Skip to content

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 <= 105
  • 0 <= cards[i] <= 106

Solutions

Solution 1: Hash Table

We initialize the answer as \(+\infty\). We traverse the array, and for each number \(x\), if \(\textit{last}[x]\) exists, it means \(x\) has a matching pair of cards. In this case, we update the answer to \(\textit{ans} = \min(\textit{ans}, i - \textit{last}[x] + 1)\). Finally, if the answer is \(+\infty\), we return \(-1\); otherwise, we return the answer.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the array.

1
2
3
4
5
6
7
8
9
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
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;
}

Comments