Looking for assistance in creating a function?

I am seeking assistance with creating a function for my app.

Imagine I have an array of 5 objects.

var pool = [
    {"color":"Pink"},
    {"color":"Pink"},
    {"color":"Green"},
    {"color":"Orange"},
    {"color":"Orange"}
];

Additionally, I have another array that specifies tasks to complete.

var tasks = [
    [
        {
            "amount": 1,
            "color": "Orange"
        },
        {
            "amount": 1,
            "color": "Pink"
        }
    ],
    [
        {
            "amount": 2,
            "color": "Green"
        },
        {
            "amount": 1,
            "color": "Orange"
        }
    ],
    [
        {
            "amount": 1,
            "color": "Orange"
        }
    ]
];

The tasks array consists of 3 arrays. These arrays represent the following tasks:

  1. Remove 1 Pink object or 1 Orange object from the pool.
  2. Remove 2 Green objects or 1 Orange object from the pool.
  3. Remove 1 Orange object from the pool.

I require a function that can assess if there are enough objects in order to complete all 3 tasks sequentially.

For the first task, we need to confirm if we have either 1 Pink object or 1 Orange object in the pool. With 2 Pink and 2 Orange objects available, the task is feasible.

Regarding the second task, we must check for 2 Green objects or 1 Orange object to remove. Since there is only 1 Green object, removing 2 does not work. Consequently, we opt to remove 1 Orange object as it still allows us to tackle the remaining tasks.

In the third task, the objective is to eliminate 1 Orange object. Even though there are 2 available, one was previously designated for removal in the second task. This means we cannot select an Orange object for the initial task either.

To summarize, we ascertain that both Pink objects can be removed for the first task. The second task requires removal of an Orange object, while the final task addresses the other Orange object.

Prior to commencing any tasks, a function needs to be developed to manage this logic effectively.

The existing function called canDoTasks1 lacks sufficient complexity as it disregards whether previous tasks depend on specific objects.

function canDoTasks1() {
    outerloop:
    for (var i = 0; i < tasks.length; i++) {
        var mandatory_task = tasks[i];
        for (var j = 0; j < mandatory_task.length; j++) {
            var optional_task = mandatory_task[j];
            var amount = countRemainingObjects(pool, optional_task.color);
            if (amount >= optional_task.amount) {
                continue outerloop;
            }
        }
        return false;
    }
    return true;
}

function countRemainingObjects(arr, color) {
    var total = 0;
    for (var i = 0; i < arr.length; i++) {
        if (arr[i].color == color) {
            total++;
        }
    }
    return total;
}

A revised function named canDoTasks2 is presented below, aiming to provide more precise object removal guidance for each task execution.

function canDoTasks2() {
    var pool_copy = pool.slice();
    outerloop:
    for (var i = 0; i < tasks.length; i++) {
        var mandatory_task = tasks[i];
        for (var j = 0; j < mandatory_task.length; j++) {
            var optional_task = mandatory_task[j];
            var amount = countRemainingObjects(pool_copy, optional_task.color);
            if (amount >= optional_task.amount) {
                removeFromPool(pool_copy, optional_task.color);
                continue outerloop;
            }
        }
        return false;
    }
    return true;
}

function removeFromPool(arr, color) {
    for (var i = 0; i < arr.length; i++) {
        if (arr[i].color == color) {
            arr.splice(i, 1);
            return;
        }
    }
}

Answer №1

One possible approach is to perform a depth-first search.

The function findTheWay() organizes the pool into a dictionary that tracks the availability of each color. This information is then used to conduct a depth-first search through the tasks list to find the optimal sequence for completing all tasks.

findTheWay() outputs an array containing indexes pointing to the steps in the task list.

const pool = [
  {"color":"Pink"},
  {"color":"Pink"},
  {"color":"Green"},
  {"color":"Orange"},
  {"color":"Orange"}
];

const tasks = [
  [
    { "amount": 1, "color": "Orange" },
    { "amount": 1, "color": "Pink" }
  ],
  [
    { "amount": 2, "color": "Green" },
    { "amount": 1, "color": "Orange" }
  ],
  [
    { "amount": 1, "color": "Orange" }
  ]
];

const findTheWay = (items, tasks) => {
  // Code logic for finding the way goes here
};

const thisIsTheWay = findTheWay(pool, tasks);
console.log(thisIsTheWay);
console.log(thisIsTheWay.map((v,i) => tasks[i][v]));


Iterator/Iterable

If you're looking to efficiently find multiple solutions to the problem, consider implementing an iterator/iterable based solution as an extension to the original code. Note that additional adjustments are required for this functionality.

Below is a modified version of the initial solution that supports iteration and provides insights on handling multiple solutions using iterators and iterables.

As the number of tasks and options increase, the time taken to find all solutions significantly rises. For instance, adding new tasks with multiple choices escalates processing time.

In your specific example set, only one solution exists.

// Same pool and tasks setup as before

// Implementing iterator/iterable
const theWays = (items, tasks) => {
  // Logic for creating iterator based solutions goes here
};

// Function to find first solution
const findTheWay = (items, tasks) => {
  const results = theWays(items, tasks).next();
  return results.done ? null : results.value;
}

// Function to find all solutions
const findTheWays = (items, tasks) => Array.from(theWays(items, tasks));

// Demo
const thisIsTheWay = findTheWay(pool, tasks);
console.log('First solution');
console.log(thisIsTheWay);
console.log(thisIsTheWay.map((v,i) => tasks[i][v]));

const theseAreTheWays = findTheWays(pool, tasks);
console.log(`All solutions: ${theseAreTheWays.length} found.`);
console.log(theseAreTheWays);
console.log(theseAreTheWays.map(v => v.map((v,i) => tasks[i][v])));

Answer №2

Implementing an array to keep track of items that need to be removed per task is a useful approach.

const removeItems = (items, task, removed) => {
    const cacheItems = [].concat(items);
  const type = task.type;
  let quantity = task.quantity;

  for (let i = items.length - 1; i >= 0; --i) {
    const item = items[i];
    if (item.type === type) {
      items.splice(i, 1);
      quantity--;
      if (quantity === 0) {
        removed.push(task);
        return true;
      }
    }
  }
  
  items.splice(0, items.length); // reset items
  cacheItems.forEach(item => items.push(item)); 
  return false;
}

const canFinishTask = (tasks, pool, removed, index = 0) => {
  const cacheItems = [].concat(pool);
  const currentTask = tasks[index];

  if (currentTask === undefined) { 
    return true;
  }

  for (let i = 0; i < currentTask.length; i++) {
    const taskItem = currentTask[i];
    if (removeItems(cacheItems, taskItem, removed) &&
      canFinishTask(tasks, cacheItems, removed, index + 1)) {
      return true;
    }
  }
  return false;
}

const removedTasks = []

console.log(canFinishTask([
  [{
      "quantity": 1,
      "type": "Red"
    },
    {
      "quantity": 1,
      "type": "Blue"
    }
  ],
  [{
      "quantity": 2,
      "type": "Green"
    },
    {
      "quantity": 1,
      "type": "Green"
    }
  ],
  [{
    "quantity": 1,
    "type": "Red"
  }]
], [{
    "type": "Blue"
  },
  {
    "type": "Blue"
  },
  {
    "type": "Green"
  },
  {
    "type": "Red"
  },
  {
    "type": "Red"
  }
], removedTasks));

console.log(removedTasks);

Answer №3

At this point, my current solution involves verifying the feasibility of all tasks when using all available options instead of selecting one specific option. This approach only provides a partial resolution to the problem.

Ensuring utilization of all tasks & options

function checkTaskFeasibility (availableOptions, taskList) {
  
  // Converting the pool into an object to track the quantity of each color
  let poolData = {};
  for (let element of availableOptions) {
    if (poolData[element.color]) {
      poolData[element.color] == poolData[element.color] + 1;
    } else {
      poolData[element.color] = 1;
    }
  }
 
  // Summing up the required colors from the tasks
  let taskColors= {};
  for (let group in taskList) {
    for (let task in group) {
      if (taskColors[task.color]) {
        taskColors[task.color] = taskColors[task.color] + task.amount;
      } else {
        taskColors[task.color] = task.amount;
      }
    }
  }

  function isColorAvailableInPool (color, quantity) {
    if (!poolData[color]) { return false }
    return poolData[color] >= quantity
  }

  // Ensuring all tasks can be completed using any available option
  for (let color of Object.keys(taskColors)) {
    let viability = isColorAvailableInPool(color, taskColors[color]); 
    if (!viability) { return false }
  }

  return true

}

console.log( checkTaskFeasibility(pool, tasks) );

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