Update readme.md

src/main/java/g2301_2400/s2322_minimum_score_after_removals_on_a_tree/readme.md

@@ -11,7 +11,7 @@ Remove two **distinct** edges of the tree to form three connected components. Fo
 1.  Get the XOR of all the values of the nodes for **each** of the three components respectively.
 2.  The **difference** between the **largest** XOR value and the **smallest** XOR value is the **score** of the pair.
 
-*   For example, say the three components have the node values: `[4,5,7]`, `[1,9]`, and `[3,3,3]`. The three XOR values are `4 ^ 5 ^ 7 = <ins>**6**</ins>`, `1 ^ 9 = <ins>**8**</ins>`, and `3 ^ 3 ^ 3 = <ins>**3**</ins>`. The largest XOR value is `8` and the smallest XOR value is `3`. The score is then `8 - 3 = 5`.
+*   For example, say the three components have the node values: `[4,5,7]`, `[1,9]`, and `[3,3,3]`. The three XOR values are <code>4 ^ 5 ^ 7 = <ins>**6**</ins></code>, <code>1 ^ 9 = <ins>**8**</ins></code>, and <code>3 ^ 3 ^ 3 = <ins>**3**</ins></code>. The largest XOR value is `8` and the smallest XOR value is `3`. The score is then `8 - 3 = 5`.
 
 Return _the **minimum** score of any possible pair of edge removals on the given tree_.
 
@@ -23,17 +23,7 @@ Return _the **minimum** score of any possible pair of edge removals on the given
 
 **Output:** 9
 
-**Explanation:** The diagram above shows a way to make a pair of removals.
-
-- The 1<sup>st</sup> component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10.
-
-- The 2<sup>nd</sup> component has node [0] with value [1]. Its XOR value is 1 = 1.
-
-- The 3<sup>rd</sup> component has node [2] with value [5]. Its XOR value is 5 = 5.
-
-The score is the difference between the largest and smallest XOR value which is 10 - 1 = 9.
-
-It can be shown that no other pair of removals will obtain a smaller score than 9. 
+**Explanation:** The diagram above shows a way to make a pair of removals. - The 1<sup>st</sup> component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10. - The 2<sup>nd</sup> component has node [0] with value [1]. Its XOR value is 1 = 1. - The 3<sup>rd</sup> component has node [2] with value [5]. Its XOR value is 5 = 5. The score is the difference between the largest and smallest XOR value which is 10 - 1 = 9. It can be shown that no other pair of removals will obtain a smaller score than 9. 
 
 **Example 2:**
 
@@ -43,17 +33,7 @@ It can be shown that no other pair of removals will obtain a smaller score than
 
 **Output:** 0
 
-**Explanation:** The diagram above shows a way to make a pair of removals.
-
-- The 1<sup>st</sup> component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0.
-
-- The 2<sup>nd</sup> component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0.
-
-- The 3<sup>rd</sup> component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0.
-
-The score is the difference between the largest and smallest XOR value which is 0 - 0 = 0.
-
-We cannot obtain a smaller score than 0. 
+**Explanation:** The diagram above shows a way to make a pair of removals. - The 1<sup>st</sup> component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0. - The 2<sup>nd</sup> component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0. - The 3<sup>rd</sup> component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0. The score is the difference between the largest and smallest XOR value which is 0 - 0 = 0. We cannot obtain a smaller score than 0. 
 
 **Constraints:**
 
@@ -64,4 +44,4 @@ We cannot obtain a smaller score than 0.
 *   `edges[i].length == 2`
 *   <code>0 <= a<sub>i</sub>, b<sub>i</sub> < n</code>
 *   <code>a<sub>i</sub> != b<sub>i</sub></code>
-*   `edges` represents a valid tree.
+*   `edges` represents a valid tree.