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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
#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
Updated on: 05-May-2020
306 Views
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