Problem 155

README.md

@@ -46,7 +46,7 @@ As I work through this list I figure I would make a GitHub repo with my solution
 35. [Implement Trie (Prefix Tree) #208](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/implement-trie-208.md)
 36. [Coin Change #322](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/coin-change-322.md)
 37. [Product of Array Except Self #238](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/product-of-array-238.md)
-38. Min Stack #155
+38. [Min Stack #155](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/min-stack-155.md)
 39. Validate Binary Search Tree #98
 40. Number of Islands #200
 41. Rotting Oranges #994
@@ -129,6 +129,7 @@ In order to practice with similar data structures I'll be placing each problem i
 - [Implement Queue using Stacks #232](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/implement-queue-stacks-232.md)
 - [Maximum Depth of Binary Tree #104](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/depth-binary-tree-104.md)
 - [Evaluate Reverse Polish Notation #150](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/polish-notation-150.md)
+- [Min Stack #155](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/medium/min-stack-155.md)
 
 ### Linked Lists
 

medium/min-stack-155.md

@@ -0,0 +1,104 @@
+# Min Stack
+
+Page on leetcode: https://leetcode.com/problems/min-stack/
+
+## Problem Statement
+
+Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
+
+Implement the MinStack class:
+
+- MinStack() initializes the stack object.
+- void push(int val) pushes the element val onto the stack.
+- void pop() removes the element on the top of the stack.
+- int top() gets the top element of the stack.
+- int getMin() retrieves the minimum element in the stack.
+
+You must implement a solution with O(1) time complexity for each function.
+
+### Constraints
+
+- -2<sup>31</sup> <= val <= 2<sup>31</sup> - 1
+- Methods pop, top and getMin operations will always be called on non-empty stacks.
+- At most 3 \* 10<sup>4</sup> calls will be made to push, pop, top, and getMin
+
+### Example
+
+```
+["MinStack","push","push","push","getMin","pop","top","getMin"]
+[[],[-2],[0],[-3],[],[],[],[]]
+
+Output
+[null,null,null,null,-3,null,0,-2]
+
+Explanation
+MinStack minStack = new MinStack();
+minStack.push(-2);
+minStack.push(0);
+minStack.push(-3);
+minStack.getMin(); // return -3
+minStack.pop();
+minStack.top();    // return 0
+minStack.getMin(); // return -2
+```
+
+## Solution
+
+- pop, push and peek (top) are constant with a normal stack
+- minheap can read min val in constant time
+- Do a stack with a linked list instead? How can I reassign min?
+
+### Initial Pseudocode
+
+1. Initialize array that will be a stack
+2. Push: push to array, then sort?
+3. Pop: pop off array, then sort?
+4. Top: pop off return then push back on
+
+### Optimized Solution
+
+This solution use O(1) for all operations by utilizing two stacks. You can see an explanation of this approach here: https://www.youtube.com/watch?v=qkLl7nAwDPo
+
+```javascript
+const MinStack = function () {
+  this.stack = [];
+  this.minStack = [];
+};
+
+/**
+ * @param {number} val
+ * @return {void}
+ */
+MinStack.prototype.push = function (val) {
+  this.stack.push(val);
+  const minStackSize = this.minStack.length;
+  // update minStack with current min value
+  if (minStackSize === 0 || this.minStack[minStackSize - 1] > val) {
+    this.minStack.push(val);
+  } else {
+    this.minStack.push(this.minStack[minStackSize - 1]);
+  }
+};
+
+/**
+ * @return {void}
+ */
+MinStack.prototype.pop = function () {
+  this.stack.pop();
+  this.minStack.pop();
+};
+
+/**
+ * @return {number}
+ */
+MinStack.prototype.top = function () {
+  return this.stack[this.stack.length - 1];
+};
+
+/**
+ * @return {number}
+ */
+MinStack.prototype.getMin = function () {
+  return this.minStack[this.minStack.length - 1];
+};
+```