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3485. Longest Common Prefix of K Strings After Removal

Description

You are given an array of strings words and an integer k.

For each index i in the range [0, words.length - 1], find the length of the longest common prefix among any k strings (selected at distinct indices) from the remaining array after removing the ith element.

Return an array answer, where answer[i] is the answer for ith element. If removing the ith element leaves the array with fewer than k strings, answer[i] is 0.

 

Example 1:

Input: words = ["jump","run","run","jump","run"], k = 2

Output: [3,4,4,3,4]

Explanation:

  • Removing index 0 ("jump"):
    • words becomes: ["run", "run", "jump", "run"]. "run" occurs 3 times. Choosing any two gives the longest common prefix "run" (length 3).
  • Removing index 1 ("run"):
    • words becomes: ["jump", "run", "jump", "run"]. "jump" occurs twice. Choosing these two gives the longest common prefix "jump" (length 4).
  • Removing index 2 ("run"):
    • words becomes: ["jump", "run", "jump", "run"]. "jump" occurs twice. Choosing these two gives the longest common prefix "jump" (length 4).
  • Removing index 3 ("jump"):
    • words becomes: ["jump", "run", "run", "run"]. "run" occurs 3 times. Choosing any two gives the longest common prefix "run" (length 3).
  • Removing index 4 ("run"):
    • words becomes: ["jump", "run", "run", "jump"]. "jump" occurs twice. Choosing these two gives the longest common prefix "jump" (length 4).

Example 2:

Input: words = ["dog","racer","car"], k = 2

Output: [0,0,0]

Explanation:

  • Removing any index results in an answer of 0.

 

Constraints:

  • 1 <= k <= words.length <= 105
  • 1 <= words[i].length <= 104
  • words[i] consists of lowercase English letters.
  • The sum of words[i].length is smaller than or equal 105.

Solutions

Solution 1

Thinking

For each deleted word we want the LCP of any \(k\) remaining strings. The total length is \(\le 10^5\), so each query cannot rebuild the trie.

The maximum depth of a trie node with count \(\ge k\) is the global answer. Deleting a word only hurts ancestors whose count is exactly \(k\).

A segment tree keyed by depth stores how many nodes still have count \(\ge k\). We decrement the fragile depths of the deleted word, query the max surviving depth, and roll back. If \(n-1<k\), every answer is \(0\).

1


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class Solution {
    static class TrieNode {
        int count = 0;
        int depth = 0;
        int[] children = new int[26];

        TrieNode() {
            for (int i = 0; i < 26; ++i) children[i] = -1;
        }
    }

    static class SegmentTree {
        int n;
        int[] tree;
        int[] globalCount;

        SegmentTree(int n, int[] globalCount) {
            this.n = n;
            this.globalCount = globalCount;
            this.tree = new int[4 * (n + 1)];
            for (int i = 0; i < tree.length; i++) tree[i] = -1;
            build(1, 1, n);
        }

        void build(int idx, int l, int r) {
            if (l == r) {
                tree[idx] = globalCount[l] > 0 ? l : -1;
                return;
            }
            int mid = (l + r) / 2;
            build(idx * 2, l, mid);
            build(idx * 2 + 1, mid + 1, r);
            tree[idx] = Math.max(tree[idx * 2], tree[idx * 2 + 1]);
        }

        void update(int idx, int l, int r, int pos, int newVal) {
            if (l == r) {
                tree[idx] = newVal > 0 ? l : -1;
                return;
            }
            int mid = (l + r) / 2;
            if (pos <= mid) {
                update(idx * 2, l, mid, pos, newVal);
            } else {
                update(idx * 2 + 1, mid + 1, r, pos, newVal);
            }
            tree[idx] = Math.max(tree[idx * 2], tree[idx * 2 + 1]);
        }

        int query() {
            return tree[1];
        }
    }

    public int[] longestCommonPrefix(String[] words, int k) {
        int n = words.length;
        int[] ans = new int[n];
        if (n - 1 < k) return ans;

        ArrayList<TrieNode> trie = new ArrayList<>();
        trie.add(new TrieNode());

        for (String word : words) {
            int cur = 0;
            for (char c : word.toCharArray()) {
                int idx = c - 'a';
                if (trie.get(cur).children[idx] == -1) {
                    trie.get(cur).children[idx] = trie.size();
                    TrieNode node = new TrieNode();
                    node.depth = trie.get(cur).depth + 1;
                    trie.add(node);
                }
                cur = trie.get(cur).children[idx];
                trie.get(cur).count++;
            }
        }

        int maxDepth = 0;
        for (int i = 1; i < trie.size(); ++i) {
            if (trie.get(i).count >= k) {
                maxDepth = Math.max(maxDepth, trie.get(i).depth);
            }
        }

        int[] globalCount = new int[maxDepth + 1];
        for (int i = 1; i < trie.size(); ++i) {
            TrieNode node = trie.get(i);
            if (node.count >= k && node.depth <= maxDepth) {
                globalCount[node.depth]++;
            }
        }

        List<List<Integer>> fragileList = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            fragileList.add(new ArrayList<>());
        }

        for (int i = 0; i < n; ++i) {
            int cur = 0;
            for (char c : words[i].toCharArray()) {
                int idx = c - 'a';
                cur = trie.get(cur).children[idx];
                if (trie.get(cur).count == k) {
                    fragileList.get(i).add(trie.get(cur).depth);
                }
            }
        }

        int segSize = maxDepth;
        if (segSize >= 1) {
            SegmentTree segTree = new SegmentTree(segSize, globalCount);
            for (int i = 0; i < n; ++i) {
                if (n - 1 < k) {
                    ans[i] = 0;
                } else {
                    for (int d : fragileList.get(i)) {
                        segTree.update(1, 1, segSize, d, globalCount[d] - 1);
                    }
                    int res = segTree.query();
                    ans[i] = res == -1 ? 0 : res;
                    for (int d : fragileList.get(i)) {
                        segTree.update(1, 1, segSize, d, globalCount[d]);
                    }
                }
            }
        }

        return ans;
    }
}

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class Solution {
public:
    struct TrieNode {
        int count = 0;
        int depth = 0;
        int children[26] = {0};
    };

    class SegmentTree {
    public:
        int n;
        vector<int> tree;
        vector<int>& globalCount;
        SegmentTree(int n, vector<int>& globalCount)
            : n(n)
            , globalCount(globalCount) {
            tree.assign(4 * (n + 1), -1);
            build(1, 1, n);
        }
        void build(int idx, int l, int r) {
            if (l == r) {
                tree[idx] = globalCount[l] > 0 ? l : -1;
                return;
            }
            int mid = (l + r) / 2;
            build(idx * 2, l, mid);
            build(idx * 2 + 1, mid + 1, r);
            tree[idx] = max(tree[idx * 2], tree[idx * 2 + 1]);
        }
        void update(int idx, int l, int r, int pos, int newVal) {
            if (l == r) {
                tree[idx] = newVal > 0 ? l : -1;
                return;
            }
            int mid = (l + r) / 2;
            if (pos <= mid)
                update(idx * 2, l, mid, pos, newVal);
            else
                update(idx * 2 + 1, mid + 1, r, pos, newVal);
            tree[idx] = max(tree[idx * 2], tree[idx * 2 + 1]);
        }
        int query() {
            return tree[1];
        }
    };

    vector<int> longestCommonPrefix(vector<string>& words, int k) {
        int n = words.size();
        vector<int> ans(n, 0);
        if (n - 1 < k) return ans;
        vector<TrieNode> trie(1);
        for (const string& word : words) {
            int cur = 0;
            for (char c : word) {
                int idx = c - 'a';
                if (trie[cur].children[idx] == 0) {
                    trie[cur].children[idx] = trie.size();
                    trie.push_back({0, trie[cur].depth + 1});
                }
                cur = trie[cur].children[idx];
                trie[cur].count++;
            }
        }
        int maxDepth = 0;
        for (int i = 1; i < trie.size(); ++i) {
            if (trie[i].count >= k) {
                maxDepth = max(maxDepth, trie[i].depth);
            }
        }
        vector<int> globalCount(maxDepth + 1, 0);
        for (int i = 1; i < trie.size(); ++i) {
            if (trie[i].count >= k && trie[i].depth <= maxDepth) {
                globalCount[trie[i].depth]++;
            }
        }
        vector<vector<int>> fragileList(n);
        for (int i = 0; i < n; ++i) {
            int cur = 0;
            for (char c : words[i]) {
                int idx = c - 'a';
                cur = trie[cur].children[idx];
                if (trie[cur].count == k) {
                    fragileList[i].push_back(trie[cur].depth);
                }
            }
        }
        int segSize = maxDepth;
        if (segSize >= 1) {
            SegmentTree segTree(segSize, globalCount);
            for (int i = 0; i < n; ++i) {
                if (n - 1 < k) {
                    ans[i] = 0;
                } else {
                    for (int d : fragileList[i]) {
                        segTree.update(1, 1, segSize, d, globalCount[d] - 1);
                    }
                    int res = segTree.query();
                    ans[i] = res == -1 ? 0 : res;
                    for (int d : fragileList[i]) {
                        segTree.update(1, 1, segSize, d, globalCount[d]);
                    }
                }
            }
        }
        return ans;
    }
};

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