Added problem AVERAGE

First commit and PR

Author: siddath

Average/Average-Question.txt

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+Problem Name: Average
+
+You are given a sequence of integers a1, a2, ..., aN. An element ak is said to be an average element if there are indices i, j (with i ≠ j) such that ak = (ai + aj) / 2.
+
+In the sequence
+
+3  7  10  22  17  15
+
+for i=1, j=5 and k=3, we get ak = (ai + aj)/2. Thus a3 = 10 is an average element in this sequence. You can check that a3 is the only average element in this sequence.
+
+Consider the sequence
+
+3  7  10  3  18
+
+With i=1, j=4 and k=1 we get ak = (ai +aj)/2. Thus a1=3 is an average element. We could also choose i=1, j=4 and k=4 and get ak=(ai +aj)/2. You can check that a1 and a4 are the only average elements of this sequence.
+
+On the other hand, the sequence
+
+3  8  11  17  30
+
+has no average elements.
+
+Your task is to count the number of average elements in the given sequence.
+
+Input format
+
+The first line contains a single integer N indicating the number of elements in the sequence. This is followed by N lines containing one integer each (Line i+1 contains ai). (You may assume that ai + aj would not exceed MAXINT for any i and j).
+
+Output format
+
+The output must consist of a single line containing a single integer k indicating the number of average elements in the given sequence.
+
+Test Data:
+
+You may assume that N ≤ 10000. Further, you may assume that in 30% of the inputs N ≤ 200 and that in 60% of the inputs N ≤ 5000.
+
+Example:
+
+We illustrate the input and output format using the above examples:
+
+Sample Input 1:
+
+6
+3
+7
+10
+17
+22
+15
+
+Sample Output 1:
+
+1
+
+Sample Input 2:
+
+5
+3
+7
+10
+3
+18
+
+Sample Output 2:
+
+2
+
+Sample Input 3;
+
+5
+3
+8
+11
+17
+30
+
+Sample Output 3:
+
+0
+

Average/Solution

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+#include<iostream>
+#include<algorithm>
+int arr[100000];
+using namespace std;
+int main()
+{
+        int n;
+        cin>>n;
+        for(int h=0;h<n;h++)
+        cin>>arr[h];
+        sort(arr,arr+n);
+        int count=0;
+        for(int i=0;i<n;i++)
+        {
+                if(arr[i]==arr[i-1] || arr[i]==arr[i+1])
+                count++;
+                else
+                {
+                        int l=i-1,h=i+1;
+                        while(l>=0 && h<n)
+                        {
+                                int sum=arr[l]+arr[h];
+                                if(sum==arr[i]*2)
+                                {count++;break;}
+                                else if(sum<arr[i]*2)
+                                ++h;
+                                else
+                                --l;
+                        }
+                }
+        }
+        cout<<count<<endl;
+        return 0;
+}