feat: add solution 1787. Make the XOR of All Segments Equal to Zero

Author: tzuyi0817

solutions/1787. Make the XOR of All Segments Equal to Zero/README.md

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+# [1787. Make the XOR of All Segments Equal to Zero](https://leetcode.com/problems/make-the-xor-of-all-segments-equal-to-zero)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given an array <code>nums</code>​​​ and an integer <code>k</code>​​​​​. The <font face="monospace">XOR</font> of a segment <code>[left, right]</code> where <code>left &lt;= right</code> is the <code>XOR</code> of all the elements with indices between <code>left</code> and <code>right</code>, inclusive: <code>nums[left] XOR nums[left+1] XOR ... XOR nums[right]</code>.</p>
+
+<p>Return <em>the minimum number of elements to change in the array </em>such that the <code>XOR</code> of all segments of size <code>k</code>​​​​​​ is equal to zero.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> nums = [1,2,0,3,0], k = 1
+<strong>Output:</strong> 3
+<strong>Explanation: </strong>Modify the array from [<u><strong>1</strong></u>,<u><strong>2</strong></u>,0,<u><strong>3</strong></u>,0] to from [<u><strong>0</strong></u>,<u><strong>0</strong></u>,0,<u><strong>0</strong></u>,0].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> nums = [3,4,5,2,1,7,3,4,7], k = 3
+<strong>Output:</strong> 3
+<strong>Explanation: </strong>Modify the array from [3,4,<strong><u>5</u></strong>,<strong><u>2</u></strong>,<strong><u>1</u></strong>,7,3,4,7] to [3,4,<strong><u>7</u></strong>,<strong><u>3</u></strong>,<strong><u>4</u></strong>,7,3,4,7].
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre><strong>Input:</strong> nums = [1,2,4,1,2,5,1,2,6], k = 3
+<strong>Output:</strong> 3
+<strong>Explanation: </strong>Modify the array from [1,2,<strong><u>4,</u></strong>1,2,<strong><u>5</u></strong>,1,2,<strong><u>6</u></strong>] to [1,2,<strong><u>3</u></strong>,1,2,<strong><u>3</u></strong>,1,2,<strong><u>3</u></strong>].</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= nums.length &lt;= 2000</code></li>
+	<li><code>​​​​​​0 &lt;= nums[i] &lt; 2<sup>10</sup></code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Dynamic Programming`**
+
+- Time complexity: <em>O(n<sup>2</sup>)</em>
+- Space complexity: <em>O(nk)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+const minChanges = function (nums, k) {
+  const MAX_XOR = 1024;
+  const n = nums.length;
+  const counts = Array.from({ length: k }, () => new Map());
+  const dp = Array.from({ length: k }, () => new Array(MAX_XOR).fill(n));
+
+  const getGroupCount = index => Math.floor(n / k) + (n % k > index ? 1 : 0);
+
+  for (let index = 0; index < n; index++) {
+    const num = nums[index];
+    const group = counts[index % k];
+    const count = group.get(num) ?? 0;
+
+    group.set(num, count + 1);
+  }
+
+  for (let xor = 0; xor < MAX_XOR; xor++) {
+    const count = counts[k - 1].get(xor) ?? 0;
+
+    dp[k - 1][xor] = getGroupCount(k - 1) - count;
+  }
+
+  for (let index = k - 2; index >= 0; index--) {
+    const groupCount = getGroupCount(index);
+    const nextMin = Math.min(...dp[index + 1]);
+
+    for (let xor = 0; xor < MAX_XOR; xor++) {
+      dp[index][xor] = groupCount + nextMin;
+
+      for (const [num, count] of counts[index]) {
+        const cost = groupCount - count;
+
+        dp[index][xor] = Math.min(dp[index][xor], dp[index + 1][xor ^ num] + cost);
+      }
+    }
+  }
+
+  return dp[0][0];
+};
+```

solutions/1787. Make the XOR of All Segments Equal to Zero/solution.js

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+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+const minChanges = function (nums, k) {
+  const MAX_XOR = 1024;
+  const n = nums.length;
+  const counts = Array.from({ length: k }, () => new Map());
+  const dp = Array.from({ length: k }, () => new Array(MAX_XOR).fill(n));
+
+  const getGroupCount = index => Math.floor(n / k) + (n % k > index ? 1 : 0);
+
+  for (let index = 0; index < n; index++) {
+    const num = nums[index];
+    const group = counts[index % k];
+    const count = group.get(num) ?? 0;
+
+    group.set(num, count + 1);
+  }
+
+  for (let xor = 0; xor < MAX_XOR; xor++) {
+    const count = counts[k - 1].get(xor) ?? 0;
+
+    dp[k - 1][xor] = getGroupCount(k - 1) - count;
+  }
+
+  for (let index = k - 2; index >= 0; index--) {
+    const groupCount = getGroupCount(index);
+    const nextMin = Math.min(...dp[index + 1]);
+
+    for (let xor = 0; xor < MAX_XOR; xor++) {
+      dp[index][xor] = groupCount + nextMin;
+
+      for (const [num, count] of counts[index]) {
+        const cost = groupCount - count;
+
+        dp[index][xor] = Math.min(dp[index][xor], dp[index + 1][xor ^ num] + cost);
+      }
+    }
+  }
+
+  return dp[0][0];
+};