Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1: Input: [7, 1, 5, 3, 6, 4] Output: 5
| <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> | |
| <html xmlns="http://www.w3.org/1999/xhtml"> | |
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| <title>HTML Email Template</title> | |
| </head> | |
| <body style="margin:0;background-color:#cccccc;"> |
People
 :bowtie: |
😄 :smile: |
😆 :laughing: |
|---|---|---|
😊 :blush: |
😃 :smiley: |
:relaxed: |
😏 :smirk: |
😍 :heart_eyes: |
😘 :kissing_heart: |
😚 :kissing_closed_eyes: |
😳 :flushed: |
😌 :relieved: |
😆 :satisfied: |
😁 :grin: |
😉 :wink: |
😜 :stuck_out_tongue_winking_eye: |
😝 :stuck_out_tongue_closed_eyes: |
😀 :grinning: |
😗 :kissing: |
😙 :kissing_smiling_eyes: |
😛 :stuck_out_tongue: |
| class Solution { | |
| HashMap<Character, Character> mappings; | |
| public Solution() { | |
| this.mappings = new HashMap<Character, Character>(); | |
| this.mappings.put(')', '('); | |
| this.mappings.put('}', '{'); | |
| this.mappings.put(']', '['); | |
| } |
| /*Is palindorme*/ | |
| public boolean isPalindrome(String s) { | |
| char[] = s.toCharArray(); | |
| for (int i = 0, j = c.length - 1; i< j;) { | |
| if (!Character.isLetterOrDigit(c[i])) i++; | |
| else if (!Character.isLetterOrDigit(c[i])) j--; | |
| else if(Character.toLowerCase(c[i++]) != Character.toLowerCase(c[j--])) | |
| return false; | |
| } |
| /* | |
| place two pointers at i = 0 & j = arr.length-1 while i < j | |
| calculate prod & compare max , if arr[start] > arr[end] | |
| end -= 1 else start += 1 | |
| O(N) runtime | |
| O(1) space | |
| */ | |
| public int containerMostWater(int[] arr) { | |
| if arr.length < 2 | |
| return -1; |
| public int factorial(int n) { | |
| if (n == 0) | |
| return 1; | |
| else | |
| return n * factorial(n-1); | |
| } | |
| public int rangeSum(int[] array, int start, int end) { | |
| if (start > end) | |
| return 0; |
| /* | |
| Linked list stores a single instance variable that is a pointer to the first Node in the list | |
| Insert into a Linked List: | |
| Step 1: Create new node | |
| Step 2: Set next to current head | |
| Step 3: Set the new node to be the head | |
| Remove from a Linked List: | |
| Step 1: Set the pointer from head to the next node in the list |