生產領料、退料頻繁

Problem statement:

問題陳述：

Given an array of integers, find and print the maximum number of integers you can select from the array such that the absolute difference between any two of the chosen integers is less than or equal to 1.

給定一個整數數組，查找并打印可以從數組中選擇的最大整數數，以使任意兩個選定整數之間的絕對差小于或等于1。

Examples:

例子：

    Input array:
6, 4, 6, 8, 5, 9, 9, 9, 10, 3
Output: 
4

Example with explanation:

帶有說明的示例：

The input array is: 6, 4, 6, 8, 5, 9, 9, 9, 10, 3

輸入數組為：6，4，6，8，5，9，9，9，10，3

No we can pick up different sets for which the absolute difference between any two numbers in the set is 1.

不，我們不能選擇集合中任意兩個數字的絕對差為1的不同集合。

Possible sets are:

可能的設置有：

    {6, 6, 5}
{4, 5}
{8, 9, 9, 9}
{9, 9, 9, 10}
{4, 3}

Thus maximum no of picked up elements is: 4

因此，拾取元素的最大數量為： 4

Algorithm:

算法：

This problem can be implemented with help of map data structure. We have used STL for map implementation. (For details regarding STL map, C++ STL Map)

可以借助地圖數據結構來實現此問題。 我們已經使用STL來實現地圖。 (有關STL映射， C ++ STL映射的詳細信息)

FUNCTION pickingNumbers(input array)
1.  Declare   map<int,int>m to store key with their frequencies;
2.  Build the map.
For  i=0:length of array
m[array[i]]++;
3.  Declare max as INT_MIN;
4.  Declare map<int,int>::iteratorit;
5.  For(it=m.begin();it!=m.end();it++)
IF (it+1==m.end()) //last case
IF(it->second>max)
max=it->second;
END IF
ELSE IF(it->first+1==(it+1)->first){ //absolute difference is 1
IF((it->second +(it+1)->second)>max)
max=it->second +(it+1)->second;
END IF
ELSE
IF(it->second>max) //absolute difference 0 case
max=it->second;
END IF
END IF-ELSE
END FOR
6.  return max;
END FUNCTION

Algorithm is pretty simple.

算法非常簡單。

We first extract the unique numbers and store their frequencies. Then we simply check for two unique number's additive frequency or any one unique number's frequency itself and return the greater one.

我們首先提取唯一數字并存儲其頻率。 然后，我們只需檢查兩個唯一數字的加法頻率或任何一個唯一數字的頻率本身，然后返回較大的一個。

Let's solve the above example.

讓我們解決以上示例。

    The input array is: 6, 4, 6, 8, 5, 9, 9, 9, 10, 3
Map m:
Key 			Frequency
3				1
4				1
5				1
6				2
8				1
9				3
10				1
So if we do all the iterations then each iteration, 
maxgets to be updated(or not, keeps last value)
From this map, we can see max is 4 
1+3 //one 8 and three 9
3+1 //three 9 and one 10
Now lets think that we append six 12 to the array
Thus input is now: 6, 4, 6, 8, 5, 9, 9, 9, 10, 3, 12, 12, 12, 12, 12, 12
Map m:
Key 			Frequency
3				1
4				1
5				1
6				2
8				1
9				3
10				1
12				6
Now the max will be 6 //absolute difference 0 case
Since the subset will be {12, 12, 12, 12, 12, 12}

C++ implementation:

C ++實現：

#include <bits/stdc++.h>
using namespace std;
int pickingNumbers(vector<int> a) 
{
map<int,int> m;
for(int i=0;i<a.size();i++)
m[a[i]]++;
int max=INT_MIN;
map<int,int>::iterator it;
for(it=m.begin();it!=m.end();it++){
//std::next(it) points to it+1
if((std::next(it))==m.end()){
if(it->second>max)
max=it->second;
}
else if(it->first+1==(std::next(it))->first){
if((it->second +(std::next(it))->second)>max)
max=it->second +(std::next(it))->second;
}
else{
if(it->second>max)
max=it->second;
}
}
return max;
}
int main(){
int n,item;
cout<<"Enter number of elements in the array\n";
cin>>n;
vector<int> a;
cout<<"enter numbers\n";
for(int i=0;i<n;i++){
cin>>item;
a.push_back(item);
}
cout<<"Maximum no of such numbers can be picked: "<<pickingNumbers(a)<<endl;
return 0;
}

Output

輸出量

First run:
Enter number of elements in the array
10
enter numbers
6 4 6 8 5 9 9 9 10 3
Maximum no of such numbers can be picked: 4
Second run:
Enter number of elements in the array
16
enter numbers
6 4 6 8 5 9 9 9 10 3 12 12 12 12 12 12
Maximum no of such numbers can be picked: 6

翻譯自: https://www.includehelp.com/icp/picking-numbers.aspx

生產領料、退料頻繁