Added 1 JS solution for problem 155

Author: imaginate

javascript/155-Min-Stack.js

@@ -0,0 +1,56 @@
+//////////////////////////////////////////////////////////////////////////////
+// Two Stacks
+// Time: O(1)
+// Space: O(n)
+// This solution uses two stacks to save the total values and the minimum
+// values. Per each new value if it is less than or equal to the current
+// minimum it is pushed to both stacks. We save duplicate minimum values to
+// avoid the conundrum of inquiring whether the minimum value can be removed
+// (i.e. when the minimum value equals the top value can it be removed or are
+// there duplicate values in the main stack).
+//////////////////////////////////////////////////////////////////////////////
+
+class MinStack {
+    /**
+     * @constructor
+     */
+    constructor() {
+        this.mainStack = [];
+        this.minStack = [];
+    }
+
+    /**
+     * @param {number} val
+     * @return {void}
+     */
+    push(val) {
+        this.mainStack.push(val);
+        if (!this.minStack.length || val <= this.minStack[this.minStack.length - 1]) {
+    	    this.minStack.push(val);
+        }
+    }
+
+    /**
+     * @return {void}
+     */
+    pop() {
+        const val = this.mainStack.pop();
+        if (val === this.minStack[this.minStack.length - 1]) {
+    	    this.minStack.pop();
+        }
+    }
+
+    /**
+     * @return {number}
+     */
+    top() {
+        return this.mainStack[this.mainStack.length - 1];
+    }
+
+    /**
+     * @return {number}
+     */
+    getMin() {
+        return this.minStack[this.minStack.length - 1];
+    }
+}