#plea # Count the number of unique paths from a[0][0] to a[m-1][n-1] # We are allowed to move either right or down from a cell in the matrix. # Approaches- # (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards, # add the path count of both recursions and return count. # (ii) Dynamic Programming- Start from a[0][0].Store the count in a count # matrix. Return count[m-1][n-1] # T(n)- O(mn), S(n)- O(mn) # def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1] print(count[m - 1][n - 1]) def main(): m, n = map(int, input('Enter two positive integers: ').split()) count_paths(m, n). .
#plea
# Count the number of unique paths from a[0][0] to a[m-1][n-1]
# We are allowed to move either right or down from a cell in the matrix.
# Approaches-
# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,
# add the path count of both recursions and return count.
# (ii) Dynamic
# matrix. Return count[m-1][n-1]
# T(n)- O(mn), S(n)- O(mn)
#
def count_paths(m, n):
if m < 1 or n < 1:
return -1
count = [[None for j in range(n)] for i in range(m)]
# Taking care of the edge cases- matrix of size 1xn or mx1
for i in range(n):
count[0][i] = 1
for j in range(m):
count[j][0] = 1
for i in range(1, m):
for j in range(1, n):
# Number of ways to reach a[i][j] = number of ways to reach
# a[i-1][j] + a[i][j-1]
count[i][j] = count[i - 1][j] + count[i][j - 1]
print(count[m - 1][n - 1])
def main():
m, n = map(int, input('Enter two positive integers: ').split())
count_paths(m, n). .
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