Update 300_longest_increasing_subsequence.md

Author: Joseph Luce

leetcode/medium/300_longest_increasing_subsequence.md

@@ -8,26 +8,26 @@
 If we were to start from the left to the right, we would have seen the longest subsequence on the left side as we are going to the right.
 Inorder for us to know what those longest subsequences were, we will need a way to store that, hello dynamic programming. 
 
-For each element, we would need look at the numbers less than the current element on the left side.
+For each element, we would need look at the numbers on the left that are less than the current element.
 Now that we know which numbers those are, we can then look at their corresponding longest subsequence in the dynamic programming array.
 From this list, get the longest subsequence.
-This will tell us what to set as our longest subsequence for this current element + 1.
 
 We are basically building the longest increasing subsequence from the bottom up.
 
 **Example:**
-
-I(Input): [10,9,2,5,3,7,101,18]
-
-DP: [1,1,1,1,1,1,1,1]
-
-1. DP[1] = 1 + max([]) **(I[1] not > I[0])**
-2. DP[2] = 1 + max([]) **(I[2] not > I[0],I[1])**
-3. DP[3] = 1 + max(DP[2]) **(I[3] > I[2] and I[3] not > I[0],I[1])**
-4. DP[4] = 1 + max(DP[2]) **(I[4] > I[2] and I[4] not > I[0],I[1],I[3])**
-5. DP[5] = 1 + max(DP[1], DP[2], DP[3]) **(I[5] > I[1],I[2],I[3] and I[5] not > I[0])**
-6. DP[6] = 1 + max(DP[0], DP[1], DP[2], DP[3], DP[4], DP[5]) **(I[6] > all prev numbers)**
-7. DP[7] = 1 + max(DP[0], DP[1], DP[2], DP[3], DP[4], DP[5]) **(I[7] > all prev numbers except I[6])**
+```
+Input(I): [10,9,2,5,3,7,101,18]
+
+DP: 
+10   [1, 1, 1, 1, 1, 1, 1, 1]
+9    [1, 1, 1, 1, 1, 1, 1, 1] -> DP[1] = 1 + max([]) (I[1] not > I[0])
+2    [1, 1, 1, 1, 1, 1, 1, 1] -> DP[2] = 1 + max([]) (I[2] not > I[0],I[1])
+5    [1, 1, 1, 2, 1, 1, 1, 1] -> DP[3] = 1 + max(DP[2]) (I[3] > I[2] and I[3] not > I[0],I[1])
+3    [1, 1, 1, 2, 2, 1, 1, 1] -> DP[4] = 1 + max(DP[2]) (I[4] > I[2] and I[4] not > I[0],I[1],I[3])
+7    [1, 1, 1, 2, 2, 3, 1, 1] -> DP[5] = 1 + max(DP[1], DP[2], DP[3]) (I[5] > I[1],I[2],I[3] and I[5] not > I[0])
+101  [1, 1, 1, 2, 2, 3, 4, 1] -> DP[6] = 1 + max(DP[0], DP[1], DP[2], DP[3], DP[4], DP[5]) (I[6] > all prev numbers)
+18   [1, 1, 1, 2, 2, 3, 4, 4] -> DP[7] = 1 + max(DP[0], DP[1], DP[2], DP[3], DP[4], DP[5]) (I[7] > all prev numbers except I[6])
+```
 
 ```
 class Solution: