Merge pull request #244 from capricemoto/d2q1

soln of q1 of d2;

Author: Dheeraj-06

Problems/Mathematics/Day-02/sol/Samarth/sol1.cpp

@@ -0,0 +1,59 @@
+#include <bits/stdc++.h>
+#define endl '\n'
+typedef long long ll;
+
+using namespace std;
+
+/*
+PROBLEM STATEMENT:
+Hossam woke up bored, so he decided to create an interesting array with his friend Hazem.
+
+Now, they have an array a
+of n positive integers, Hossam will choose a number ai and Hazem will choose a number aj
+
+.
+
+Count the number of interesting pairs (ai,aj)
+
+that meet all the following conditions:
+
+1≤i,j≤n
+i≠j
+The absolute difference |ai−aj| must be equal to the maximum absolute difference over all the pairs in the array. More formally, |ai−aj|=max1≤p,q≤n|ap−aq|.
+
+APPROACH:
+just count the freq of max &min elements and by basic combinotorics we get ans 2*fmax*fmin
+but if all elements are equal i.e fmax ==fmin then ans is n*(n-1) by basic combinototics too.
+
+Time Complexity: O(n)
+Space complexity:O(n)
+SUBMISSION LINK:https://codeforces.com/problemset/submission/1771/355306251;
+*/
+
+
+
+
+void solve(){
+	int n;cin>>n;int a[n];int max=INT_MIN,min=INT_MAX,fmax=0,fmin=0;
+	for(int i=0;i<n;i++){
+		cin>>a[i];
+		if(a[i]>max){
+			max=a[i];fmax=0;}
+		if(a[i]<min){
+			min=a[i];fmin=0;}
+		if(a[i]==max)fmax++;
+		if(a[i]==min)fmin++;
+	}
+	ll ans;
+	if(max==min)ans=((long long)(n)*(long long)(n-1));
+	else ans=2*(ll)fmax*(ll)fmin;
+	cout<<ans<<endl;
+}
+
+int main(){
+    ios::sync_with_stdio(0); cin.tie(0);
+	int t;cin>>t;
+	while(t--)
+		solve();
+	
+}