Solve daily problem with easy code

Author: SartHak-0-Sach

211-29th_July-Count_number_of_teams/approach-explanation.txt

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+Intuition
+Since we can use the same soldier for multiple teams, we just need to find number of ratings less than index i , mumber of ratings greater than index i which are on the left side of the index i and similarly umber of ratings less than index i , mumber of ratings greater than index i which are on the right side of the index i, multiply and add it to the answer.
+
+Approach
+Iterate over the array.
+For every index i calculate the following:
+lowerThanLeft
+higherThanLeft
+lowerThanRight
+higherThanRight
+Multiply lowerThanLeft and higherThanRight, similarly multiply higherThanLeft and lowerThanRight and add it to answer.
+Complexity
+Time complexity:
+O(n^2)
+Space complexity:
+O(1)

211-29th_July-Count_number_of_teams/code.cpp

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+class Solution
+{
+public:
+    int numTeams(vector<int> &rating)
+    {
+        const int m = rating.size();
+        int ans = 0;
+        for (int i = 1; i < m - 1; ++i)
+        {
+            int lowerThanLeft = 0;
+            int higherThanLeft = 0;
+            int lowerThanRight = 0;
+            int higherThanRight = 0;
+            for (int j = 0; j < i; ++j)
+            {
+                if (rating[j] < rating[i])
+                {
+                    ++lowerThanLeft;
+                }
+                else if (rating[j] > rating[i])
+                {
+                    ++higherThanLeft;
+                }
+            }
+            for (int j = i + 1; j < m; ++j)
+            {
+                if (rating[j] < rating[i])
+                {
+                    ++lowerThanRight;
+                }
+                else if (rating[j] > rating[i])
+                {
+                    ++higherThanRight;
+                }
+            }
+            ans += ((lowerThanLeft * higherThanRight) + (higherThanLeft * lowerThanRight));
+        }
+        return ans;
+    }
+};