Added 1 JS solution for problem 155

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];
+ }
+}