/**
 * @license
 * Copyright 2017 Google LLC
 * SPDX-License-Identifier: BSD-3-Clause
 */

import {
  getCommittedValue,
  insertPart,
  removePart,
  setChildPartValue,
  setCommittedValue,
} from '../directive-helpers.js';
import {directive, Directive, PartInfo, PartType} from '../directive.js';
import {ChildPart, noChange} from '../lit-html.js';

export type KeyFn<T> = (item: T, index: number) => unknown;
export type ItemTemplate<T> = (item: T, index: number) => unknown;

// Helper for generating a map of array item to its index over a subset
// of an array (used to lazily generate `newKeyToIndexMap` and
// `oldKeyToIndexMap`)
const generateMap = (list: unknown[], start: number, end: number) => {
  const map = new Map<unknown, number>();
  for (let i = start; i <= end; i++) {
    map.set(list[i], i);
  }
  return map;
};

class RepeatDirective extends Directive {
  private _itemKeys?: unknown[];

  constructor(partInfo: PartInfo) {
    super(partInfo);
    if (partInfo.type !== PartType.CHILD) {
      throw new Error('repeat() can only be used in text expressions');
    }
  }

  private _getValuesAndKeys<T>(
    items: Iterable<T>,
    keyFnOrTemplate: KeyFn<T> | ItemTemplate<T>,
    template?: ItemTemplate<T>
  ) {
    let keyFn: KeyFn<T> | undefined;
    if (template === undefined) {
      template = keyFnOrTemplate;
    } else if (keyFnOrTemplate !== undefined) {
      keyFn = keyFnOrTemplate as KeyFn<T>;
    }
    const keys = [];
    const values = [];
    let index = 0;
    for (const item of items) {
      keys[index] = keyFn ? keyFn(item, index) : index;
      values[index] = template!(item, index);
      index++;
    }
    return {
      values,
      keys,
    };
  }

  render<T>(items: Iterable<T>, template: ItemTemplate<T>): Array<unknown>;
  render<T>(
    items: Iterable<T>,
    keyFn: KeyFn<T> | ItemTemplate<T>,
    template: ItemTemplate<T>
  ): Array<unknown>;
  render<T>(
    items: Iterable<T>,
    keyFnOrTemplate: KeyFn<T> | ItemTemplate<T>,
    template?: ItemTemplate<T>
  ) {
    return this._getValuesAndKeys(items, keyFnOrTemplate, template).values;
  }

  update<T>(
    containerPart: ChildPart,
    [items, keyFnOrTemplate, template]: [
      Iterable<T>,
      KeyFn<T> | ItemTemplate<T>,
      ItemTemplate<T>
    ]
  ) {
    // Old part & key lists are retrieved from the last update
    // TODO: deal with directive being swapped out?
    const oldParts = getCommittedValue(
      containerPart
    ) as Array<ChildPart | null>;
    const {values: newValues, keys: newKeys} = this._getValuesAndKeys(
      items,
      keyFnOrTemplate,
      template
    );

    if (!oldParts) {
      this._itemKeys = newKeys;
      return newValues;
    }

    const oldKeys = (this._itemKeys ??= []);

    // New part list will be built up as we go (either reused from
    // old parts or created for new keys in this update). This is
    // saved in the above cache at the end of the update.
    const newParts: ChildPart[] = [];

    // Maps from key to index for current and previous update; these
    // are generated lazily only when needed as a performance
    // optimization, since they are only required for multiple
    // non-contiguous changes in the list, which are less common.
    let newKeyToIndexMap!: Map<unknown, number>;
    let oldKeyToIndexMap!: Map<unknown, number>;

    // Head and tail pointers to old parts and new values
    let oldHead = 0;
    let oldTail = oldParts.length - 1;
    let newHead = 0;
    let newTail = newValues.length - 1;

    // Overview of O(n) reconciliation algorithm (general approach
    // based on ideas found in ivi, vue, snabbdom, etc.):
    //
    // * We start with the list of old parts and new values (and
    //   arrays of their respective keys), head/tail pointers into
    //   each, and we build up the new list of parts by updating
    //   (and when needed, moving) old parts or creating new ones.
    //   The initial scenario might look like this (for brevity of
    //   the diagrams, the numbers in the array reflect keys
    //   associated with the old parts or new values, although keys
    //   and parts/values are actually stored in parallel arrays
    //   indexed using the same head/tail pointers):
    //
    //      oldHead v                 v oldTail
    //   oldKeys:  [0, 1, 2, 3, 4, 5, 6]
    //   newParts: [ ,  ,  ,  ,  ,  ,  ]
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6] <- reflects the user's new
    //                                      item order
    //      newHead ^                 ^ newTail
    //
    // * Iterate old & new lists from both sides, updating,
    //   swapping, or removing parts at the head/tail locations
    //   until neither head nor tail can move.
    //
    // * Example below: keys at head pointers match, so update old
    //   part 0 in-place (no need to move it) and record part 0 in
    //   the `newParts` list. The last thing we do is advance the
    //   `oldHead` and `newHead` pointers (will be reflected in the
    //   next diagram).
    //
    //      oldHead v                 v oldTail
    //   oldKeys:  [0, 1, 2, 3, 4, 5, 6]
    //   newParts: [0,  ,  ,  ,  ,  ,  ] <- heads matched: update 0
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]    and advance both oldHead
    //                                      & newHead
    //      newHead ^                 ^ newTail
    //
    // * Example below: head pointers don't match, but tail
    //   pointers do, so update part 6 in place (no need to move
    //   it), and record part 6 in the `newParts` list. Last,
    //   advance the `oldTail` and `oldHead` pointers.
    //
    //         oldHead v              v oldTail
    //   oldKeys:  [0, 1, 2, 3, 4, 5, 6]
    //   newParts: [0,  ,  ,  ,  ,  , 6] <- tails matched: update 6
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]    and advance both oldTail
    //                                      & newTail
    //         newHead ^              ^ newTail
    //
    // * If neither head nor tail match; next check if one of the
    //   old head/tail items was removed. We first need to generate
    //   the reverse map of new keys to index (`newKeyToIndexMap`),
    //   which is done once lazily as a performance optimization,
    //   since we only hit this case if multiple non-contiguous
    //   changes were made. Note that for contiguous removal
    //   anywhere in the list, the head and tails would advance
    //   from either end and pass each other before we get to this
    //   case and removals would be handled in the final while loop
    //   without needing to generate the map.
    //
    // * Example below: The key at `oldTail` was removed (no longer
    //   in the `newKeyToIndexMap`), so remove that part from the
    //   DOM and advance just the `oldTail` pointer.
    //
    //         oldHead v           v oldTail
    //   oldKeys:  [0, 1, 2, 3, 4, 5, 6]
    //   newParts: [0,  ,  ,  ,  ,  , 6] <- 5 not in new map: remove
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]    5 and advance oldTail
    //         newHead ^           ^ newTail
    //
    // * Once head and tail cannot move, any mismatches are due to
    //   either new or moved items; if a new key is in the previous
    //   "old key to old index" map, move the old part to the new
    //   location, otherwise create and insert a new part. Note
    //   that when moving an old part we null its position in the
    //   oldParts array if it lies between the head and tail so we
    //   know to skip it when the pointers get there.
    //
    // * Example below: neither head nor tail match, and neither
    //   were removed; so find the `newHead` key in the
    //   `oldKeyToIndexMap`, and move that old part's DOM into the
    //   next head position (before `oldParts[oldHead]`). Last,
    //   null the part in the `oldPart` array since it was
    //   somewhere in the remaining oldParts still to be scanned
    //   (between the head and tail pointers) so that we know to
    //   skip that old part on future iterations.
    //
    //         oldHead v        v oldTail
    //   oldKeys:  [0, 1, -, 3, 4, 5, 6]
    //   newParts: [0, 2,  ,  ,  ,  , 6] <- stuck: update & move 2
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]    into place and advance
    //                                      newHead
    //         newHead ^           ^ newTail
    //
    // * Note that for moves/insertions like the one above, a part
    //   inserted at the head pointer is inserted before the
    //   current `oldParts[oldHead]`, and a part inserted at the
    //   tail pointer is inserted before `newParts[newTail+1]`. The
    //   seeming asymmetry lies in the fact that new parts are
    //   moved into place outside in, so to the right of the head
    //   pointer are old parts, and to the right of the tail
    //   pointer are new parts.
    //
    // * We always restart back from the top of the algorithm,
    //   allowing matching and simple updates in place to
    //   continue...
    //
    // * Example below: the head pointers once again match, so
    //   simply update part 1 and record it in the `newParts`
    //   array.  Last, advance both head pointers.
    //
    //         oldHead v        v oldTail
    //   oldKeys:  [0, 1, -, 3, 4, 5, 6]
    //   newParts: [0, 2, 1,  ,  ,  , 6] <- heads matched: update 1
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]    and advance both oldHead
    //                                      & newHead
    //            newHead ^        ^ newTail
    //
    // * As mentioned above, items that were moved as a result of
    //   being stuck (the final else clause in the code below) are
    //   marked with null, so we always advance old pointers over
    //   these so we're comparing the next actual old value on
    //   either end.
    //
    // * Example below: `oldHead` is null (already placed in
    //   newParts), so advance `oldHead`.
    //
    //            oldHead v     v oldTail
    //   oldKeys:  [0, 1, -, 3, 4, 5, 6] <- old head already used:
    //   newParts: [0, 2, 1,  ,  ,  , 6]    advance oldHead
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]
    //               newHead ^     ^ newTail
    //
    // * Note it's not critical to mark old parts as null when they
    //   are moved from head to tail or tail to head, since they
    //   will be outside the pointer range and never visited again.
    //
    // * Example below: Here the old tail key matches the new head
    //   key, so the part at the `oldTail` position and move its
    //   DOM to the new head position (before `oldParts[oldHead]`).
    //   Last, advance `oldTail` and `newHead` pointers.
    //
    //               oldHead v  v oldTail
    //   oldKeys:  [0, 1, -, 3, 4, 5, 6]
    //   newParts: [0, 2, 1, 4,  ,  , 6] <- old tail matches new
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]   head: update & move 4,
    //                                     advance oldTail & newHead
    //               newHead ^     ^ newTail
    //
    // * Example below: Old and new head keys match, so update the
    //   old head part in place, and advance the `oldHead` and
    //   `newHead` pointers.
    //
    //               oldHead v oldTail
    //   oldKeys:  [0, 1, -, 3, 4, 5, 6]
    //   newParts: [0, 2, 1, 4, 3,   ,6] <- heads match: update 3
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]    and advance oldHead &
    //                                      newHead
    //                  newHead ^  ^ newTail
    //
    // * Once the new or old pointers move past each other then all
    //   we have left is additions (if old list exhausted) or
    //   removals (if new list exhausted). Those are handled in the
    //   final while loops at the end.
    //
    // * Example below: `oldHead` exceeded `oldTail`, so we're done
    //   with the main loop.  Create the remaining part and insert
    //   it at the new head position, and the update is complete.
    //
    //                   (oldHead > oldTail)
    //   oldKeys:  [0, 1, -, 3, 4, 5, 6]
    //   newParts: [0, 2, 1, 4, 3, 7 ,6] <- create and insert 7
    //   newKeys:  [0, 2, 1, 4, 3, 7, 6]
    //                     newHead ^ newTail
    //
    // * Note that the order of the if/else clauses is not
    //   important to the algorithm, as long as the null checks
    //   come first (to ensure we're always working on valid old
    //   parts) and that the final else clause comes last (since
    //   that's where the expensive moves occur). The order of
    //   remaining clauses is is just a simple guess at which cases
    //   will be most common.
    //
    // * TODO(kschaaf) Note, we could calculate the longest
    //   increasing subsequence (LIS) of old items in new position,
    //   and only move those not in the LIS set. However that costs
    //   O(nlogn) time and adds a bit more code, and only helps
    //   make rare types of mutations require fewer moves. The
    //   above handles removes, adds, reversal, swaps, and single
    //   moves of contiguous items in linear time, in the minimum
    //   number of moves. As the number of multiple moves where LIS
    //   might help approaches a random shuffle, the LIS
    //   optimization becomes less helpful, so it seems not worth
    //   the code at this point. Could reconsider if a compelling
    //   case arises.

    while (oldHead <= oldTail && newHead <= newTail) {
      if (oldParts[oldHead] === null) {
        // `null` means old part at head has already been used
        // below; skip
        oldHead++;
      } else if (oldParts[oldTail] === null) {
        // `null` means old part at tail has already been used
        // below; skip
        oldTail--;
      } else if (oldKeys[oldHead] === newKeys[newHead]) {
        // Old head matches new head; update in place
        newParts[newHead] = setChildPartValue(
          oldParts[oldHead]!,
          newValues[newHead]
        );
        oldHead++;
        newHead++;
      } else if (oldKeys[oldTail] === newKeys[newTail]) {
        // Old tail matches new tail; update in place
        newParts[newTail] = setChildPartValue(
          oldParts[oldTail]!,
          newValues[newTail]
        );
        oldTail--;
        newTail--;
      } else if (oldKeys[oldHead] === newKeys[newTail]) {
        // Old head matches new tail; update and move to new tail
        newParts[newTail] = setChildPartValue(
          oldParts[oldHead]!,
          newValues[newTail]
        );
        insertPart(containerPart, newParts[newTail + 1], oldParts[oldHead]!);
        oldHead++;
        newTail--;
      } else if (oldKeys[oldTail] === newKeys[newHead]) {
        // Old tail matches new head; update and move to new head
        newParts[newHead] = setChildPartValue(
          oldParts[oldTail]!,
          newValues[newHead]
        );
        insertPart(containerPart, oldParts[oldHead]!, oldParts[oldTail]!);
        oldTail--;
        newHead++;
      } else {
        if (newKeyToIndexMap === undefined) {
          // Lazily generate key-to-index maps, used for removals &
          // moves below
          newKeyToIndexMap = generateMap(newKeys, newHead, newTail);
          oldKeyToIndexMap = generateMap(oldKeys, oldHead, oldTail);
        }
        if (!newKeyToIndexMap.has(oldKeys[oldHead])) {
          // Old head is no longer in new list; remove
          removePart(oldParts[oldHead]!);
          oldHead++;
        } else if (!newKeyToIndexMap.has(oldKeys[oldTail])) {
          // Old tail is no longer in new list; remove
          removePart(oldParts[oldTail]!);
          oldTail--;
        } else {
          // Any mismatches at this point are due to additions or
          // moves; see if we have an old part we can reuse and move
          // into place
          const oldIndex = oldKeyToIndexMap.get(newKeys[newHead]);
          const oldPart = oldIndex !== undefined ? oldParts[oldIndex] : null;
          if (oldPart === null) {
            // No old part for this value; create a new one and
            // insert it
            const newPart = insertPart(containerPart, oldParts[oldHead]!);
            setChildPartValue(newPart, newValues[newHead]);
            newParts[newHead] = newPart;
          } else {
            // Reuse old part
            newParts[newHead] = setChildPartValue(oldPart, newValues[newHead]);
            insertPart(containerPart, oldParts[oldHead]!, oldPart);
            // This marks the old part as having been used, so that
            // it will be skipped in the first two checks above
            oldParts[oldIndex as number] = null;
          }
          newHead++;
        }
      }
    }
    // Add parts for any remaining new values
    while (newHead <= newTail) {
      // For all remaining additions, we insert before last new
      // tail, since old pointers are no longer valid
      const newPart = insertPart(containerPart, newParts[newTail + 1]);
      setChildPartValue(newPart, newValues[newHead]);
      newParts[newHead++] = newPart;
    }
    // Remove any remaining unused old parts
    while (oldHead <= oldTail) {
      const oldPart = oldParts[oldHead++];
      if (oldPart !== null) {
        removePart(oldPart);
      }
    }

    // Save order of new parts for next round
    this._itemKeys = newKeys;
    // Directly set part value, bypassing it's dirty-checking
    setCommittedValue(containerPart, newParts);
    return noChange;
  }
}

export interface RepeatDirectiveFn {
  <T>(
    items: Iterable<T>,
    keyFnOrTemplate: KeyFn<T> | ItemTemplate<T>,
    template?: ItemTemplate<T>
  ): unknown;
  <T>(items: Iterable<T>, template: ItemTemplate<T>): unknown;
  <T>(
    items: Iterable<T>,
    keyFn: KeyFn<T> | ItemTemplate<T>,
    template: ItemTemplate<T>
  ): unknown;
}

/**
 * A directive that repeats a series of values (usually `TemplateResults`)
 * generated from an iterable, and updates those items efficiently when the
 * iterable changes based on user-provided `keys` associated with each item.
 *
 * Note that if a `keyFn` is provided, strict key-to-DOM mapping is maintained,
 * meaning previous DOM for a given key is moved into the new position if
 * needed, and DOM will never be reused with values for different keys (new DOM
 * will always be created for new keys). This is generally the most efficient
 * way to use `repeat` since it performs minimum unnecessary work for insertions
 * and removals.
 *
 * The `keyFn` takes two parameters, the item and its index, and returns a unique key value.
 *
 * ```js
 * ${repeat(this.items, (item) => item.id, (item, index) =>
     html`<li>${index}: ${item.name}</li>`)}
 * ```
 *
 * **Important**: If providing a `keyFn`, keys *must* be unique for all items in a
 * given call to `repeat`. The behavior when two or more items have the same key
 * is undefined.
 *
 * If no `keyFn` is provided, this directive will perform similar to mapping
 * items to values, and DOM will be reused against potentially different items.
 */
export const repeat = directive(RepeatDirective) as RepeatDirectiveFn;

/**
 * The type of the class that powers this directive. Necessary for naming the
 * directive's return type.
 */
export type {RepeatDirective};