Sync LeetCode submission Runtime - 119 ms (16.67%), Memory - 24 MB (33.33%)

Author: koralkulacoglu

LeetCode/4300-minimum-operations-to-make-array-modulo-alternating-i/README.md

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+<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>
+
+<p>In one operation, you can <strong>increase</strong> or <strong>decrease</strong> any element of <code>nums</code> by 1.</p>
+
+<p>An array is called <strong>modulo alternating</strong> if there exist two <strong>distinct</strong> integers <code>x</code> and <code>y</code> (<code>0 &lt;= x, y &lt; k</code>) such that:</p>
+
+<ul>
+	<li>For every <strong>even</strong> index <code>i</code>, <code>nums[i] % k == x</code></li>
+	<li>For every <strong>odd</strong> index <code>i</code>, <code>nums[i] % k == y</code></li>
+</ul>
+
+<p>Return the <strong>minimum</strong> number of operations required to make <code>nums</code> <strong>modulo alternating</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,4,2,8], k = 3</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Let&#39;s choose <code>x = 1</code> for even indices and <code>y = 2</code> for odd indices.</li>
+	<li>Perform the following operations:
+	<ul>
+		<li>Increment <code>nums[1] = 4</code> by 1, giving <code>nums = [1, 5, 2, 8]</code>.</li>
+		<li>Decrement <code>nums[2] = 2</code> by 1, giving <code>nums = [1, 5, 1, 8]</code>.</li>
+	</ul>
+	</li>
+	<li>Now, for even indices, <code>nums[i] % k = 1</code>, and for odd indices, <code>nums[i] % k = 2</code>.</li>
+	<li>Thus, the total number of operations required is 2.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,1,1], k = 3</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Incrementing <code>nums[1]</code> by 1 gives <code>nums = [1, 2, 1]</code>, which satisfies the condition with <code>x = 1</code> and <code>y = 2</code>.</li>
+	<li>Thus, the total number of operations required is 1.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>2 &lt;= k &lt;= 100</code></li>
+</ul>

LeetCode/4300-minimum-operations-to-make-array-modulo-alternating-i/solution.cpp

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+class Solution {
+public:
+    int minOperations(vector<int>& nums, int k) {
+        int mx, mn;
+        int n = nums.size();
+        int ans = 1e9;
+        for (int x=0; x<k; x++) {
+            for (int y=0; y<k; y++) {
+                if (x == y) continue;
+                int cur = 0;
+                for (int i=0; i<n; i++) {
+                    int val = nums[i] % k;
+                    if (i%2) {
+                        mn = min(y, val);
+                        mx = max(y, val);
+                    }
+                    else {
+                        mn = min(x, val);
+                        mx = max(x, val);
+                    }
+                    int ops = min(mx - mn, mn + k - mx);
+                    cur += ops;
+                }
+                ans = min(ans, cur);
+            }
+        }
+        return ans;
+    }
+};