count number od teams

Author: ishan.gupta

count-number-of-teams.cpp

@@ -0,0 +1,69 @@
+using namespace std;
+#include <iostream>
+#include <bits/stdc++.h>
+
+// Approach used - Thought process is that since n is 10^3 means it should be solve in 
+// N^2 or NlogN or N^2logN complexity. So for this I have used time complexity of N^2
+// and it is still better than lot of other submissions.
+// What I did is I looped on all the values and while looping, at certain index 'i' I thought
+// of making two arrays - one on the right side of index i and other on left side of index i.
+// so eventually array will be splitted like - 
+// [12,5,3,9,7,1,6,13,10,17]
+//           ^               -> suppose I am at value 7. 
+// [12,5,3,9]   7   [1,6,13,10,17]
+//  left arr  indx    right arr
+
+// do a second loop on left and right array and make a count of all the values greater and smaller 
+// than rating[i] i.e 7 in this case.
+
+// [   12    , 5,3,    9       ]
+//    llow=0        lhigh=i-1
+
+
+// rating[i] - 7
+
+
+//[    1    ,  6,13,10,    17    ]
+//  rlow=i+1            rhigh=n-1   n is size of arr.
+
+// cnt+=(llessThan*rgreaterThan)+(lgreaterThan*rlessThan); will give the count of all the pairs
+class Solution {
+public:
+    int numTeams(vector<int>& rating) {
+        int cnt=0;
+        for( int i=0;i<rating.size();i++){
+            int val = rating[i];
+            int llow = 0;
+            int lhigh = i-1;
+            int rlow = i+1;
+            int rhigh = rating.size()-1;
+            int llessThan=0;
+            int lgreaterThan=0;
+            int rlessThan=0;
+            int rgreaterThan=0;
+            if(lhigh>=0){
+                for(int j=llow;j<=lhigh;j++){
+                    if(rating[j]<val){
+                        llessThan++;
+                    }
+                    if(rating[j]>val){
+                        lgreaterThan++;
+                    }
+                }
+            }
+            if(rlow<=rating.size()-1) {          
+                for(int k=rlow;k<=rhigh;k++){
+                    if(rating[k]<val){
+                        rlessThan++;
+                    }
+                    if(rating[k]>val){
+                        rgreaterThan++;
+                    }
+                }
+            }
+
+            cnt+=(llessThan*rgreaterThan)+(lgreaterThan*rlessThan);
+        }
+        return cnt;
+    }
+};