11.1

Chapter11-Sorting&Searching/11.1-MergeArrays.js

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+/*
+Problem: You are given two sorted arrays, A and B, where A has a large anough buffer at the end to hold B. Write a method to merge B into A in sorted order.
+*/
+
+/** ex:
+A-[1,5,10,20]
+B- [0,2,6,7,11,25]
+output - [ 0, 1, 2, 5, 6, 7, 10, 11, 20, 25 ]
+**/
+
+//javascript has no native array insert method so create one here
+Array.prototype.insert = function(index, item) {
+  return this.splice(index, 0, item)
+}
+
+//binary search through array to find where to insert value
+function insertVal(low, high, key, array) {
+	//key is higher than highest val - insert @ end
+  if (key > array[high]){
+  	return high + 1;
+  }
+  //key is lower than lowest in array - insert @ beginning
+  if (low === 0 && key < array[0]) {
+  	return -1;
+  }
+  //comparison point
+  var midPoint = Math.ceil((high - low) / 2 + low);
+
+  if (key < array[midPoint]) {
+  	if (key > array[midPoint - 1]) {
+  		return midPoint - 1;
+  	}
+    return insertVal(low, midPoint, key, array);
+  } else if (key > array[midPoint]) {
+  	if (key < array[midPoint + 1]) {
+  		return midPoint;
+  	}
+    return insertVal(midPoint, high, key, array);
+  }
+}
+
+function mergeTwoArrays(A, B) {
+  let low = 0;
+  let high = A.length -1;
+  B.forEach(val => {
+    let index = insertVal(low, high, val, A);
+    low = index;
+    high ++;
+    return A.insert(index + 1, val);
+  })
+  return A;
+}
+
+//Runtime: n log log n

Chapter9-Recursion&Dynamic Programming/9.1-HoppingSteps.js

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+/*
+Problem: A child is running up a staircase with n steps, and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs.
+*/
+
+/** ex: 4 steps should return 7 ways:
+1,1,1,1; 2,2; 2,1,1; 1,2,1; 1,1,2; 1,3; 3,1
+**/
+
+function numberOfWays(steps) {
+  if (steps < 0) return 0;
+  else if (steps === 0) return 1;
+  //num of steps will equal num of steps to previous step plus one
+  else {
+    return numberOfWays(steps - 1) + numberOfWays(steps - 2) + numberOfWays(steps - 3);
+  }
+}