Comment for explanation

Author: PhysicsX

Cpp/maxValueProfit.cpp

@@ -2,6 +2,41 @@
 #include <algorithm>
 #include <iostream>
 
+// To solve "Prefix sum" can be used.
+// So the question is that there is investors these are making investment
+// in the predefined range in array. At the end, maximum profit is desired.
+/*
+
+    1 2 10 => 1 and 2 are the coordinates of the indices. 10 is the profit
+
+            Total profit
+            0  0  0  0  0
+    1 2 10  10 10   
+    2 4 5   10 15 5  5
+    3 5 12  10 10 17 17 12
+
+    At the end max profit is 17.
+
+    In brute force, there can be double for loop to check each input vector and add them to result vector.
+    This will lead time complexit for O(n^2)
+
+    To fix this time complexity we can use "Prefix sum"
+    Instead of adding each rounds lets add only investment value as a start and end.
+    For the first round between 1 and 2 there is 10 profit.
+    So, in the investment vector investment[start] += 10; and investment[end-1] -= 10;
+    start = 1; end = 2;
+    This is because instead of adding each profit in each step, we can track the positons
+    and at the end we can add them together (prefix sum) to find the maximum profit/investment
+
+    In simple example lets assume between 2 and 6 we need to add 5
+    vector should be [0 5 5 5 5 5 0 0]
+    instead of wrtting like this. lets use start and end points
+    [0 5 0 0 0 0 0 -5] (vector size should be round+1). 
+    then lets use prefix sum to get same result
+    vec[1] = 0; vec[2] = vec[1] + vec[2]; vec[3] = vec[0] + vec[1] + vec[2]; .....
+    Result = s[0 5 5 5 5 5 5 0].
+*/
+
 long maxValue(int n, std::vector<std::vector<int>>& rounds) {
     std::vector<long> investments(n + 1, 0); // Initialize the investment array with n+1 elements, all set to 0