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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).