Solution: Shortest Path in Binary Matrix

Leetcode Solutions (161 Part Series)

1 Solution: Next Permutation
2 Solution: Trim a Binary Search Tree
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3 Leetcode Solutions Index
4 Solution: Minimize Deviation in Array
5 Solution: Vertical Order Traversal of a Binary Tree
6 Solution: Count Ways to Make Array With Product
7 Solution: Smallest String With A Given Numeric Value
8 Solution: Linked List Cycle
9 Solution: Path With Minimum Effort
10 Solution: Concatenation of Consecutive Binary Numbers
11 Solution: Minimum Operations to Make a Subsequence
12 Solution: Longest Harmonious Subsequence
13 Solution: Simplify Path
14 Solution: Building Boxes
15 Solution: Decode XORed Permutation
16 Solution: Binary Tree Right Side View
17 Solution: Find Kth Largest XOR Coordinate Value
18 Solution: Change Minimum Characters to Satisfy One of Three Conditions
19 Solution: Shortest Distance to a Character
20 Solution: Peeking Iterator
21 Solution: Convert BST to Greater Tree
22 Solution: Copy List with Random Pointer
23 Solution: Valid Anagram
24 Solution: Number of Steps to Reduce a Number to Zero
25 Solution: Shortest Path in Binary Matrix
26 Solution: Is Graph Bipartite?
27 Solution: Maximum Score From Removing Substrings (ver. 1)
28 Solution: Maximum Score From Removing Substrings (ver. 2)
29 Solution: Sort the Matrix Diagonally
30 Solution: The K Weakest Rows in a Matrix (ver. 1)
31 Solution: The K Weakest Rows in a Matrix (ver. 2)
32 Solution: Letter Case Permutation
33 Solution: Container With Most Water
34 Solution: Arithmetic Slices
35 Solution: Minimum Remove to Make Valid Parentheses
36 Solution: Roman to Integer
37 Solution: Broken Calculator
38 Solution: Find the Most Competitive Subsequence
39 Solution: Longest Word in Dictionary through Deleting
40 Solution: Search a 2D Matrix II
41 Solution: Score of Parentheses
42 Solution: Shortest Unsorted Continuous Subarray
43 Solution: Validate Stack Sequences
44 Solution: Divide Two Integers (ver. 1)
45 Solution: Divide Two Integers (ver. 2)
46 Solution: Maximum Frequency Stack
47 Solution: Distribute Candies
48 Solution: Set Mismatch (ver. 1)
49 Solution: Set Mismatch (ver. 2)
50 Solution: Missing Number
51 Solution: Intersection of Two Linked Lists
52 Solution: Average of Levels in Binary Tree
53 Solution: Short Encoding of Words (ver. 1)
54 Solution: Design HashMap (ver. 1)
55 Solution: Short Encoding of Words (ver. 2)
56 Solution: Design HashMap (ver. 2)
57 Solution: Remove Palindromic Subsequences
58 Solution: Add One Row to Tree
59 Solution: Integer to Roman
60 Solution: Coin Change
61 Solution: Check If a String Contains All Binary Codes of Size K
62 Solution: Binary Trees With Factors
63 Solution: Swapping Nodes in a Linked List
64 Solution: Encode and Decode TinyURL
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67 Solution: Wiggle Subsequence
68 Solution: Keys and Rooms
69 Solution: Design Underground System
70 Solution: Reordered Power of 2
71 Solution: Vowel Spellchecker
72 Solution: 3Sum With Multiplicity
73 Solution: Advantage Shuffle
74 Solution: Pacific Atlantic Water Flow
75 Solution: Word Subsets
76 Solution: Palindromic Substrings
77 Solution: Reconstruct Original Digits from English
78 Solution: Flip Binary Tree To Match Preorder Traversal
79 Solution: Russian Doll Envelopes
80 Solution: Stamping The Sequence
81 Solution: Palindrome Linked List
82 Solution: Ones and Zeroes
83 Solution: Longest Valid Parentheses
84 Solution: Design Circular Queue
85 Solution: Global and Local Inversions
86 Solution: Minimum Operations to Make Array Equal
87 Solution: Determine if String Halves Are Alike
88 Solution: Letter Combinations of a Phone Number
89 Solution: Verifying an Alien Dictionary
90 Solution: Longest Increasing Path in a Matrix
91 Solution: Deepest Leaves Sum
92 Solution: Beautiful Arrangement II
93 Solution: Flatten Nested List Iterator
94 Solution: Partition List
95 Solution: Fibonacci Number
96 Solution: Remove All Adjacent Duplicates in String II
97 Solution: Number of Submatrices That Sum to Target
98 Solution: Remove Nth Node From End of List
99 Solution: Combination Sum IV
100 Solution: N-ary Tree Preorder Traversal
101 Solution: Triangle
102 Solution: Brick Wall
103 Solution: Count Binary Substrings
104 Solution: Critical Connections in a Network
105 Solution: Rotate Image
106 Solution: Furthest Building You Can Reach
107 Solution: Power of Three
108 Solution: Unique Paths II
109 Solution: Find First and Last Position of Element in Sorted Array
110 Solution: Powerful Integers
111 Solution: Prefix and Suffix Search
112 Solution: Course Schedule III
113 Solution: Running Sum of 1d Array
114 Solution: Non-decreasing Array
115 Solution: Jump Game II
116 Solution: Convert Sorted List to Binary Search Tree
117 Solution: Delete Operation for Two Strings
118 Solution: Super Palindromes
119 Solution: Construct Target Array With Multiple Sums
120 Solution: Count Primes
121 Solution: Maximum Points You Can Obtain from Cards
122 Solution: Range Sum Query 2D – Immutable
123 Solution: Ambiguous Coordinates
124 Solution: Flatten Binary Tree to Linked List
125 Solution: Valid Number
126 Solution: Binary Tree Cameras
127 Solution: Longest String Chain
128 Solution: Find Duplicate File in System
129 Solution: Minimum Moves to Equal Array Elements II
130 Solution: Binary Tree Level Order Traversal
131 Solution: Find and Replace Pattern
132 Solution: N-Queens
133 Solution: To Lower Case
134 Solution: Evaluate Reverse Polish Notation
135 Solution: Partitioning Into Minimum Number Of Deci-Binary Numbers
136 Solution: Maximum Product of Word Lengths
137 Solution: Maximum Erasure Value
138 Solution: N-Queens II
139 Solution: Maximum Gap
140 Solution: Search Suggestions System
141 Solution: Max Area of Island
142 Solution: Interleaving String
143 Solution: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts
144 Solution: Open the Lock
145 Solution: Maximum Performance of a Team
146 Solution: Longest Consecutive Sequence
147 Solution: Min Cost Climbing Stairs
148 Solution: Construct Binary Tree from Preorder and Inorder Traversal
149 Solution: Jump Game VI
150 Solution: My Calendar I
151 Solution: Stone Game VII
152 Solution: Minimum Number of Refueling Stops
153 Solution: Palindrome Pairs
154 Solution: Maximum Units on a Truck
155 Solution: Matchsticks to Square
156 Solution: Generate Parentheses
157 Solution: Number of Subarrays with Bounded Maximum
158 Solution: Swim in Rising Water
159 Solution: Pascal’s Triangle
160 Solution: Out of Boundary Paths
161 Solution: Redundant Connection

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Leetcode Problem #1091 (Medium): Shortest Path in Binary Matrix

Description:

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
C_1 is at location (0, 0) (ie. has value grid[0][0])
C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right. If such a path does not exist, return -1.

Examples:

Example 1:
Input:	[[0,1],[1,0]]
Output:	2
Visual:

Example 2:
Input:	[[0,0,0],[1,1,0],[1,1,0]]
Output:	4
Visual:

Constraints:

1 <= grid.length == grid[0].length <= 100
grid[r][c] is 0 or 1

Idea:

When we’re asked about finding the “shortest path”, the first thing that should come to mind is a breadth-first solution (BFS) approach. In a standard graph BFS solution, we set up a queue (q) and fill it with our starting position (grid[0][0]). Then we keep pulling entries from q, figuring out the next moves from that position, and input those next moves back into q.

When we’re ready to start, we can change grid[0][0] to 1, then as we reach new cells, we can store the distance to that cell in the cell at the same time we add it to the queue. The distance will simply be one more than the distance to the cell we’re moving from. This will also eliminate duplicate queue entries by changing visited cells to a non-zero number.

Through the nature of a BFS approach to graph traversal (with non-weighted edges), the first time we reach the end location (grid[n][n]) will represent the best possible distance.

Since 0 <= i, j <= 100, both i and j will fit into 7 bits each, so we can utilize bit manipulation to store both in one integer. With a bitwise left shift (<<) we can move the value of j to the left by 7 bits before adding it to i to allow for both to fit in 14 bits of an integer.

Bitwise shift example:

   i  =  93 (base 10)  =  1011101 (base 2)
   j  =  75 (base 10)  =  1001011 (base 2)

   j << 7  =  1001011<<<<<<<     // Bitwise left shift moves the value left
           =  10010110000000     // And fills in the space with 0s

   i:                           1011101 
        j << 7:       +  10010110000000
                        ----------------
   i + (j << 7):      =  10010111011101 (base 2)
                      =            9693 (base 10)

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To read i from the first 7 bits of our stored integer again, you can use bitwise AND (&) and a bitmask of 1111111. The easiest way to get a bitmask of 1111111 is to shift a single bit to the left by 7 (1 << 7 = 10000000) and then subtract 1, rolling it back to all 1s.

Bitmask example:

   1 << 7:               10000000
                      -         1
                        ----------
   (1 << 7) - 1:      =   1111111

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The bitwise AND will only keep any bits that have a 1 in both numbers, thus stripping away anything except the first 7 bits of data.

Bitwise AND example:

      10010111011101
   &         1111111
     ----------------
   =         1011101

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To read the j value from our integer, we can just shift it to the right by 7 bits, which will throw away the first 7 bits of data corresponding to the i value.

If q becomes empty without finding a path to the end, then return -1.

Implementation:

If either the starting point or the ending point are a 1, then we quickly return -1.

To check which moves can be made, we can just iterate over a three-value range for each i and j, and to make sure that they remain in bounds, we can apply a max and min to the range.

Javascript Code:

var shortestPathBinaryMatrix = function(grid) {
    let n = grid.length - 1, q = [0]
    if (grid[0][0] || grid[n][n]) return -1
    grid[0][0] = 1
    while (q.length) {
        let curr = q.shift(), i = curr & (1 << 7) - 1, j = curr >> 7
        if (i === n && j === n) return grid[n][n]
        for (let a = Math.max(i-1,0); a <= Math.min(i+1,n); a++)
            for (let b = Math.max(j-1,0); b <= Math.min(j+1,n); b++)
                if (grid[a][b] === 0)
                    grid[a][b] = grid[i][j] + 1, q.push(a + (b << 7))
    }
    return -1
};

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Python Code:

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        n = len(grid)-1
        if grid[0][0] or grid[n][n]: return -1
        q, grid[0][0] = [0], 1
        while len(q):
            curr = q.pop(0)
            i, j = curr & ((1 << 7) - 1), curr >> 7
            if i == n and j == n: return grid[n][n]
            for a in range(max(i-1,0),min(i+2,n+1)):
                for b in range(max(j-1,0),min(j+2,n+1)):
                    if grid[a][b] == 0:
                        grid[a][b] = grid[i][j] + 1
                        q.append(a + (b << 7))
        return -1

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Java Code:

class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        int n = grid.length - 1;
        Queue<Integer> q = new ArrayDeque<Integer>();
        q.add(0);
        if (grid[0][0] == 1 || grid[n][n] == 1) return -1;
        grid[0][0] = 1;
        while (q.size() > 0) {
            int curr = q.remove(), i = curr & (1 << 7) - 1, j = curr >> 7;
            if (i == n && j == n) return grid[n][n];
            for (int a = Math.max(i-1,0); a <= Math.min(i+1,n); a++)
                for (int b = Math.max(j-1,0); b <= Math.min(j+1,n); b++)
                    if (grid[a][b] == 0) {
                        grid[a][b] = grid[i][j] + 1;
                        q.add(a + (b << 7));
                    }
        }
        return -1;
    }
}

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C++ Code:

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int m = grid.size(), n = m - 1;
        std::queue<int> q;
        q.push(0);
        if (grid[0][0] == 1 || grid[n][n] == 1) return -1;
        grid[0][0] = 1;
        while (q.size() > 0) {
            int curr = q.front();
            q.pop();
            int i = curr & (1 << 7) - 1, j = curr >> 7;
            if (i == n && j == n) return grid[n][n];
            for (int a = std::max(i-1,0); a <= std::min(i+1,n); a++)
                for (int b = std::max(j-1,0); b <= std::min(j+1,n); b++)
                    if (grid[a][b] == 0) {
                        grid[a][b] = grid[i][j] + 1;
                        q.push(a + (b << 7));
                    }
        }
        return -1;
    }
};

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Leetcode Solutions (161 Part Series)

原文链接：Solution: Shortest Path in Binary Matrix

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