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Original file line number Diff line number Diff line change
@@ -0,0 +1,45 @@
'''
making a backtracking based sol which uses all the permutations to fit a number at a pos using backtracking.
link:-https://leetcode.com/problems/construct-the-lexicographically-largest-valid-sequence/?envType=daily-question&envId=2025-02-16
'''
class Solution:
def constructDistancedSequence(self, n: int) -> List[int]:
ans = [0] * (2 * n - 1)
use = set()

def back(i):
if i == len(ans):
return True
if ans[i]: # Skip filled positions
return back(i + 1)

for num in range(n, 0, -1):
if num in use:
continue
if num > 1 and (i + num >= len(ans) or ans[i + num]):
continue

# Place the number
use.add(num)
ans[i] = num
if num > 1:
ans[i + num] = num

# Recurse to next available position
if back(i + 1):
return True

# Backtrack
use.remove(num)
ans[i] = 0
if num > 1:
ans[i + num] = 0

return False

back(0)
return ans
if __name__ == "__main__":
s= Solution()
num = int(input())
print(s.constructDistancedSequence(num))
27 changes: 27 additions & 0 deletions Python/furthest-building-you-can-reach.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,27 @@
'''
the condition was a person can use ladder or brick to climb a building but the sucessive building's height can be greater than or less than current buiiding.
no. of bricks and ladder given,ladder length can be anything i.e if next buliding in 2 units greater than and we have no bricks so we can use ladder.
so i used maxheap to know where should i use ladder so that it will not waste my resource and we can reach the farthest building.
'''
# import heapq
class Solution:
def furthestBuilding(self, heights: list[int], bricks: int, ladders: int) -> int:
heap = []
for i in range(len(heights)-1):
diff = heights[i+1]-heights[i]
print(diff)
if diff<0:
continue
print("hello",diff)
bricks-=diff
print("b",bricks)
heapq.heappush(heap,diff)
print(heap)
if bricks<0:
if ladders==0:
return i
ladders-=1
print("bricks",bricks,"ladder",ladders)
bricks+=heapq.heappop(heap)

return len(heights)-1