Hand of Straights in C++

Suppose Rima has a hand of cards, given as an array of integers. Now she wants to shuffle the cards into groups so that each group is size W, and consists of W consecutive cards. We have to check whether it is possible or not.

So if the cards are [1,2,3,6,2,3,4,7,8], and W = 3, then the answer will be true, as she can rearrange them like [1,2,3],[2,3,4],[6,7,8]

To solve this, we will follow these steps −

• Define a map m, and store frequency of each element in hands into m
• while size of hand is not 0
  • prev := 0
  • it := pointer to the first key-value pair in m
  • for i in range 0 to W – 1
    • while value of it is 0, it := point to next pair
    • if i > 0 and key of it – 1 = prev or i = 0, then
      • decrease it value by 1
      • prev := key of it
    • otherwise return false
    • it := point to next pair
  • n := n – W
• return true.

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool isNStraightHand(vector<int>& hand, int W) {
      map <int, int> m;
      int n = hand.size();
      if(n % W != 0) return false;
      for(int i = 0; i < n; i++){
         m[hand[i]]++;
      }
      while(n){
         map <int, int> :: iterator it = m.begin();
         int prev = 0;
         for(int i = 0; i < W; i++){
            while(it->second == 0) it++;
            if((i > 0 && it->first - 1 == prev) || i == 0){
               it->second--;
               prev = it->first;
            }else{
               return false;
            }
            it++;
         }
         n -= W;
      }
      return true;
   }
};
main(){
   vector<int> v = {1,2,3,6,2,3,4,7,8};
   Solution ob;
   cout << (ob.isNStraightHand(v, 3));
}

Input

[1,2,3,6,2,3,4,7,8]
3

Output

1