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:

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

Example 2:

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

Constraints:

• 1 <= hand.length <= 104

• 0 <= hand[i] <= 109

• 1 <= groupSize <= hand.length

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:

C++

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;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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