[ppxyn1] WEEK 02 solutions

[ppxyn1]

답안 제출 문제

• Valid Anagram #218
• Climbing Stairs #230
• Product of Array Except Self #239
• 3Sum #241
• Validate Binary Search Tree #251

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[samcho0608]

probably better to remove print statement to reduce IO time

Suggested change

print(s_dic, t_dic)

[samcho0608]

you can simplify it a bit

Suggested change

	if n <= 2:
	return n
	dp = [0] * (n+1)
	dp[2], dp[3] = 2, 3
	#for i in range(4, n): error when n=4
	for i in range(4, n+1):
	dp[i] = dp[i-1] + dp[i-2]
	return dp[n]
	dp = [0] * (n+1)
	dp[0], dp[1] = 1, 1
	for i in range(2, n + 1):
	dp[i] = dp[i-1] + dp[i-2]
	return dp[n]

[samcho0608]

my rule of thumb is checking if current value(aka value at index) can be determined using the previous.

for instance, both #230 and #264 are both similar in that desired value at index can be determined using previous values.

• Climbing Stairs #230 :
  • f(i) = # of possible combinations to reach i
  • can take 1 or 2 steps means that f(i) can be reached using:
    • f(i-1)—taking 1 step—and
    • f(i-2)—taking 2 steps—
  • thus, f(i) = f(i-1) + f(i-2)
• House Robber #264 :
  • f(i) = max possible amount that can be robbed to reach i
  • cannot rob adjacent houses means that f(i) can be calculated using the max between:
    • f(i-1)—choosing not to rob house[i]—and
    • f(i-2)—choosing to rob house[i]
  • f(i) = max(f(i-1), f(i-2)) + house[i]

[ppxyn1]

Thank you for your detailed explanation! it was really helpful:)
I haven’t solved that many problems yet, so this might be why I have one more question: isn’t it easy to miss that a value can be determined using previous values when the step size is larger? For instance, dp[6] could be obtained from dp[1] and dp[5].

[samcho0608]

@ppxyn1

That's why you gotta focus on the current affected by previous part (kinda like Induction proof from discrete math courses)

Begin with smaller n's and see if there is a pattern.

Take #230 with max step size up to 4 as an example(6 is too big for a written explanation of patterns).

1. n=0: There's only one way (0)
2. n=1: There's only one way (0->1)
3. n=2: 2 because (0 -> 2 or 0 -> 1 -> 2)
   • f(0): 0 -> 2
   • f(1): 0 -> 1 -> 2
   • Already kinda seeing the pattern since we're seeing f(2) = f(0) + f(1)
4. n=3: 4
   • f(0): 0 -> 3
   • f(1): 0 -> 1 -> 3
   • f(2):
     • 0 -> 2 -> 3
     • 0 -> 1 -> 2 -> 3
5. n=4: 8
   • f(0): 0 -> 4
   • f(1): 0 -> 1 -> 4
   • f(2):
     • 0 -> 2 -> 4
     • 0 -> 1 -> 2 -> 4
   • f(3):
     • 0 -> 3 -> 4
     • 0 -> 1 -> 3 -> 4
     • 0 -> 2 -> 3 -> 4
     • 0 -> 1 -> 2 -> 3 -> 4

So the pattern is:
f(n) = f(n-1) + f(n-2) + .... f(n-k), where k is the step size

[ppxyn1]

Appreciate it!

[dalestudy]

현재 주차가 종료되어 자동으로 승인되었습니다. PR을 병합해주세요!