/**
 * @license
 * Copyright 2017 Google LLC
 * SPDX-License-Identifier: BSD-3-Clause
 */
import { noChange } from '../lit-html.js';
import { directive, Directive, PartType } from '../directive.js';
import { insertPart, getCommittedValue, removePart, setCommittedValue, setChildPartValue, } from '../directive-helpers.js';
// 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, start, end) => {
    const map = new Map();
    for (let i = start; i <= end; i++) {
        map.set(list[i], i);
    }
    return map;
};
class RepeatDirective extends Directive {
    constructor(partInfo) {
        super(partInfo);
        if (partInfo.type !== PartType.CHILD) {
            throw new Error('repeat() can only be used in text expressions');
        }
    }
    _getValuesAndKeys(items, keyFnOrTemplate, template) {
        let keyFn;
        if (template === undefined) {
            template = keyFnOrTemplate;
        }
        else if (keyFnOrTemplate !== undefined) {
            keyFn = keyFnOrTemplate;
        }
        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(items, keyFnOrTemplate, template) {
        return this._getValuesAndKeys(items, keyFnOrTemplate, template).values;
    }
    update(containerPart, [items, keyFnOrTemplate, template]) {
        // Old part & key lists are retrieved from the last update (which may
        // be primed by hydration)
        const oldParts = getCommittedValue(containerPart);
        const { values: newValues, keys: newKeys } = this._getValuesAndKeys(items, keyFnOrTemplate, template);
        // We check that oldParts, the committed value, is an Array as an
        // indicator that the previous value came from a repeat() call. If
        // oldParts is not an Array then this is the first render and we return
        // an array for lit-html's array handling to render, and remember the
        // keys.
        if (!Array.isArray(oldParts)) {
            this._itemKeys = newKeys;
            return newValues;
        }
        // In SSR hydration it's possible for oldParts to be an array but for us
        // to not have item keys because the update() hasn't run yet. We set the
        // keys to an empty array. This will cause all oldKey/newKey comparisons
        // to fail and execution to fall to the last nested brach below which
        // reuses the oldPart.
        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 = [];
        // 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;
        let oldKeyToIndexMap;
        // 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.
        //
        // * 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] = 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;
    }
}
/**
 * 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
 * html`
 *   <ol>
 *     ${repeat(this.items, (item) => item.id, (item, index) => {
 *       return html`<li>${index}: ${item.name}</li>`;
 *     })}
 *   </ol>
 * `
 * ```
 *
 * **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);
//# sourceMappingURL=repeat.js.map