Attempting to reverse a linked list recursively by taking it as an input without altering the original copy, and producing a fresh output

Question

Attempting to reverse a linked list recursively by taking it as an input without altering the original copy, and producing a fresh output

I have encountered an issue with my current implementation where it only works properly for lists with a length of two or less. When the list is longer, I lose the nodes in the middle. How can I modify the return object after the recursive call to handle lists of any length while also preserving the nodes in the middle?

In this code snippet, Type List refers to the head of the list and ConsCell represents a node in the list

type List = null | ConsCell;

type ConsCell = {
  readonly first: number;
  readonly rest: List;
}

function reverse(list: List): List {

}

Your task is to implement the reverse function so that it returns the reversed version of the input list.

Here are some examples:

input: null (empty list) output: null

input: { first: 1, rest: null } output: { first: 1, rest: null }

input: { first: 1, rest: { first: 2, rest: { first: 3, rest: null } } } output: { first: 3, rest: { first: 2, rest: { first: 1, rest: null } } }

Remember, you should create a new list as the output without modifying the input list

Current solution:

function reverse(list: List): List {

  if(!list) {
    return null;
  }
  
  else if(list.rest == null) {
    return {
      first: list.first,
      rest: list.rest

    };
  }

  else {

  let oldRest: List = reverse(list.rest);

  return {
    first: oldRest.first,
      rest: {
        first: list.first,
        rest: null
      }
    }
  }
}

typescript algorithm recursion linked-list singly-linked-list

Answer 1

Answer №1

It seems that achieving this task is quite challenging due to the specified constraints you have attempted to work with:

recursive
single pass
no helper or inner function
can't modify function signature (parameters and return types)
not using extra space

The main issue lies in the need to manage the new head created at the bottom of the call stack while also having access to the most recent previous node for setting the next node.

Various options are available, each breaking one of the above constraints and most being suboptimal, except for the first option:

Performing an iterative approach, which is more efficient and less prone to stack overflow.
Conducting a two-pass linear operation: copying the linked list in the first pass, then reversing it with any in-place linked list reversal algorithm in the second pass.
Utilizing an inner or helper function that can return both nodes simultaneously (illustrated below).
Employing an inner function that makes modifications to the head within the closure (shown below).
Maintaining state in default parameters hoping they won't be altered by the top-level caller (demonstrated below).
Opting for a quadratic method where the head is returned and additional nodes are attached by iteratively walking to its tail.

Below is Option 3, involving packing up two return values with a helper or inner function:

// Code snippet illustrating reverse linked list operation
// Note: TypeScript annotations not used for simplicity

const reverseLL = ll => {
  
  const reverse = ll => {
    if (!ll) {
      return null;
    }
    else if (!ll.next) {
      const head = {val: ll.val};
      return {head, prev: head};
    }

    const {head, prev} = reverse(ll.next);
    prev.next = {val: ll.val, next: null};
    return {head, prev: prev.next};
  };

  return reverse(ll)?.head ?? null;
};

console.log(reverseLL(null));
console.log(reverseLL({val: 1}));
console.log(reverseLL({val: 1, next: {val: 2}}));
console.log(reverseLL({val: 1, next: {val: 2, next: {val: 3}}}));

// Sample input linked list object
const ll = {
  val: 1,
  next: {
    val: 2,
    next: {
      val: 3,
      next: {val: 4}
    }
  }
};

console.log(reverseLL(ll));

An alternative using closure to track the head is demonstrated as Option 4:

// Code snippet showcasing reverse linked list operation while utilizing closure
// Note: TypeScript annotations not used for simplicity

const reverseLL = ll => {
  let head = null;

  const reverse = ll => {
    if (!ll) {
      return null;
    }
    else if (!ll.next) {
      return head = { val: ll.val, next: null };
    }

    const prev = reverse(ll.next);
    prev.next = { val: ll.val, next: null };
    return prev.next;
  };

  reverse(ll);
  return head;
};

console.log(reverseLL(null));
console.log(reverseLL({ val: 1}));
console.log(reverseLL({ val: 1, next: { val: 2 }}));
console.log(reverseLL({ val: 1, next: { val: 2, next: { val: 3 }} }));

// Sample input linked list object
const ll = {
  val: 1,
  next: {
    val: 2,
    next: {
      val: 3,
      next: { val: 4 }
    }
  }
};

console.log(reverseLL(ll));

Option 5 involves keeping state in parameters expected to be ignored by the client. Though not preferred, it becomes necessary when inner or helper functions are prohibited and optional parameters can be added.

// Additional option demonstrating state management with parameters
// Note: TypeScript annotations not used for simplicity

const reverseLL = (ll, head={}, firstCall=true) => {
  if (!ll) {
    return null;
  }
  else if (!ll.next) {
    head.val = ll.val;
    head.next = null;
    return head;
  }

  const prev = reverseLL(ll.next, head, false);
  prev.next = {val: ll.val, next: null};
  return firstCall ? head : prev.next;
};

console.log(reverseLL(null));
console.log(reverseLL({val: 1}));
console.log(reverseLL({val: 1, next: {val: 2}}));
console.log(reverseLL({val: 1, next: {val: 2, next: {val: 3}}}));

// Sample input linked list object
const ll = {
  val: 1,
  next: {
    val: 2,
    next: {
      val: 3,
      next: {val: 4}
    }
  }
};

console.log(reverseLL(ll));

Please excuse the key renaming and absence of TypeScript usage for runnable examples. The code can easily be converted accordingly.

Answer 2

It seems that achieving this task is quite challenging due to the specified constraints you have attempted to work with:

recursive
single pass
no helper or inner function
can't modify function signature (parameters and return types)
not using extra space

The main issue lies in the need to manage the new head created at the bottom of the call stack while also having access to the most recent previous node for setting the next node.

Various options are available, each breaking one of the above constraints and most being suboptimal, except for the first option:

Performing an iterative approach, which is more efficient and less prone to stack overflow.
Conducting a two-pass linear operation: copying the linked list in the first pass, then reversing it with any in-place linked list reversal algorithm in the second pass.
Utilizing an inner or helper function that can return both nodes simultaneously (illustrated below).
Employing an inner function that makes modifications to the head within the closure (shown below).
Maintaining state in default parameters hoping they won't be altered by the top-level caller (demonstrated below).
Opting for a quadratic method where the head is returned and additional nodes are attached by iteratively walking to its tail.

Below is Option 3, involving packing up two return values with a helper or inner function:

// Code snippet illustrating reverse linked list operation
// Note: TypeScript annotations not used for simplicity

const reverseLL = ll => {
  
  const reverse = ll => {
    if (!ll) {
      return null;
    }
    else if (!ll.next) {
      const head = {val: ll.val};
      return {head, prev: head};
    }

    const {head, prev} = reverse(ll.next);
    prev.next = {val: ll.val, next: null};
    return {head, prev: prev.next};
  };

  return reverse(ll)?.head ?? null;
};

console.log(reverseLL(null));
console.log(reverseLL({val: 1}));
console.log(reverseLL({val: 1, next: {val: 2}}));
console.log(reverseLL({val: 1, next: {val: 2, next: {val: 3}}}));

// Sample input linked list object
const ll = {
  val: 1,
  next: {
    val: 2,
    next: {
      val: 3,
      next: {val: 4}
    }
  }
};

console.log(reverseLL(ll));

An alternative using closure to track the head is demonstrated as Option 4:

// Code snippet showcasing reverse linked list operation while utilizing closure
// Note: TypeScript annotations not used for simplicity

const reverseLL = ll => {
  let head = null;

  const reverse = ll => {
    if (!ll) {
      return null;
    }
    else if (!ll.next) {
      return head = { val: ll.val, next: null };
    }

    const prev = reverse(ll.next);
    prev.next = { val: ll.val, next: null };
    return prev.next;
  };

  reverse(ll);
  return head;
};

console.log(reverseLL(null));
console.log(reverseLL({ val: 1}));
console.log(reverseLL({ val: 1, next: { val: 2 }}));
console.log(reverseLL({ val: 1, next: { val: 2, next: { val: 3 }} }));

// Sample input linked list object
const ll = {
  val: 1,
  next: {
    val: 2,
    next: {
      val: 3,
      next: { val: 4 }
    }
  }
};

console.log(reverseLL(ll));

Option 5 involves keeping state in parameters expected to be ignored by the client. Though not preferred, it becomes necessary when inner or helper functions are prohibited and optional parameters can be added.

// Additional option demonstrating state management with parameters
// Note: TypeScript annotations not used for simplicity

const reverseLL = (ll, head={}, firstCall=true) => {
  if (!ll) {
    return null;
  }
  else if (!ll.next) {
    head.val = ll.val;
    head.next = null;
    return head;
  }

  const prev = reverseLL(ll.next, head, false);
  prev.next = {val: ll.val, next: null};
  return firstCall ? head : prev.next;
};

console.log(reverseLL(null));
console.log(reverseLL({val: 1}));
console.log(reverseLL({val: 1, next: {val: 2}}));
console.log(reverseLL({val: 1, next: {val: 2, next: {val: 3}}}));

// Sample input linked list object
const ll = {
  val: 1,
  next: {
    val: 2,
    next: {
      val: 3,
      next: {val: 4}
    }
  }
};

console.log(reverseLL(ll));

Please excuse the key renaming and absence of TypeScript usage for runnable examples. The code can easily be converted accordingly.

Answer 3

Answer №2

Exploring the world of programming with constraints can be an enjoyable experience. Hopefully, @ggorlen's insights will shed some light on the bigger picture.

type List<T> = null | {
  readonly first: T
  readonly rest: List<T>
}

In this context, we focus solely on the reverse function and a single parameter for the input list, represented as t.

function reverse<T>(t: List<T>): List<T> {
  if (t == null)
    return null
  else if (t.rest == null)
    return t
  else
    return {
      first: reverse(t.rest)!.first,
      rest: reverse({
        first: t.first,
        rest: reverse(reverse(t.rest)!.rest)
      })
    }
}

Given the constraints, the code may not be optimized for time/space efficiency. However, by allowing variable assignment, we can minimize duplicated computations:

function reverse<T>(t: List<T>): List<T> {
  if (t == null) return null
  else if (t.rest == null) return t
  const r = reverse(t.rest)! // non-null guaranteed
  return {
    first: r.first,
    rest: reverse({
      first: t.first,
      rest: reverse(r.rest)
    })
  }
}

Incorporating logical or operations can further streamline the conditional statements:

function reverse<T>(t: List<T>): List<T> {
  if (t == null || t.rest == null) return t
  const r = reverse(t.rest)!
  return {
    first: r.first,
    rest: reverse({
      first: t.first,
      rest: reverse(r.rest)
    })
  }
}

Introducing abstraction enables us to define additional list operations like cons:

function cons<T>(value: T, t: List<T>): List<T> {
  return { first: value, rest: t }
}

function reverse<T>(t: List<T>): List<T> {
  if (t == null || t.rest == null) return t
  const r = reverse(t.rest)!
  return cons(
    r.first,
    reverse(cons(
      t.first,
      reverse(r.rest),
    ))
  )
}

By embracing abstraction, we can implement other list operations such as fold:

function fold<T,R>(t: List<T>, state: R, func: (value: T, state: R) => R): R {
  if (t == null)
    return state
  else
    return fold(t.rest, func(t.first, state), func)
}

const add = (a, b) => a + b

const mylist = cons(1, cons(2, cons(3, null)))

console.log("sum", fold(mylist, 0, add))

We can derive reverse as a fold operation applied to cons:

function reverse<T>(t: List<T>): List<T> {
  return fold(t, null as List<T>, cons)
}

To visualize the output, we introduce another valuable list operation called toString:

function toString<T>(t: List<T>): string {
  if (t == null)
    return "∅"
  else
    return String(t.first) + " -> " + toString(t.rest)
}

const mylist = cons(1, cons(2, cons(3, cons(4, cons(5, null)))))

console.log(toString(mylist))
console.log(toString(reverse(mylist)))
console.log(toString(mylist))

Printing the original list demonstrates that reverse operates immutably:

1 -> 2 -> 3 -> 4 -> 5 -> ∅
5 -> 4 -> 3 -> 2 -> 1 -> ∅ 
1 -> 2 -> 3 -> 4 -> 5 -> ∅

Answer 4

Exploring the world of programming with constraints can be an enjoyable experience. Hopefully, @ggorlen's insights will shed some light on the bigger picture.

type List<T> = null | {
  readonly first: T
  readonly rest: List<T>
}

In this context, we focus solely on the reverse function and a single parameter for the input list, represented as t.

function reverse<T>(t: List<T>): List<T> {
  if (t == null)
    return null
  else if (t.rest == null)
    return t
  else
    return {
      first: reverse(t.rest)!.first,
      rest: reverse({
        first: t.first,
        rest: reverse(reverse(t.rest)!.rest)
      })
    }
}

Given the constraints, the code may not be optimized for time/space efficiency. However, by allowing variable assignment, we can minimize duplicated computations:

function reverse<T>(t: List<T>): List<T> {
  if (t == null) return null
  else if (t.rest == null) return t
  const r = reverse(t.rest)! // non-null guaranteed
  return {
    first: r.first,
    rest: reverse({
      first: t.first,
      rest: reverse(r.rest)
    })
  }
}

Incorporating logical or operations can further streamline the conditional statements:

function reverse<T>(t: List<T>): List<T> {
  if (t == null || t.rest == null) return t
  const r = reverse(t.rest)!
  return {
    first: r.first,
    rest: reverse({
      first: t.first,
      rest: reverse(r.rest)
    })
  }
}

Introducing abstraction enables us to define additional list operations like cons:

function cons<T>(value: T, t: List<T>): List<T> {
  return { first: value, rest: t }
}

function reverse<T>(t: List<T>): List<T> {
  if (t == null || t.rest == null) return t
  const r = reverse(t.rest)!
  return cons(
    r.first,
    reverse(cons(
      t.first,
      reverse(r.rest),
    ))
  )
}

By embracing abstraction, we can implement other list operations such as fold:

function fold<T,R>(t: List<T>, state: R, func: (value: T, state: R) => R): R {
  if (t == null)
    return state
  else
    return fold(t.rest, func(t.first, state), func)
}

const add = (a, b) => a + b

const mylist = cons(1, cons(2, cons(3, null)))

console.log("sum", fold(mylist, 0, add))

We can derive reverse as a fold operation applied to cons:

function reverse<T>(t: List<T>): List<T> {
  return fold(t, null as List<T>, cons)
}

To visualize the output, we introduce another valuable list operation called toString:

function toString<T>(t: List<T>): string {
  if (t == null)
    return "∅"
  else
    return String(t.first) + " -> " + toString(t.rest)
}

const mylist = cons(1, cons(2, cons(3, cons(4, cons(5, null)))))

console.log(toString(mylist))
console.log(toString(reverse(mylist)))
console.log(toString(mylist))

Printing the original list demonstrates that reverse operates immutably:

1 -> 2 -> 3 -> 4 -> 5 -> ∅
5 -> 4 -> 3 -> 2 -> 1 -> ∅ 
1 -> 2 -> 3 -> 4 -> 5 -> ∅

Attempting to reverse a linked list recursively by taking it as an input without altering the original copy, and producing a fresh output

Answer №1

Answer №2

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