GfG Weekly - 175 [Rated Contest] All Solutions

[@ArnabBhadra02]

I was solving the qs in cpp and converted into python due to plag

[@ArnabBhadra02]

Soln 2:
from typing import List
class Solution:
def boundarySum(self, n : int, matrix : List[List[int]]) -> List[int]:
# code here
sums = []
for level in range((n + 1) // 2):
level_sum = 0
for i in range(level, n - level):
level_sum += matrix[level][i]
for i in range(level + 1, n - level):
level_sum += matrix[i][n - level - 1]
if n - level - 1 > level:
for i in range(n - level - 2, level - 1, -1):
level_sum += matrix[n - level - 1][i]
if n - level - 1 > level:
for i in range(n - level - 2, level, -1):
level_sum += matrix[i][level]
sums.append(level_sum)
return sums

[@hhhhhhhhhhhh0]

Brother 3rd and 4th que ke solutions bhi sed kr do

[@ArnabBhadra02]

3rd kuch tc fail ho raha he
Dusra compiler pe try karna hoga bahut wrong submission ho gya he
Thodi wait karlo trying to figure it out

[@saurabhmaurya2805]

@@ArnabBhadra02 hint - make each index i as a boundary index its mean that you can pick atmost x drinks of type A from left of index i and atmost y drinks of type B from right of index i (inclusive). and try to find maximum health.

[@ArnabBhadra02]

@@hhhhhhhhhhhh0 4th Done
Check code and subscribe

[@ArnabBhadra02]

Soln 1:
from typing import List
class Solution:
def maximumProduct(self, n : int, arr : List[int], l : int, r : int) -> int:
mx = float('-inf')
mn = float('inf')
for i in range(l-1, r):
mx = max(mx, arr[i])
mn = min(mn, arr[i])
mx2 = float('-inf')
mn2 = float('inf')
for i in range(0, l-1):
mx2 = max(mx2, arr[i])
mn2 = min(mn2, arr[i])
for i in range(r, n):
mx2 = max(mx2, arr[i])
mn2 = min(mn2, arr[i])
product1 = mx * mx2
product2 = mn * mn2
product3 = mx * mn2
product4 = mn * mx2
return max(product1, product2, product3, product4)

[@nishudwivedi3478]

Sol 3 and sol 4 please

[@saurabhmaurya2805]

from typing import List

from collections import defaultdict
class Solution:
def longestIncreasingPath(self, n: int, edges: List[List[int]], values: List[int]) -> int:
# Create the adjacency list for the tree
tree = defaultdict(list)
for u, v in edges:
tree[u].append(v)
tree[v].append(u)
# Memoization table to store results of longest path starting from each node
memo = {}
# Depth-First Search to find the longest increasing path from node u
def dfs(u, parent):
if u in memo:
return memo[u]
max_path = 1 # The minimum path length is 1 (the node itself)
# Explore neighbors
for v in tree[u]:
if v != parent and values[v-1] > values[u-1]:
max_path = max(max_path, 1 + dfs(v, u))
memo[u] = max_path
return max_path
# Try starting DFS from each node and track the maximum increasing path length
longest_path = 0
for i in range(1, n + 1):
longest_path = max(longest_path, dfs(i, -1))
return longest_path

[@ArnabBhadra02]

Uploaded the 4th onr

[@saurabhmaurya2805]

3 wala de do