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Formatted question description: https://leetcode.ca/all/2260.html

2260. Minimum Consecutive Cards to Pick Up

  • Difficulty: Medium.
  • Related Topics: Array, Hash Table, Sliding Window.
  • Similar Questions: Longest Substring Without Repeating Characters.

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public int minimumCardPickup(int[] cards) {
        int mindiff = Integer.MAX_VALUE;
        Map<Integer, Integer> map = new HashMap<>();
        int n = cards.length;
        for (int i = 0; i < n; i++) {
            if (map.containsKey(cards[i])) {
                int j = map.get(cards[i]);
                mindiff = Math.min(mindiff, i - j + 1);
            }
            map.put(cards[i], i);
        }
        if (mindiff == Integer.MAX_VALUE) {
            return -1;
        }
        return mindiff;
    }
}

############

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:
    def minimumCardPickup(self, cards: List[int]) -> int:
        m = {}
        ans = 10**6
        for i, c in enumerate(cards):
            if c in m:
                ans = min(ans, i - m[c] + 1)
            m[c] = i
        return -1 if ans == 10**6 else ans

############

# 2260. Minimum Consecutive Cards to Pick Up
# https://leetcode.com/problems/minimum-consecutive-cards-to-pick-up/

class Solution:
    def minimumCardPickup(self, cards: List[int]) -> int:
        res = float('inf')
        last = {}
        
        for j, x in enumerate(cards):
            if x in last:
                res = min(res, j - last[x] + 1)
                
            last[x] = j
        
        
        return -1 if res == float('inf') else res


  • 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;
    }
};

  • 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;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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