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# Minimum Consecutive Cards to Pick Up LeetCode Solution

## Problem – Minimum Consecutive Cards to Pick Up LeetCode Solution

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`

### Minimum Consecutive Cards to Pick Up LeetCode Solution in C++

``````int minimumCardPickup(vector<int>& cards) {
int last[1000001] = {}, res = INT_MAX;
for (int i = 0; i < cards.size(); ++i) {
if (last[cards[i]])
res = min(res, i - last[cards[i]] + 2);
last[cards[i]] = i + 1;
}
return res == INT_MAX ? -1 : res;
}
``````

### Minimum Consecutive Cards to Pick Up LeetCode Solution in Java

``````class Solution
{
public int minimumCardPickup(int[] cards)
{
Map<Integer,Integer> map = new HashMap<>();
int min = Integer.MAX_VALUE;
for(int i = 0; i < cards.length; i++)
{
if(map.containsKey(cards[i]))
min = Math.min(i-map.get(cards[i])+1,min); // Check if the difference in indices is smaller than minimum
map.put(cards[i],i); // Update the last found index of the card
}
return min == Integer.MAX_VALUE?-1:min; // Repetition found or not
}
}
``````

### Minimum Consecutive Cards to Pick Up LeetCode Solution in Python

``````class Solution:
def minimumCardPickup(self, cards: List[int]) -> int:
minPick = float('inf')
seen = {}
for i, n in enumerate(cards):
if n in seen:
if i - seen[n] + 1 < minPick:
minPick = i - seen[n] + 1
seen[n] = i
if minPick == float('inf'):
return -1
return minPick
``````
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