C editorial updated

Author: harasees-singh

coDecode_2022/Editorial/C_DecreaseToZero.md

@@ -1,6 +1,7 @@
 # <center><u>[C. Decrease to Zero](https://www.hackerrank.com/contests/codecode-pec/challenges/decrease-to-zero)</u></center>
 ---
-The condition $0 < X*A_i ≤ A_i^2$ implies that X can have any integer value from 0 to $abs(A_i)$ and its sign will be the same as the sign of $A_i$.
+
+The condition $0 < x*A_i ≤ A_i^2$ implies that $x$ can have any integer value from 0 to $|A_i|$ and its sign will be the same as the sign of $A_i$. In other words the sign of $A_i$ can be ignored and you can replace $A_i$ with $|A_i|$ for every $i$.
 
 Now, if the array only contains 1's, then every time the player will choose it and make it zero. So, the answer will be dependent on the parity of n.
 
@@ -11,10 +12,25 @@ Else, the player getting a number > 1 for the first time wins the game.
 
 Suppose, a player gets a number $x$ on his/her turn such that $x > 1$. Then, the player can make it either 1 or 0 to confirm his/her win.
 
-For example consider the following cases
-Case 1: A = [3, 1, 1, 1, 4, 5] -> [0, 1, 1, 1, 4, 5] -> [0, 0, 1, 1, 4, 5] -> [0, 0, 0, 1, 4, 5] -> [0, 0, 0, 0, 4, 5] -> [0, 0, 0, 0, 1, 5] -> [0, 0, 0, 0, 0, 5] -> [0, 0, 0, 0, 0, 0]
-So, the above sequence of operation shows that Alice wins when there are an odd number of 1's after $x$
+`For Example` consider the following cases
+Case 1: $A = [3, 1, 1, 1, 4, 5]$
+* $[0, 1, 1, 1, 4, 5]$ Alice picks 3 
+* $[0, 0, 1, 1, 4, 5]$ Bob is forced to pick 1
+* $[0, 0, 0, 1, 4, 5]$ Alice picks 1
+* $[0, 0, 0, 0, 4, 5]$ Bob is forced to pick 1 
+* $[0, 0, 0, 0, 1, 5]$ Alice reaches a number with value $> 1$ again.
+<!-- * $[0, 0, 0, 0, 0, 5]$ Bob is forced to pick 1 
+* $[0, 0, 0, 0, 0, 0]$ Alice winds by picking 5 -->
+
+So, the above sequence of operation shows that Alice can always find some set of moves that results her to jump from a number $> 1$ to another number $> 1$ by skipping odd number of $1s$.
+
+Case 2: $A =  [3, 1, 1, 4, 5]$ 
+* $[1, 1, 1, 4, 5]$ Alice picks 2
+* $[0, 1, 1, 4, 5]$ Bob is forced to pick 1
+* $[0, 0, 1, 4, 5]$ Alice picks 1
+* $[0, 0, 0, 4, 5]$ Bob is forced to pick 1
+* $[0, 0, 0, 1, 5]$ Alice reaches a number with value $> 1$ again.
+
+Therefore, if there are even number of $1s$ between 2 numbers (both $> 1$) then the player can always find a way to jump from 1 to the other.
 
-Case 2: A =  [3, 1, 1, 4, 5] -> [1, 1, 1, 4, 5] -> [0, 1, 1, 4, 5] -> [0, 0, 1, 4, 5] -> [0, 0, 0, 4, 5] -> [0, 0, 0, 1, 5] -> [0, 0, 0, 0, 5] -> [0, 0, 0, 0, 0]
-Again Alice wins, even though there were even number of 1's after $x$.
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