Update D-Query Explanation.txt

Author: MathProgrammer

Explanations/Explanations- 3/D-Query Explanation.txt

@@ -1,79 +1,79 @@
 
-Let us say we have an array of  $1$�s and $ 0$�s. How do we find the sum of intervals of this array and support quick updates ? 
-Segment tree. This is just a simple sum tree. Now, let us reduce the above problem to this question. 
-
+Let us say we have an array of  $1$’s and $ 0$’s. How do we find the sum of intervals of this array and support quick updates ? 
+Segment tree. This is just a simple sum tree. Now, let us reduce the above problem to this question. 
 
 
-(Firstly, observe that the number of distinct elements in the range $[L, R]$ is equal to the number of elements who�s last occurence $[1, R]$ is  $>= L$. )
-
 
+(Firstly, observe that the number of distinct elements in the range $[L, R]$ is equal to the number of elements who’s last occurence $[1, R]$ is  $>= L$. )
 
-Let us maintain an array of length $ N$.
 
-$A[i] = 1$, if position $i$ is the last occurence of some element.
 
-$A[i] = 0$, otherwise.
-
-Now, one by one, we will insert elements into this tree. 
+Let us maintain an array of length $ N$.
 
+$A[i] = 1$, if position $i$ is the last occurence of some element.
 
-Let us say the current element is $v$
-We make the $A[L[v]] = 0$  and $A[i] = 1$
+$A[i] = 0$, otherwise.
+
+Now, one by one, we will insert elements into this tree. 
+
+
+Let us say the current element is $v$
+We make the $A[L[v]] = 0$  and $A[i] = 1$
 
 Where$ L[v] $represents the last occurence of $v $
-
+
 Prior to this insertion.
-
+
 Update the sum tree accordingly. 
 
-
-
+
+
 Now, check if any query ends at $i$. 
-
-For all queries with $R = i$
+
+For all queries with $R = i$
 
 There is sufficient information to answer it. 
 
 We know the last occurence of every element upto $R$. 
 We just need to check how many of them are after $L$
 
-
-Since, the array $A$ holds $1$ if it is the last occurence, we just find the sum $[L, R]$
 
-The number of  $1$�s in that range gives us the number of elements who�s last occurence  $>= L$
-
+Since, the array $A$ holds $1$ if it is the last occurence, we just find the sum $[L, R]$
+
+The number of  $1$’s in that range gives us the number of elements who’s last occurence  $>= L$
+
 
 
 Now, how do we efficiently get all queries with $R = i $efficiently without degrading to $O(NQ)$
 
-
-
+
+
 Treat everything as an event. 
 
-There are two kinds of events - Sum events and Query events.
-
-
-For insertion event, we need  ---> Position, value. 
+There are two kinds of events - Insertion events and Query events.
+
+
+For insertion event, we need  ---> Position, value. 
+
 
-
 For query event, we need ---> Left, right, query_no.
-
-
-Sort all the events by the following criteria  ---> (Position, for insertion), (Right, for queries)
-
 
-This is $O((N + Q) log(N + Q))$
-
+
+Sort all the events by the following criteria  ---> (Position, for insertion), (Right, for queries)
+
+
+This is $O((N + Q) log(N + Q))$
+
 Then go through the events one by one. 
 
-
-If it�s an insertion, set $A[L[v]] = 0$ and $A[i] = 1$
-
+
+If it’s an insertion, set $A[L[v]] = 0$ and $A[i] = 1$
+
 with appropriate updates on the sum tree. 
 
-
-
-If it�s a query event, return the number of $1$�s (sum) in the range $[L, R]$
+
+
+If it’s a query event, return the number of $1$’s (sum) in the range $[L, R]$
 
 ----------------------------------------------------------------------------
 
@@ -201,4 +201,4 @@ int main()
         printf("%d\n", answer[i]);
 
     return 0;
-}
+}