Nine Chapter
  • Introduction
    • Summary
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      • 二叉树八股文:递归改迭代
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    • Introduction
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  • 4.Depth First Search+BackTracking
    • Summary
      • FloodFill 算法
    • 136.Palindrome Partitioning
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    • 77.Combinations (M)
    • 78.Subsets (M)
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    • 46.🌟Permutations
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    • 490.The Maze (M)
    • 51.N-Queens (H)
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    • 698. Partition to K Equal Sum Subsets (M)
    • 22. Generate Parentheses (M)
    • 岛屿问题
      • 200.Number of Islands (M)
      • 1254. Number of Closed Islands (M)
      • 1020. Number of Enclaves (M)
      • 695. Max Area of Island (M)
      • 1905. Count Sub Islands (M)
      • 694. Number of Distinct Islands
    • 131. Palindrome Partitioning (M)
    • 967. Numbers With Same Consecutive Differences (M)
    • 79. Word Search (M)
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    • 472. Concatenated Words (H)
    • Page 2
    • 291. Word Pattern II
    • 17. Letter Combinations of a Phone Number (M)
  • 5.LinkedList
    • Summary
      • 单链表的倒数第 k 个节点
      • Merge two/k sorted LinkedList
      • Middle of the Linked List
      • 判断链表是否包含环
      • 两个链表是否相交 Intersection of Two Linked Lists
      • 递归反转链表
      • 如何判断回文链表
    • 599.Insert into a Cyclic Sorted List
    • 21.Merge Two Sorted Lists (E)
    • 23.Merge k Sorted Lists (H)
    • 105.Copy List with Random Pointer
    • 141.Linked List Cycle (E)
    • 142.Linked List Cycle II (M)
    • 148.Sort List (M)
    • 86.Partition List (M)
    • 83.Remove Duplicates from Sorted List(E)
    • 82.Remove Duplicates from Sorted List II (M)
    • 206.Reverse Linked List (E)
    • 92.Reverse Linked List II (M)
    • 143.Reorder List (M)
    • 19.Remove Nth Node From End of List (E)
    • 170.Rotate List
    • 🤔25.Reverse Nodes in k-Group (H)
    • 452.Remove Linked List Elements
    • 167.Add Two Numbers
    • 221.Add Two Numbers II
    • 876. Middle of the Linked List (E)
    • 160. Intersection of Two Linked Lists (E)
    • 234. Palindrome Linked List (E)
    • 2130. Maximum Twin Sum of a Linked List (M)
  • 6.Array
    • Summary
      • 前缀和思路PrefixSum
      • 差分数组 Difference Array
      • 双指针Two Pointers
      • 滑动窗口算法算法
      • Sliding windows II
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      • 排序算法
      • 快速选择算法
    • 604.Window Sum
    • 138.Subarray Sum
    • 41.Maximum Subarray
    • 42.Maximum Subarray II
    • 43.Maximum Subarray III
    • 620.Maximum Subarray IV
    • 621.Maximum Subarray V
    • 6.Merge Two Sorted Arrays
    • 88.Merge Sorted Array
    • 547.Intersection of Two Arrays
    • 548.Intersection of Two Arrays II
    • 139.Subarray Sum Closest
    • 65.Median of two Sorted Arrays
    • 636.132 Pattern
    • 402.Continuous Subarray Sum
    • 303. Range Sum Query - Immutable (E)
    • 304.Range Sum Query 2D - Immutable (M)
    • 560. Subarray Sum Equals K (M)
    • 370. Range Addition(M)
    • 1109. Corporate Flight Bookings(M)
    • 1094. Car Pooling (M)
    • 76. Minimum Window Substring(H)
    • 567. Permutation in String (M)
    • 438. Find All Anagrams in a String(M)
    • 3. Longest Substring Without Repeating Characters (M)
    • 380. Insert Delete GetRandom O(1) (M)
    • 710. Random Pick with Blacklist (H)
    • 528. Random Pick with Weight (M)
    • 26. Remove Duplicates from Sorted Array (E)
    • 27. Remove Element (E)
    • 283. Move Zeroes (E)
    • 659. Split Array into Consecutive Subsequences (M)
    • 4. Median of Two Sorted Arrays (H)
    • 48. Rotate Image (M)
    • 54. Spiral Matrix (M)
    • 59. Spiral Matrix II (M)
    • 918. Maximum Sum Circular Subarray
    • 128. Longest Consecutive Sequence (M)
    • 238. Product of Array Except Self (M)
    • 1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit (M)
    • 1151. Minimum Swaps to Group All 1's Together (M)
    • 2134. Minimum Swaps to Group All 1's Together II
    • 2133. Check if Every Row and Column Contains All Numbers
    • 632. Smallest Range Covering Elements from K Lists (H)
    • 36. Valid Sudoku (M)
    • 383. Ransom Note
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  • 7.Two pointers
    • Summary
      • Two Sum
      • 2Sum 3Sum 4Sum 问题
    • 1.Two Sum I
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    • 609.Two Sum - Less than or equal to target
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    • 587.Two Sum - Unique pairs
    • 533.Two Sum - Closest to target
    • 443.Two Sum - Greater than target
    • 653. Two Sum IV - Input is a BST (M)
    • 57.3Sum
    • 59.3Sum Closest
    • 58.4Sum
    • 148.Sort Colors
    • 143.Sort Colors II
    • 31.Partition Array
    • 625.Partition Array II
    • 382.Triangle Count
      • 611. Valid Triangle Number
    • 521.Remove Duplicate Numbers in Array
    • 167. Two Sum II - Input Array Is Sorted (E)
    • 870. Advantage Shuffle (M)
    • 9. Palindrome Number (E)
    • 125. Valid Palindrome(E)
    • 5. Longest Palindromic Substring (M)
    • 42. Trapping Rain Water
    • 11. Container With Most Water (M)
    • 658. Find K Closest Elements (M)
    • 392. Is Subsequence
  • 8.Data Structure
    • Summary
      • 数据结构的存储方式
      • 单调栈
      • 单调队列
      • 二叉堆 Binary Heap
      • TreeMap
      • TreeSet
      • 🌟Trie
      • Trie Application
    • 155. Min Stack (E)
    • 716. Max Stack (E)
    • 1648. Sell Diminishing-Valued Colored Balls
    • 232. Implement Queue using Stacks (E)
    • 225. Implement Stack using Queues(E)
    • 84.Largest Rectangle in Histogram
    • 128.Hash Function
    • Max Tree
    • 544.Top k Largest Numbers
    • 545.Top k Largest Numbers II
    • 613.High Five
    • 606.Kth Largest Element II
    • 5.Kth Largest Element
    • 129.Rehashing
    • 4.Ugly Number II
    • 517.Ugly Number
    • 28. Implement strStr()
    • 594.strStr II
    • 146.LRU Cache
    • 460.LFU Cache
    • 486.Merge k Sorted Arrays
    • 130.Heapify
    • 215. Kth Largest Element in an Array (M)
    • 612.K Closest Points
    • 692. Top K Frequent Words
    • 347.Top K Frequent Elements
    • 601.Flatten 2D Vector
    • 540.Zigzag Iterator
    • 541.Zigzag Iterator II
    • 423.Valid Parentheses
    • 488.Happy Number
    • 547.Intersection of Two Arrays
    • 548.Intersection of Two Arrays II
    • 627.Longest Palindrome
    • 638.Strings Homomorphism
    • 138.Subarray Sum
    • 647.Substring Anagrams
    • 171.Anagrams
    • 739. Daily Temperatures(M)
    • 496. Next Greater Element I (E)
    • 503. Next Greater Element II(M)
    • 316. Remove Duplicate Letters(M) & 1081. Smallest Subsequence of Distinct Characters
    • 239. Sliding Window Maximum (H)
    • 355. Design Twitter (M)
    • 895. Maximum Frequency Stack (H)
    • 20. Valid Parentheses (E)
    • 921. Minimum Add to Make Parentheses Valid (M)
    • 1541. Minimum Insertions to Balance a Parentheses String (M)
    • 32. Longest Valid Parentheses (H)
    • Basic Calculator (*)
      • 224. Basic Calculator
      • 227. Basic Calculator II (M)
    • 844. Backspace String Compare
    • 295. Find Median from Data Stream
    • 208. Implement Trie (Prefix Tree)
    • 461.Kth Smallest Numbers in Unsorted Array
    • 1152.Analyze user website visit pattern
    • 811. Subdomain Visit Count (M)
    • 71. Simplify Path (M)
    • 362. Design Hit Counter
  • 9.Dynamic Programming
    • Summary
      • 最优子结构 Optimal Sustructure
      • 子序列解题模板
      • 空间压缩
      • 背包问题
        • Untitled
      • 股票买卖问题
      • KMP
    • 109.Triangle
    • 110.Minimum Path Sum
    • 114.Unique Paths
    • 115.Unique Paths II
    • 70.Climbing Stairs
    • 272.Climbing StairsII
    • 116.Jump Game
    • 117.Jump Game II
    • 322.Coin Change
    • 518. Coin Change 2 ()
    • Backpack I~VI
      • LintCode 563.Backpack V (M)
    • Best Time to Buy and Sell Stock(*)
      • 121. Best Time to Buy and Sell Stock
      • 122. Best Time to Buy and Sell Stock II (M)
      • 123. Best Time to Buy and Sell Stock III (H)
      • 188. Best Time to Buy and Sell Stock IV (H)
      • 309. Best Time to Buy and Sell Stock with Cooldown (M)
      • 714. Best Time to Buy and Sell Stock with Transaction Fee (M)
    • 394.Coins in a line
    • 395.Coins in a Line II
    • 509. Fibonacci Number (E)
    • 931. Minimum Falling Path Sum (M)
    • 494. Target Sum (M)
    • 72. Edit Distance (H)
    • 300.Longest Increasing Subsequence
    • 1143. Longest Common Subsequence (M)
    • 718. Maximum Length of Repeated Subarray
    • 583. Delete Operation for Two Strings (M)
    • 712. Minimum ASCII Delete Sum for Two Strings(M)
    • 53. Maximum Subarray (E)
    • 516. Longest Palindromic Subsequence (M)
    • 1312. Minimum Insertion Steps to Make a String Palindrome (H)
    • 416. Partition Equal Subset Sum (M)
    • 64. Minimum Path Sum(M)
    • 651. 4 Keys Keyboards (M)
    • House Robber (*)
      • 198. House Robber (M)
      • 213. House Robbber II
      • 337. House Robber III (M)
    • Word Break (*)
      • 139.Word Break (M)
    • 140. Word Break II (H)
    • 828. Count Unique Characters of All Substrings of a Given String (H)
    • 174. Dungeon Game (H)
    • 1567. Maximum Length of Subarray With Positive Product (M)
  • 10. Graph
    • Introduction
      • 有向图的环检测
      • 拓扑排序
      • 二分图判定
      • Union-Find
      • 最小生成树(Minimum Spanning Tree)算法
        • KRUSKAL 最小生成树算法
        • Prim 最小生成树算法
      • Dijkstra 最短路径算法
      • BFS vs DFS
    • 797. All Paths From Source to Target (M)
    • 785. Is Graph Bipartite? (M)
    • 886. Possible Bipartition (M)
    • 130. Surrounded Regions (M)
    • 990. Satisfiability of Equality Equations (M)
    • 721. Accounts Merge (M)
    • 323. Number of Connected Components in an Undirected Graph (M)
    • 261. Graph Valid Tree
    • 1135. Connecting Cities With Minimum Cost
    • 1584. Min Cost to Connect All Points (M)
    • 277. Find the Celebrity (M)
    • 743. Network Delay Time (M)
    • 1631. Path With Minimum Effort (M)
    • 1514. Path with Maximum Probability (M)
    • 589.Connecting Graph
    • 🌟787. Cheapest Flights Within K Stops (M)
    • 2050. Parallel Courses III (H)
    • 1293. Shortest Path in a Grid with Obstacles Elimination (H)
    • 864. Shortest Path to Get All Keys (H)
    • 269. Alien Dictionary (H)
    • 1192. Critical Connections in a Network (H)
    • 529. Minesweeper (M)
  • 11.Math
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  1. 6.Array

710. Random Pick with Blacklist (H)

https://leetcode.com/problems/random-pick-with-blacklist/

You are given an integer n and an array of unique integers blacklist. Design an algorithm to pick a random integer in the range [0, n - 1] that is not in blacklist. Any integer that is in the mentioned range and not in blacklist should be equally likely to be returned.

Optimize your algorithm such that it minimizes the number of calls to the built-in random function of your language.

Implement the Solution class:

  • Solution(int n, int[] blacklist) Initializes the object with the integer n and the blacklisted integers blacklist.

  • int pick() Returns a random integer in the range [0, n - 1] and not in blacklist.

Example 1:

Input
["Solution", "pick", "pick", "pick", "pick", "pick", "pick", "pick"]
[[7, [2, 3, 5]], [], [], [], [], [], [], []]
Output
[null, 0, 4, 1, 6, 1, 0, 4]

Explanation
Solution solution = new Solution(7, [2, 3, 5]);
solution.pick(); // return 0, any integer from [0,1,4,6] should be ok. Note that for every call of pick,
                 // 0, 1, 4, and 6 must be equally likely to be returned (i.e., with probability 1/4).
solution.pick(); // return 4
solution.pick(); // return 1
solution.pick(); // return 6
solution.pick(); // return 1
solution.pick(); // return 0
solution.pick(); // return 4

Constraints:

  • 1 <= n <= 109

  • 0 <= blacklist.length <- min(105, n - 1)

  • 0 <= blacklist[i] < n

  • All the values of blacklist are unique.

  • At most 2 * 104 calls will be made to pick.

Solution:

Version 2:

pick 函数会被多次调用,每次调用都要在区间 [0,N) 中「等概率随机」返回一个「不在 blacklist 中」的整数。

这应该不难理解吧,比如给你输入 N = 5, blacklist = [1,3],那么多次调用 pick 函数,会等概率随机返回 0, 2, 4 中的某一个数字。

而且题目要求,在 pick 函数中应该尽可能少调用随机数生成函数 rand()。

这句话什么意思呢,比如说我们可能想出如下拍脑袋的解法:

int pick() {
    int res = rand() % N;
    while (res exists in blacklist) {
        // 重新随机一个结果
        res = rand() % N;
    }
    return res;
}

这个函数会多次调用 rand() 函数,执行效率竟然和随机数相关,不是一个漂亮的解法。

聪明的解法类似上一道题,我们可以将区间 [0,N) 看做一个数组,然后将 blacklist 中的元素移到数组的最末尾,同时用一个哈希表进行映射:

根据这个思路,我们可以写出第一版代码(还存在几处错误):

class Solution {
public:
    int sz;
    unordered_map<int, int> mapping;
    
    Solution(int N, vector<int>& blacklist) {
        // 最终数组中的元素个数
        sz = N - blacklist.size();
        // 最后一个元素的索引
        int last = N - 1;
        // 将黑名单中的索引换到最后去
        for (int b : blacklist) {
            mapping[b] = last;
            last--;
        }
    }
};

如上图,相当于把黑名单中的数字都交换到了区间 [sz, N) 中,同时把 [0, sz) 中的黑名单数字映射到了正常数字。

根据这个逻辑,我们可以写出 pick 函数:

int pick() {
    // 随机选取一个索引
    int index = rand() % sz;
    // 这个索引命中了黑名单,
    // 需要被映射到其他位置
    if (mapping.count(index)) {
        return mapping[index];
    }
    // 若没命中黑名单,则直接返回
    return index;
}

这个 pick 函数已经没有问题了,但是构造函数还有两个问题。

第一个问题,如下这段代码:

int last = N - 1;
// 将黑名单中的索引换到最后去
for (int b : blacklist) {
    mapping[b] = last;
    last--;
}

我们将黑名单中的 b 映射到 last,但是我们能确定 last 不在 blacklist 中吗?

比如下图这种情况,我们的预期应该是 1 映射到 3,但是错误地映射到 4:

在对 mapping[b] 赋值时,要保证 last 一定不在 blacklist 中,可以如下操作:

// 构造函数
Solution(int N, vector<int>& blacklist) {
    sz = N - blacklist.size();
    // 先将所有黑名单数字加入 map
    for (int b : blacklist) { 
        // 这里赋值多少都可以
        // 目的仅仅是把键存进哈希表
        // 方便快速判断数字是否在黑名单内
        mapping[b] = 666;
    }

    int last = N - 1;
    for (int b : blacklist) {
        // 跳过所有黑名单中的数字
        while (mapping.count(last)) {
            last--;
        }
        // 将黑名单中的索引映射到合法数字
        mapping[b] = last;
        last--;
    }
}

第二个问题,如果 blacklist 中的黑名单数字本身就存在区间 [sz, N) 中,那么就没必要在 mapping 中建立映射,比如这种情况:

我们根本不用管 4,只希望把 1 映射到 3,但是按照 blacklist 的顺序,会把 4 映射到 3,显然是错误的。

我们可以稍微修改一下,写出正确的解法代码:

class Solution {
public:
    int sz;
    unordered_map<int, int> mapping;
    
    Solution(int N, vector<int>& blacklist) {
        sz = N - blacklist.size();
        for (int b : blacklist) {
            mapping[b] = 666;
        }

        int last = N - 1;
        for (int b : blacklist) {
            // 如果 b 已经在区间 [sz, N)
            // 可以直接忽略
            if (b >= sz) {
                continue;
            }
            while (mapping.count(last)) {
                last--;
            }
            mapping[b] = last;
            last--;
        }
    }

    // 见上文代码实现
    int pick() {}
};

至此,这道题也解决了,总结一下本文的核心思想:

1、如果想高效地,等概率地随机获取元素,就要使用数组作为底层容器。

2、如果要保持数组元素的紧凑性,可以把待删除元素换到最后,然后 pop 掉末尾的元素,这样时间复杂度就是 O(1) 了。当然,我们需要额外的哈希表记录值到索引的映射。

3、对于第二题,数组中含有「空洞」(黑名单数字),也可以利用哈希表巧妙处理映射关系,让数组在逻辑上是紧凑的,方便随机取元素。

JAVA Version:

HashMap Remapping

  1. Split these N numbers into two parts from 0 to N - M - 1 and from N - M to N - 1

  2. Now, we remap the blacklisted numbers on the left part to the white ones

    1. Find all the white numbers on the right

    2. Loop through the blacklist and remap the number that are less than N - M to the numbers above

  3. We can simply generate an random number and pick from the map, and return the default k if it's not present in the map

  4. Time complexity O(B)

  5. Space complexity O(B)

class Solution {
    private Random random;
    private Map<Integer, Integer> map;
    private int L;
    public Solution(int N, int[] blacklist) {
        this.random = new Random();
        this.map = new HashMap<>();
        this.L = N - blacklist.length;
        Set<Integer> set = new HashSet<>();
        for (int i = L; i < N; i++) set.add(i);
        for (int b : blacklist) set.remove(b);
        Iterator<Integer> it = set.iterator();
        for (int b : blacklist) {
            if (b < L) {
                map.put(b, it.next());
            }
        }
    }

    public int pick() {
        int k = random.nextInt(L);
        return map.getOrDefault(k, k);
    }
}
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