Runtime: 308 ms (Top 27.53%) | Memory: 40.6 MB (Top 11.24%)

Author: AnasImloul

scripts/algorithms/M/Maximum Split of Positive Even Integers/Maximum Split of Positive Even Integers.cpp

@@ -1,7 +1,8 @@
+// Runtime: 308 ms (Top 27.53%) | Memory: 40.6 MB (Top 11.24%)
 class Solution {
 public:
     using ll = long long;
-    
+
     ll bs(ll low , ll high, ll fs){
 
         ll ans = 1;
@@ -20,42 +21,42 @@ class Solution {
         }
         return ans;
     }
-    
+
     vector<long long> maximumEvenSplit(long long finalSum) {
         // ****some base cases / corner cases****
-		if(finalSum&1)  return {}; 
-        if(finalSum==4) return {4}; 
-        if(finalSum==8) return {2,6}; 
-		
-        vector<ll> ans; 
-        
-		// assume that we are giving indices to even numbers
-		// EVEN NUMBERS -> 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 ..............
-		// THEIR INDICES-> 1 , 2 , 3 , 4 ,  5 ,  6 ,  7 , 8  ..............
-		
-		// 'idx' is the index of that EVEN number uptil which the total sum of all even numbers <= finalSum
+        if(finalSum&1) return {};
+        if(finalSum==4) return {4};
+        if(finalSum==8) return {2,6};
+
+        vector<ll> ans;
+
+        // assume that we are giving indices to even numbers
+        // EVEN NUMBERS -> 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 ..............
+        // THEIR INDICES-> 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ..............
+
+        // 'idx' is the index of that EVEN number uptil which the total sum of all even numbers <= finalSum
         ll idx = bs(1,finalSum/2,finalSum);
-		
-		//Consequently, 'end' is that EVEN number uptil which the total sum of all even numbers <= finalSum
+
+        //Consequently, 'end' is that EVEN number uptil which the total sum of all even numbers <= finalSum
         ll start = 2, end = idx*2;
-        
-		//Now, we add all the even numbers from index 1 to index 'idx-1'
-		// 2 + 4 + 6 + 8  ........................... + (end-2) + end
-		// 1   2   3   4  ...........................    idx-1    idx
+
+        //Now, we add all the even numbers from index 1 to index 'idx-1'
+        // 2 + 4 + 6 + 8 ........................... + (end-2) + end
+        // 1 2 3 4 ........................... idx-1 idx
         for(int i = start; i<= (idx-1)*2; i+=2){
             ans.push_back(i);
         }
-		
-		// We do not add the last even number yet, so that we can modify it and add it later to make the (totalSumSoFar) == finalSum
-		// 'totalSumSoFar' can be easily calculated by using the formula ( totalSumSoFar = idx*(idx+1) )
-		
-		// increasing the last even number 'end' by the difference of (finalSum and totalSumSoFar)
+
+        // We do not add the last even number yet, so that we can modify it and add it later to make the (totalSumSoFar) == finalSum
+        // 'totalSumSoFar' can be easily calculated by using the formula ( totalSumSoFar = idx*(idx+1) )
+
+        // increasing the last even number 'end' by the difference of (finalSum and totalSumSoFar)
         if(idx*(idx+1)<finalSum){
             end = end + abs(finalSum - idx*(idx+1));
         }
-		
-		// adding the last even number after increasing it with the required factor
+
+        // adding the last even number after increasing it with the required factor
         ans.push_back(end);
-        return ans;    
+        return ans;
     }
-};
+};