update the docs for voting algo

Author: realdennis

src/voting-algorithm/readme.md

@@ -1,4 +1,5 @@
 # Summary of Voting-Algo
+Boyer-Moore Majority Vote algorithm
 
 ## Prob pattern
 1. Majority element
@@ -30,4 +31,45 @@ for(const num of nums) {
         vote --;
     }
 }
-```
+```
+
+## One-third case (majority-element-ii)
+Since we wanna find the candidate's `vote > Math.floor(n/3)`, there're **2 potential candidates** could meet it (no possibility for 3 cands, you can rethink about it).
+
+So we keep `cand1/vote1` & `cand2/vote2` and yes! still the same process.
+
+1. Replace the zero vote candidate (replace the candidate now lose authority)
+2. Accumulate
+```typescript
+ for (const num of nums) {
+        // replace
+        if (vote1 === 0 && num !== cand2) { // ---> yes we need this mutual condition
+            cand1 = num;
+            vote1 = 1;
+            continue;
+        } else if (vote2 === 0 && num !== cand1) { // --> since we don't want cand1 is same as cand2
+            cand2 = num;
+            vote2 = 1;
+            continue;
+        }
+
+        // accumulate (increment or decrement)
+        if (cand1 === num) {
+            vote1++;
+        } else if (cand2 === num) {
+            vote2++;
+        } else {
+            vote1--;
+            vote2--;
+        }
+    }
+```
+
+3. We need to re-analysis the vote of these two candidates, above process is only to pickup the tier two potential candidates (the top 2 highest votes candidate doesn't mean they all `vote > Math.floor(n/3)` ), but we need to know the specific votes.
+
+4. Yes, so now we can filter the potential candidate whiose vote is not greater `Math.floor(n/3)`.
+
+
+Note: 
+You might think "Why top 2 highest vote candidates" doesn't meaning all's vote greater than one-third of all.
+Counterexample: [1,2,3,3,3] -> you can say top 2 is `[2,3]` (or `[1,3]`) but only the candidate `3` meet the condition.