Learn and practice the Greedy Techniques and Dynamic Programming patterns to solve coding interview problems.

Educative-77 in JavaScript: Accelerate Your Coding Interview Prep

This module teaches you two of the most important problem-solving techniques in the optimization domain: Greedy Techniques and Dynamic Programming. Not all problems require an exhaustive search through the solution space. Rather, consistently choosing the locally optimal solution can be proven to yield the globally optimal solution. In such situations, Greedy Techniques result in simple, highly efficient solutions. Other problems have a well-defined structure, where the same subproblem is encountered and solved multiple times. Dynamic Programming techniques allow us to compute once and reuse the solutions to recurring subproblems.

Optimization Techniques

# Statement

You need to find the minimum number of refueling stops that a car needs to make to cover a distance, `target`. For simplicity, assume that the car has to travel from west to east in a straight line. There are various fuel stations on the way that are represented as a 2-D array of stations, i.e., `stations[i]` $= [d_i, f_i]$, where $d_i$ is the distance (in miles) of the $i^{th}$ gas station from the starting position, and $f_i$ is the amount of fuel (in liters) that it stores. Initially, the car starts with `k` liters of fuel. The car consumes one liter of fuel for every mile traveled. Upon reaching a gas station, the car can stop and refuel using all the petrol stored at the station. If it cannot reach the target, the program returns $−1$.

>**Note:** If the car reaches a station with $0$ fuel left, it can refuel from that station, and all the fuel from that station can be transferred to the car. If the car reaches the target with $0$ fuel left, it is still considered to have arrived.


**Constraints:**

- $1 \leq$ `target`, `k` $\leq 10^9$
- $0 \leq$ `stations.length` $\leq 900$
- $1 \leq d_{i} < d_{i+1}<$ `target`
- $1 \leq f_{i} < 10^9$

# Examples

Suppose the starting fuel is 9, and the target to be achieved is 33. Which of the following is the correct option for the minimum number of refueling stops?

```text                                          
                           |   Distance |   Fuel    |
                           --------------------------
                           |      5     |     8     |
                           |      7     |    11     |
                           |     10     |     3     |
                           |     13     |     7     |
                           |     21     |     6     |
                                
```

Suppose the starting fuel is 9, and the target to be achieved is 33. Which of the following is the correct option for the minimum number of refueling stops?
<pre><code class="language-text"> | Distance | Fuel |
 --------------------------
 | 5 | 8 |
 | 7 | 11 |
 | 10 | 3 |
 | 13 | 7 |
 | 21 | 6 |
 
</code></pre>

Suppose the starting fuel is 14, and the target to be achieved is 59. Which of the following is the correct option for the minimum number of refueling stops?

```text                                          
                           |   Distance |   Fuel     |
                           --------------------------
                           |      9     |     12     |
                           |     11     |      7     |
                           |     13     |     16     |
                           |     21     |     18     |
                           |     47     |      6     |             
                           |     53     |     17     |              
                           |     58     |      5     |
                                
```

Suppose the starting fuel is 14, and the target to be achieved is 59. Which of the following is the correct option for the minimum number of refueling stops?
<pre><code class="language-text"> | Distance | Fuel |
 --------------------------
 | 9 | 12 |
 | 11 | 7 |
 | 13 | 16 |
 | 21 | 18 |
 | 47 | 6 | 
 | 53 | 17 | 
 | 58 | 5 |
 
</code></pre>

Minimum Number of Refueling Stops

Minimum Number of Refueling Stops

## Try it yourself

Implement your solution in `main.js` in the following coding playground. We have provided a useful code template in the other file that you may build on to solve this problem.

export class MaxHeap {
 constructor() {
 this.heap = [];
 }

 push(val) {
 this.heap.push(val);
 this.bubbleUp(this.heap.length - 1);
 }

 bubbleUp(idx) {
 if (idx === 0) return;
 const parentIdx = Math.floor((idx - 1) / 2);
 if (this.heap[idx] > this.heap[parentIdx]) {
 [this.heap[idx], this.heap[parentIdx]] = [this.heap[parentIdx], this.heap[idx]];
 this.bubbleUp(parentIdx);
 }
 }

 pop() {
 if (this.heap.length === 1) {
 return this.heap.pop();
 }

 const max = this.heap[0];
 this.heap[0] = this.heap.pop();
 this.bubbleDown(0);
 return max;
 }

 bubbleDown(idx) {
 const leftChildIdx = 2 * idx + 1;
 const rightChildIdx = 2 * idx + 2;
 let maxChildIdx;
 if (leftChildIdx >= this.heap.length) return;

 if (rightChildIdx >= this.heap.length) {
 maxChildIdx = leftChildIdx;
 } else {
 maxChildIdx = this.heap[leftChildIdx] > this.heap[rightChildIdx] ? leftChildIdx : rightChildIdx;
 }

 if (this.heap[idx] < this.heap[maxChildIdx]) {
 [this.heap[idx], this.heap[maxChildIdx]] = [this.heap[maxChildIdx], this.heap[idx]];
 this.bubbleDown(maxChildIdx);
 }
 }

 peekMax() {
 return this.heap[0];
 }

 isEmpty() {
 return this.heap.length === 0
 }
}

Help me debug only the function Minimum Number of Refueling Stops . . I don't need the corrected code.I don't want explanations or code review. Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Please help me understand why this code failed for the last failed test case. Don't give me the corrected code. . If yes, then Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Your current task is to review only the minRefuelStops function and the other function(s) invoked in it/them. Do not review extra/helper files or functions not invoked in the solution function minRefuelStops. Do not review the code of data structures objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. Also, Answer the following questions: Is the Greedy Techniques pattern used? Take your time and review the code.

Please provide a brief time complexity analysis of the function minRefuelStops. While calculating time complexity, include the time complexity of built-in Javascript functions invoked by the code. The Answer should be to the point and in a few lines.

Please provide a brief Space complexity analysis of the function minRefuelStops. While calculating space complexity, include the space complexity of built-in Javascript functions invoked by the code. Do not include the space used for the input parameters and the space used for just returning an output(not used as a scratch space). The Answer should be to the point and in a few lines

TwoDArray

import DataStructure from "../../data/DataStructure.js";

// className: MyCustomDS
export default class TwoDArray extends DataStructure {
  constructor() {
    super();
    // class attributes
  }

  // throw parsing errors for incorrect test case format
  onError(data) {
    process.stderr.write("Parsing Error: " + data + " in TwoDArray is not parsable");
    process.exit();
  }

  toString = function () {
    // override print() for your data structure
  };

  // return value will be passed to user program
  construct(value, typeOfArr) {
    // parse test case in json string 'value' to your desired data structure
    return JSON.parse(JSON.stringify(value))
  }

  // input param contains return value of user program
  traverse(input, typeOfArr) {
    // convert your data structure 'input' to a json string representing a test case output
    return input
  }
}


def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """
 import heapq

 def min_refuel_stops(target, start_fuel, stations):
 if start_fuel >= target:
 return 0

 max_heap = []
 i, n = 0, len(stations)
 stops = 0
 max_distance = start_fuel

 while max_distance < target:
 if i < n and stations[i][0] <= max_distance:
 heapq.heappush(max_heap, -stations[i][1])
 i += 1
 elif not max_heap:
 return -1
 else:
 max_distance += -heapq.heappop(max_heap)
 stops += 1

 return stops
 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = []
 target = self.__inputs[0]
 startFuel = self.__inputs[1]
 stations = self.__inputs[2] 
 expected = min_refuel_stops(target,startFuel,stations)
 # store the 1st expected output that should be displayed to the user 
 expected_outputs.append(expected)
 actual = self.__actual_outputs[0]
 if json.dumps(expected) == json.dumps(actual):
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

import { MaxHeap } from "./maxHeap.js";

// Tip: You can use the min heap class 
// imported above

export function minRefuelStops(target, startFuel, stations){
    // Replace this placeholder return statement with your code
    return -1
}

refueling_stops.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : 3
    - 2 : 3
    - 3 : []
    output: 
    - 1 : 0
  - name: "testcase 2"
    inputs:
    - 1: 120
    - 2: 10
    - 3 : [[10, 60],[20, 25],[30, 30],[60, 40]]
    output:
    - 1 : 3
  - name: "testcase 3"
    inputs:
    - 1 : 15
    - 2 : 3
    - 3 : [[2, 5],[3, 1],[6, 3],[12, 6]]
    output: 
    - 1 : 4
  - name: "testcase 4"
    inputs:
    - 1: 570
    - 2: 140
    - 3 : [[140, 200], [160, 130], [310, 200], [330, 250]]
    output:
    - 1 : 2
  - name: "testcase 5"
    inputs:
    - 1 : 1360
    - 2 : 380
    - 3 : [[310, 160], [380, 620], [700, 89], [850, 190], [990, 360]]
    output: 
    - 1 : 2
  

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """
 def is_valid_2d_array(arr):
 if not isinstance(arr, list) or not all(isinstance(row, list) for row in arr):
 return False # Not a valid 2D array if it's empty or contains non-list elements
 
 for row in arr:
 if any(elem is None or isinstance(elem, bool) for elem in row):
 return False # Not a valid 2D array if any element in a row is None
 if not all(isinstance(elem, int) for elem in row):
 return False # Not a valid 2D array if any element in a row is not an integer
 
 return True # Valid 2D array
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 target = self.__inputs[0]
 startFuel = self.__inputs[1]
 stations = self.__inputs[2]
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not is_valid_2d_array(stations):
 self.store_result ("Not a valid 2D Integer array", ValidationFunction.FAIL)
 elif not all(len(row) == 2 for row in stations):
 self.store_result("expected rows of 'stations' to have 2 <= size <= 2", ValidationFunction.FAIL)
 elif not (1 <= target <= 10**9):
 self.store_result ("expected 'target' to have 1 <= target <= 10^9", ValidationFunction.FAIL)
 elif not (1 <= startFuel <= 10**9):
 self.store_result ("expected 'startFuel' to have 1 <= startFuel <= 10^9", ValidationFunction.FAIL)
 elif not (0 <= len(stations) <= 900):
 self.store_result ("expected 'stations' to have 0 <= size <= 900", ValidationFunction.FAIL)
 elif not all(1 <= stations[i][0] < stations[i + 1][0] < target for i in range(len(stations) - 1)):
 self.store_result ("expected 1 <= stations[i][0] < stations[i+1][0] < target", ValidationFunction.FAIL)
 elif not all(1 <= station[1] < 10**9 for station in stations):
 self.store_result ("fuel[i] should be 1 <= fuel[i] < 10^9", ValidationFunction.FAIL)
 elif len(stations) == 1 and not(1 <= stations[0][0] < target):
 self.store_result ("expected 1 <= stations[i][0] < target", ValidationFunction.FAIL)
 elif stations != [] and not target > stations[len(stations)-1][0]:
 self.store_result ("distance cannot be greater than or equal to target", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Minimum Number of Refueling Stops problem.

Greedy Techniques

Dynamic Programming

Conclusion

Minimum Number of Refueling Stops

Statement

Examples

Understand the problem

Figure it out!

Try it yourself