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leetcode/1057. Campus Bikes/README.md

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+# [1057. Campus Bikes (Medium)](https://leetcode.com/problems/campus-bikes/)
+
+<p>On a campus represented on the X-Y plane, there are <code>n</code> workers and <code>m</code> bikes, with <code>n &lt;= m</code>.</p>
+
+<p>You are given an array <code>workers</code> of length <code>n</code> where <code>workers[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> is the position of the <code>i<sup>th</sup></code> worker. You are also given an array <code>bikes</code> of length <code>m</code> where <code>bikes[j] = [x<sub>j</sub>, y<sub>j</sub>]</code> is the position of the <code>j<sup>th</sup></code> bike. All the given positions are <strong>unique</strong>.</p>
+
+<p>Assign a bike to each worker. Among the available bikes and workers, we choose the <code>(worker<sub>i</sub>, bike<sub>j</sub>)</code> pair with the shortest <strong>Manhattan distance</strong> between each other and assign the bike to that worker.</p>
+
+<p>If there are multiple <code>(worker<sub>i</sub>, bike<sub>j</sub>)</code> pairs with the same shortest <strong>Manhattan distance</strong>, we choose the pair with <strong>the smallest worker index</strong>. If there are multiple ways to do that, we choose the pair with <strong>the smallest bike index</strong>. Repeat this process until there are no available workers.</p>
+
+<p>Return <em>an array </em><code>answer</code><em> of length </em><code>n</code><em>, where </em><code>answer[i]</code><em> is the index (<strong>0-indexed</strong>) of the bike that the </em><code>i<sup>th</sup></code><em> worker is assigned to</em>.</p>
+
+<p>The <strong>Manhattan distance</strong> between two points <code>p1</code> and <code>p2</code> is <code>Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/03/06/1261_example_1_v2.png" style="width: 376px; height: 366px;">
+<pre><strong>Input:</strong> workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
+<strong>Output:</strong> [1,0]
+<strong>Explanation:</strong> Worker 1 grabs Bike 0 as they are closest (without ties), and Worker 0 is assigned Bike 1. So the output is [1, 0].
+</pre>
+
+<p><strong>Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/03/06/1261_example_2_v2.png" style="width: 376px; height: 366px;">
+<pre><strong>Input:</strong> workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
+<strong>Output:</strong> [0,2,1]
+<strong>Explanation:</strong> Worker 0 grabs Bike 0 at first. Worker 1 and Worker 2 share the same distance to Bike 2, thus Worker 1 is assigned to Bike 2, and Worker 2 will take Bike 1. So the output is [0,2,1].
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == workers.length</code></li>
+	<li><code>m == bikes.length</code></li>
+	<li><code>1 &lt;= n &lt;= m &lt;= 1000</code></li>
+	<li><code>workers[i].length == bikes[j].length == 2</code></li>
+	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt; 1000</code></li>
+	<li><code>0 &lt;= x<sub>j</sub>, y<sub>j</sub> &lt; 1000</code></li>
+	<li>All worker and bike locations are <strong>unique</strong>.</li>
+</ul>
+
+
+**Companies**:  
+[Google](https://leetcode.com/company/google)
+
+**Related Topics**:  
+[Array](https://leetcode.com/tag/array/), [Greedy](https://leetcode.com/tag/greedy/), [Sorting](https://leetcode.com/tag/sorting/)
+
+**Similar Questions**:
+* [Campus Bikes II (Medium)](https://leetcode.com/problems/campus-bikes-ii/)
+
+## Solution 1. Sorting
+
+```cpp
+// OJ: https://leetcode.com/problems/campus-bikes/
+// Author: github.com/lzl124631x
+// Time: O(MNlog(MN))
+// Space: O(MN)
+class Solution {
+public:
+    vector<int> assignBikes(vector<vector<int>>& W, vector<vector<int>>& B) {
+        int N = W.size(), M = B.size();
+        vector<array<int, 3>> v;
+        auto dist = [](auto &a, auto &b) {
+            return abs(a[0] - b[0]) + abs(a[1] - b[1]);
+        };
+        for (int i = 0; i < N; ++i) {
+            for (int j = 0; j < M; ++j) {
+                v.push_back({dist(W[i], B[j]), i, j});
+            }
+        }
+        vector<bool> used(M);
+        sort(begin(v), end(v));
+        vector<int> ans(N, -1);
+        for (int i = 0, cnt = 0; i < v.size() && cnt < N; ++i) {
+            if (used[v[i][2]] || ans[v[i][1]] != -1) continue;
+            used[v[i][2]] = true;
+            ans[v[i][1]] = v[i][2];
+            ++cnt;
+        }
+        return ans;
+    }
+};
+```