Runtime: 157 ms (Top 72.81%) | Memory: 44.7 MB (Top 76.75%)

Author: AnasImloul

scripts/algorithms/M/Maximum Equal Frequency/Maximum Equal Frequency.cpp

@@ -1,29 +1,29 @@
+// Runtime: 157 ms (Top 72.81%) | Memory: 44.7 MB (Top 76.75%)
+
 class Solution {
 public:
     int maxEqualFreq(vector<int>& nums) {
         int n=nums.size();
-        unordered_map<int,int> mp1,mp2; //one to store the frequency of each number , and one which will store the frequency of the frequency 
+        unordered_map<int,int> mp1,mp2; //one to store the frequency of each number , and one which will store the frequency of the frequency
         int ans=0;
         for(int i=0;i<n;i++)
         {
             int c=nums[i];
             mp1[c]++;
             mp2[mp1[c]]++;
-            if(i!=n-1 && mp2[mp1[c]]*mp1[c]==i+1) 
-			//when curernt length is equal to the total same frequency of numbers then we need one more 
-			//number which may be anything so doesn't matter . 
-			//so we just traverse till n-2 and that's it.
-			
-			
+            if(i!=n-1 && mp2[mp1[c]]*mp1[c]==i+1)
+            //when curernt length is equal to the total same frequency of numbers then we need one more
+            //number which may be anything so doesn't matter .
+            //so we just traverse till n-2 and that's it.
+
                 ans=i+2;
             if(mp2[mp1[c]]*mp1[c]==i)
-			//when current length is equal to the total same frequency of numbers plus one extra no.
-			//( so here taken ==i (i+1-1)) which can be 1. same as any number present 2. different number ..
-			//but doesn't matter and our condition just satisfies ... so just take the current length .
-			
-			
+            //when current length is equal to the total same frequency of numbers plus one extra no.
+            //( so here taken ==i (i+1-1)) which can be 1. same as any number present 2. different number ..
+            //but doesn't matter and our condition just satisfies ... so just take the current length .
+
                 ans=i+1;
         }
         return ans;
     }
-};
+};