Leetcode 2260: Minimum Consecutive Cards to Pick Up

Input: The input consists of a list `cards` containing integers where each integer represents the value of a card.

Example: cards = [7, 2, 3, 7, 5, 6]

Constraints:

• 1 <= cards.length <= 105

• 0 <= cards[i] <= 106

Output: Return the minimum number of consecutive cards you need to pick up to have at least one matching pair of cards. If no pair exists, return -1.

Example: Output: 4

Constraints:

Goal: The goal is to find the smallest subarray of consecutive cards that contains at least one pair of matching cards.

Steps:

• Use a hashmap to keep track of the last seen index of each card.

• As you iterate over the cards, check if the current card has been seen before.

• If a match is found, calculate the number of consecutive cards between the current and previous occurrence of the card.

• Keep track of the smallest such subarray length.

Goal: Make sure the solution works within the constraints of large input sizes.

Steps:

• The length of the `cards` array is between 1 and 100,000.

• The card values are between 0 and 1,000,000.

Assumptions:

• The array will contain at least one pair of matching cards, or none at all.

• Input: cards = [7, 2, 3, 7, 5, 6]

• Explanation: We can pick the cards [7, 2, 3, 7], which contains a matching pair of 7's. Therefore, the minimum number of consecutive cards is 4.

• Input: cards = [1, 2, 3, 4]

• Explanation: There are no matching cards, so it is impossible to have a pair of matching cards. The output is -1.

Approach: We will use a hashmap to store the last index of each card and check for matching pairs while iterating through the list.

Observations:

• We need to find matching cards and track the smallest number of consecutive cards that contain such a pair.

• Using a hashmap will allow us to check for duplicates efficiently as we iterate through the array.

Steps:

• Initialize a hashmap to store the last index of each card.

• Iterate through the list of cards.

• For each card, check if it has appeared before in the hashmap.

• If it has, calculate the distance between the current index and the previous index of the matching card.

• Keep track of the smallest such distance and return it once the iteration is complete.

Empty Inputs:

• The input will never be empty, as the length of `cards` is guaranteed to be at least 1.

Large Inputs:

• The solution should handle arrays with lengths up to 100,000 efficiently.

Special Values:

• Ensure that if no matching pairs are found, the function correctly returns -1.

Constraints:

• The card values are within the range 0 to 1,000,000, so the solution should handle large numbers effectively.

int minimumCardPickup(vector<int>& cards) {
    map<int, int> mp;
    int n = cards.size();
    int ans = n + 1;
    
    for(int i = 0; i < n; i++){
        if(mp.count(cards[i])) {
            ans = min(ans, i - mp[cards[i]] + 1);
        }
        mp[cards[i]] = i;            
    }
    
    return ans == n + 1? -1: ans;
}

1 : Function Declaration

int minimumCardPickup(vector<int>& cards) {

The function `minimumCardPickup` takes a vector of integers, `cards`, and returns an integer representing the minimum number of cards to pick up to find a duplicate. If no duplicate is found, it returns -1.

2 : Map Initialization

    map<int, int> mp;

A map `mp` is initialized to store the last index at which each card appears in the vector `cards`.

3 : Size Calculation

    int n = cards.size();

The variable `n` is set to the size of the `cards` vector, representing the total number of cards.

4 : Answer Initialization

    int ans = n + 1;

The variable `ans` is initialized to `n + 1`, which is used to store the minimum number of cards between duplicates. This is initially set to a number larger than any possible valid result.

5 : Loop Start

    for(int i = 0; i < n; i++){

A loop is started to iterate over each card in the `cards` vector by index `i`.

6 : Check for Duplicates

        if(mp.count(cards[i])) {

This checks if the current card has appeared before by checking if it exists in the map `mp`.

7 : Update Answer

            ans = min(ans, i - mp[cards[i]] + 1);

If the card has appeared before, the minimum number of cards between this and the previous occurrence is updated. The `min` function ensures that the smallest value is stored.

8 : Update Map

        mp[cards[i]] = i;

The current card and its index are added to the map `mp`, updating the last occurrence of that card.

9 : Return Result

    return ans == n + 1? -1: ans;

After the loop finishes, the function checks if `ans` was updated. If no duplicate was found, `ans` will still be `n + 1`, and the function returns -1. Otherwise, it returns the minimum number of cards between duplicates.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is linear since we only make one pass over the input array, and hashmap operations (insert and lookup) are O(1) on average.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) due to the use of a hashmap to store the indices of the cards.

Leetcode 2260: Minimum Consecutive Cards to Pick Up

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