Minimum Consecutive Cards to Pick Up

// Accepted solution for LeetCode #2260: Minimum Consecutive Cards to Pick Up
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;
    }
}

// Accepted solution for LeetCode #2260: Minimum Consecutive Cards to Pick Up
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
}

# Accepted solution for LeetCode #2260: Minimum Consecutive Cards to Pick Up
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

// Accepted solution for LeetCode #2260: Minimum Consecutive Cards to Pick Up
/**
 * [2260] Minimum Consecutive Cards to Pick Up
 *
 * You are given an integer array cards where cards[i] represents the value of the i^th 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 <= 10^5
 * 	0 <= cards[i] <= 10^6
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-consecutive-cards-to-pick-up/
// discuss: https://leetcode.com/problems/minimum-consecutive-cards-to-pick-up/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimum_card_pickup(cards: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2260_example_1() {
        let cards = vec![3, 4, 2, 3, 4, 7];

        let result = 4;

        assert_eq!(Solution::minimum_card_pickup(cards), result);
    }

    #[test]
    #[ignore]
    fn test_2260_example_2() {
        let cards = vec![1, 0, 5, 3];

        let result = -1;

        assert_eq!(Solution::minimum_card_pickup(cards), result);
    }
}

// Accepted solution for LeetCode #2260: Minimum Consecutive Cards to Pick Up
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;
}