Welcome to Subscribe On Youtube
2260. Minimum Consecutive Cards to Pick Up
Description
You are given an integer array cards where cards[i] represents the value of the ith card. A pair of cards are matching if the cards have the same value.
Return the minimum number of consecutive cards you have to pick up to have a pair of matching cards among the picked cards. If it is impossible to have matching cards, return -1.
Example 1:
Input: cards = [3,4,2,3,4,7] Output: 4 Explanation: We can pick up the cards [3,4,2,3] which contain a matching pair of cards with value 3. Note that picking up the cards [4,2,3,4] is also optimal.
Example 2:
Input: cards = [1,0,5,3] Output: -1 Explanation: There is no way to pick up a set of consecutive cards that contain a pair of matching cards.
Constraints:
1 <= cards.length <= 1050 <= cards[i] <= 106
Solutions
-
class Solution { public int minimumCardPickup(int[] cards) { Map<Integer, Integer> last = new HashMap<>(); int n = cards.length; int ans = n + 1; for (int i = 0; i < n; ++i) { if (last.containsKey(cards[i])) { ans = Math.min(ans, i - last.get(cards[i]) + 1); } last.put(cards[i], i); } return ans > n ? -1 : ans; } } -
class Solution { public: int minimumCardPickup(vector<int>& cards) { unordered_map<int, int> last; int n = cards.size(); int ans = n + 1; for (int i = 0; i < n; ++i) { if (last.count(cards[i])) { ans = min(ans, i - last[cards[i]] + 1); } last[cards[i]] = i; } return ans > n ? -1 : ans; } }; -
class Solution: def minimumCardPickup(self, cards: List[int]) -> int: last = {} ans = inf for i, x in enumerate(cards): if x in last: ans = min(ans, i - last[x] + 1) last[x] = i return -1 if ans == inf else ans -
func minimumCardPickup(cards []int) int { last := map[int]int{} n := len(cards) ans := n + 1 for i, x := range cards { if j, ok := last[x]; ok && ans > i-j+1 { ans = i - j + 1 } last[x] = i } if ans > n { return -1 } return ans } -
function minimumCardPickup(cards: number[]): number { const n = cards.length; const last = new Map<number, number>(); let ans = n + 1; for (let i = 0; i < n; ++i) { if (last.has(cards[i])) { ans = Math.min(ans, i - last.get(cards[i]) + 1); } last.set(cards[i], i); } return ans > n ? -1 : ans; }