#3664
Two-Letter Card Game
MediumArrayHash TableStringCountingEnumerationHash MapArray
Approaches
Brute ForceOptimal
Complexity Comparison
| Brute Force | Optimal Solution★ | |
|---|---|---|
| Time | O(n²) | O(n) |
| Space | O(1) | O(n) |
💡
Intuition
Time O(n)Space O(n)
Count cards with letter x in both positions and only one position. Use these counts to form pairs efficiently.
⚙️
Algorithm
3 steps- 1Step 1: Count cards with x in both positions, and those with x in only the first or second position.
- 2Step 2: Form pairs using the counts, maximizing the number of pairs formed.
- 3Step 3: Return the total number of pairs.
solution.py15 lines
1def maxPoints(cards, x):
2 cnt1 = defaultdict(int)
3 cnt2 = defaultdict(int)
4 both = 0
5 for card in cards:
6 if card[0] == x and card[1] == x:
7 both += 1
8 elif card[0] == x:
9 cnt1[card[1]] += 1
10 elif card[1] == x:
11 cnt2[card[0]] += 1
12 points = both
13 for c in cnt1:
14 points += min(cnt1[c], cnt2[c])
15 return pointsℹ
Complexity note: Single pass to count cards, leading to linear time complexity.
- 1Pairs must differ by one character.
- 2Count cards strategically to maximize pairs.
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