if v in path:
ans = max(ans, sums - path[v])
return
path[v] = sums
dfs(Edge[v], path, sums + v)
visited.add(v)
visited = set()
ans = -1
for v in range(N):
dfs(v)
return ans | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF FUNC_DEF DICT NUMBER IF VAR VAR VAR VAR NUMBER RETURN IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR RETURN ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**7)
class Solution:
def largestSumCycle(self, N, Edge):
def getsum(node):
visited = [(False) for i in range(N)]
s = node
visited[node] = True
temp = Edge[node]
while visited[temp] != True:
s += temp
visited[temp] = True
temp = Edge[temp]
return s
def detect(node, found):
if found[0] != -2:
return
vis[node] = True
path[node] = True
nextnode = Edge[node]
if nextnode != -1:
if vis[nextnode] == False:
detect(nextnode, found)
elif path[nextnode] == True and vis[nextnode] == True:
found[0] = nextnode
path[node] = False
vis = [(False) for i in range(0, N)]
path = [(False) for i in range(0, N)]
ans = -1
found = [-2]
for node in range(0, N):
if vis[node] == False:
detect(node, found)
if found[0] != -2:
s = getsum(found[0])
ans = max(ans, s)
found[0] = -2
return ans | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR FUNC_DEF IF VAR NUMBER NUMBER RETURN ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER NUMBER RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | class Solution:
def largestSumCycle(self, N, edge):
vis = [-1] * N
dp_arr = [0] * N
res = -1
for i in range(N):
if vis[i] == -1:
temp, val, stack = i, 0, []
while True:
vis[temp] = 0
dp_arr[temp] = val
stack.append(temp)
if edge[temp] != -1:
if vis[edge[temp]] == -1:
val += temp
temp = edge[temp]
elif vis[edge[temp]] == 0:
res = max(temp + val - dp_arr[edge[temp]], res)
break
else:
break
else:
break
while stack:
vis[stack.pop()] = 1
return res | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR VAR NUMBER LIST WHILE NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR NUMBER RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | class Solution:
def largestSumCycle(self, N, Edge):
visited = [False] * N
max_cycle_sum = -1
path = dict()
path_keys = []
for u in range(N):
if visited[u]:
continue
while -1 < u < N:
if u in path:
start = path[u]
cycle_sum = sum(path_keys[start:])
max_cycle_sum = max(max_cycle_sum, cycle_sum)
break
if visited[u]:
break
visited[u] = True
path[u] = len(path_keys)
path_keys.append(u)
u = Edge[u]
path.clear()
return max_cycle_sum | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR WHILE NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**6)
class Solution:
def dfs(self, Edge, i, vis, pathV, ans):
vis[i] = True
pathV[i] = True
if Edge[i] != -1:
next = Edge[i]
if not vis[next]:
self.dfs(Edge, next, vis, pathV, ans)
elif pathV[next]:
res = next
node = Edge[next]
while node != next:
res += node
node = Edge[node]
ans[0] = max(ans[0], res)
pathV[i] = False
def largestSumCycle(self, N, Edge):
vis = [False] * N
pathV = [False] * N
ans = [-1]
for i in range(N):
if not vis[i]:
self.dfs(Edge, i, vis, pathV, ans)
return ans[0] | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR WHILE VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR RETURN VAR NUMBER |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**6)
class Solution:
def largestSumCycle(self, N, Edge):
visited = [0] * N
sc = -1
for i in range(N):
sc = max(sc, self.nodes(Edge, visited, i))
return sc
def nodes(self, Edge, visited, i):
if visited[i] == 2:
return -1
elif visited[i] == 1:
sc = i
current = i
while Edge[current] != i:
current = Edge[current]
sc = sc + current
return sc
elif Edge[i] != -1:
visited[i] = 1
sc = self.nodes(Edge, visited, Edge[i])
visited[i] = 2
return sc
else:
visited[i] = 2
return -1 | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR FUNC_DEF IF VAR VAR NUMBER RETURN NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR RETURN VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR ASSIGN VAR VAR NUMBER RETURN NUMBER |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**6)
class Solution:
def dfs(self, graph, visited, u):
if visited[u] == 2:
return -1
elif visited[u] == 1:
res, cur = u, u
while graph[cur] != u:
cur = graph[cur]
res += cur
return res
elif graph[u] != -1:
visited[u] = 1
res = self.dfs(graph, visited, graph[u])
visited[u] = 2
return res
else:
visited[u] = 2
return -1
def largestSumCycle(self, N, Edge):
visited = [0] * N
res = -1
for u in range(N):
res = max(res, self.dfs(Edge, visited, u))
return res
if __name__ == "__main__":
for _ in range(int(input())):
N = int(input())
Edge = [int(i) for i in input().split()]
print(Solution().largestSumCycle(N, Edge)) | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF IF VAR VAR NUMBER RETURN NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR VAR VAR WHILE VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR RETURN VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR ASSIGN VAR VAR NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR IF VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**8)
class Solution:
def dfs(self, graph, s, cost, localV, globalV):
if s in globalV:
return -1
if s in localV:
return cost - localV[s]
localV[s] = cost
res = -1
if graph[s] != -1:
d = graph[s]
res = max(res, self.dfs(graph, d, cost + d, localV, globalV))
return res
def largestSumCycle(self, N, Edge):
res = -1
globalV = set()
for s in range(N):
localV = {}
cost = self.dfs(Edge, s, s, localV, globalV)
res = max(res, cost)
globalV.update(set(localV))
return res | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF IF VAR VAR RETURN NUMBER IF VAR VAR RETURN BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**6)
class Solution:
def largestSumCycle(self, N, Edge):
PointScore = [-1] * len(Edge)
points = set([i for i in range(N)])
maxscore = [-1]
def Walker(point, score):
if point in points:
points.remove(point)
if PointScore[point] == -1:
PointScore[point] = score
newPoint = Edge[point]
if newPoint != -1:
Walker(newPoint, score + newPoint)
PointScore[point] = -2
elif PointScore[point] != -2:
maxscore[0] = max(maxscore[0], score - PointScore[point])
while len(points):
Walker(points.pop(), 0)
return maxscore[0] | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER FUNC_DEF IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR VAR WHILE FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER RETURN VAR NUMBER |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | class Solution:
def largestSumCycle(self, N, Edge):
OFFSET = 1
sums = [-1] * N
result = -1
for i0 in range(N):
if sums[i0] >= 0:
continue
sums[i0] = OFFSET
csum = OFFSET + i0
i = Edge[i0]
i1 = i0
while i >= 0 and sums[i] < 0:
i1 = i
sums[i] = csum
csum += i
i = Edge[i]
if i >= 0 and sums[i] >= OFFSET:
result = max(result, csum - sums[i])
else:
pass
i = i0
while sums[i] > 0:
sums[i] = 0
i = Edge[i]
return result | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR WHILE VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | class Solution:
def largestSumCycle(self, N, Edge):
visited = [-1] * N
res = -1
for i in range(N):
j = i
mn = {}
sm = 0
while visited[j] == -1:
mn[j] = sm
sm += j
visited[j] = 1
j = Edge[j]
if j == -1:
break
if j in mn:
res = max(res, sm - mn[j])
break
return res | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER WHILE VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR RETURN VAR |
Given a maze with N cells. Each cell may have multiple entry points but not more than one exit(i.e entry/exit points are unidirectional doors like valves).
You are given an array Edge[] of N integers, where Edge[i] contains the cell number that can be reached from of cell i in one step. Edge[i] is -1 if the ith cell doesn't have an exit.
The task is to find the largest sum of a cycle in the maze(Sum of a cycle is the sum of the cell indexes of all cells present in that cycle).
Note:The cells are named with an integer value from 0 to N-1. If there is no cycle in the graph then return -1.
Example 1:
Input:
N = 4
Edge[] = {1, 2, 0, -1}
Output: 3
Explanation:
There is only one cycle in the graph.
(i.e 0->1->2->0)
Sum of the cell index in that cycle
= 0 + 1 + 2 = 3.
Example 2:
Input:
N = 4
Edge[] = {2, 0, -1, 2}
Output: -1
Explanation:
1 -> 0 -> 2 <- 3
There is no cycle in the graph.
Your task:
You dont need to read input or print anything. Your task is to complete the function largestSumCycle() which takes the integer N denoting the number of cells and the array Edge[] as input parameters and returns the sum of the largest sum cycle in the maze.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 ≤ N ≤ 10^{5}
-1 < Edge[i] < N
Edge[i] != i | import sys
sys.setrecursionlimit(10**6)
class Solution:
def largestSumCycle(self, N, Edge):
adjlist = [[] for i in range(N)]
for i in range(len(Edge)):
if Edge[i] != -1:
adjlist[i].append(Edge[i])
visitedarr = [(False) for i in range(N)]
ans = 0
for i in range(N):
if visitedarr[i] == False:
ancestors = set()
stack = []
tempans = self.Helper(adjlist, i, visitedarr, ancestors, stack)
ans = max(ans, tempans)
if ans:
return ans
return -1
def Helper(self, adjlist, currentnode, visitedarr, ancestors, stack):
t = 0
for x in adjlist[currentnode]:
if visitedarr[x] == True and x in ancestors:
te = currentnode
for idx in range(-1, -len(stack) - 1, -1):
if stack[idx] == x:
te += x
break
te += stack[idx]
return te
elif visitedarr[x] == False:
ancestors.add(currentnode)
visitedarr[currentnode] = True
stack.append(currentnode)
temp = self.Helper(adjlist, x, visitedarr, ancestors, stack)
t = max(t, temp)
stack.pop()
ancestors.remove(currentnode)
return t | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR RETURN VAR RETURN NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR RETURN VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
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