Added solution to math/day2/q001

Author: verifiedHuman18

Problems/Mathematics/Day-02/sol/Ayush_Saha/Solution1.class

@@ -0,0 +1,61 @@
+// Link to submission: https://codeforces.com/contest/1771/submission/355440128
+
+/*
+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| = max|ap − aq|, 1 ≤ p, q≤ n.
+*/
+
+/*
+Brief Explanation:
+We find the minimum and the maximum element in the entire array and count the number of ordered pairs of min and max in the array.
+We count the frequency of miminum and maximum elements and use the formula of combinatorics to compute number of pairs.
+if min != max, we have count = freqMin C 1 * freqMax C 1 * 2
+if min == max, we have count = freqMax P 2
+*/
+
+import java.util.*;
+
+public class Solution1 {
+    public static void main(String[] args) {
+        Scanner sc = new Scanner (System.in);
+        int t = sc.nextInt();
+        while (t-- > 0) {
+            long min = 100000;
+            long max = 0;
+            long freqMin = 0;
+            long freqMax = 0;
+            int n = sc.nextInt();
+            long[] arr = new long[n];
+            for (int i = 0; i < n; i++) {
+                arr[i] = sc.nextLong();
+                if (arr[i] < min)
+                    min = arr[i];
+                if (arr[i] > max)
+                    max = arr[i];
+            }
+            for (int i = 0; i < n; i++) {
+                if (arr[i] == min)
+                    freqMin++;
+                if (arr[i] == max)
+                    freqMax++;
+            }
+
+            long count = 0;
+            if (min == max)
+                count = freqMax * (freqMax - 1);
+            else
+                count = freqMin * freqMax * 2;
+            System.out.println(count);
+        }
+        sc.close();
+    }
+}
+
+// Time Complexity: O(n)
+// Space Complexity: O(n)