# 846. Hand of Straights

## Problem Statement

\
Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size `groupSize`, and consists of `groupSize` consecutive cards.

Given an integer array `hand` where `hand[i]` is the value written on the `ith` card and an integer `groupSize`, return `true` if she can rearrange the cards, or `false` otherwise.

 

**Example 1:**

<pre><code><strong>Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3
</strong><strong>Output: true
</strong><strong>Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: hand = [1,2,3,4,5], groupSize = 4
</strong><strong>Output: false
</strong><strong>Explanation: Alice's hand can not be rearranged into groups of 4.
</strong>
</code></pre>

&#x20;

**Constraints:**

* `1 <= hand.length <= 104`
* `0 <= hand[i] <= 109`
* `1 <= groupSize <= hand.length`

&#x20;

**Note:** This question is the same as 1296: <https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/>

## Intuition

```
Apprach:
We keep track of minimum element using minheap
And then from that minimum element try to find
Using loop that if val+k elements exist

Now, if freq of element goes to zero, we pop in heap


But if 
1 -> 1
2-> 0 

Now, 1 has freq, but we will need 2, so we return false
```

### Links

<https://leetcode.com/problems/hand-of-straights/description/>

### Video Links

<https://www.youtube.com/watch?v=amnrMCVd2YI&t=8s&ab_channel=NeetCode>

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    bool isNStraightHand(vector<int>& hand, int size) {
        if(hand.size() % size != 0)
            return false;
        
        unordered_map<int,int> mp;
        priority_queue<int, vector<int>, greater<int>> pq;
        for(auto &it: hand)
            mp[it]++;

        for(auto &it: mp)
            pq.push(it.first);

        while(!pq.empty()){
            int val = pq.top();
            for(int i=0; i<size; i++){
                if(mp.find(val+i) == mp.end())
                    return false;

                mp[val+i]--;
                if(mp[val+i] == 0){
                    if(pq.top() != i+val)
                        return false;

                    pq.pop();
                }
            }
        }

        return true;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###


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