1802

Author: lzl124631x

leetcode/1802. Maximum Value at a Given Index in a Bounded Array/README.md

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+# [1802. Maximum Value at a Given Index in a Bounded Array (Medium)](https://leetcode.com/problems/maximum-value-at-a-given-index-in-a-bounded-array/)
+
+<p>You are given three positive integers <code>n</code>, <code>index</code> and <code>maxSum</code>. You want to construct an array <code>nums</code> <strong>(0-indexed) </strong>that satisfies the following conditions:</p>
+
+<ul>
+	<li><code>nums.length == n</code></li>
+	<li><code>nums[i]</code> is a <strong>positive</strong> integer where <code>0 &lt;= i &lt; n</code>.</li>
+	<li><code>abs(nums[i] - nums[i+1]) &lt;= 1</code> where <code>0 &lt;= i &lt; n-1</code>.</li>
+	<li>The sum of all the elements of <code>nums</code> does not exceed <code>maxSum</code>.</li>
+	<li><code>nums[index]</code> is <strong>maximized</strong>.</li>
+</ul>
+
+<p>Return <code>nums[index]</code> of the constructed array.</p>
+
+<p>Note that <code>abs(x)</code> equals <code>x</code> if <code>x &gt;= 0</code>, and <code>-x</code> otherwise.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre><strong>Input:</strong> n = 4, index = 2,  maxSum = 6
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The arrays [1,1,<strong>2</strong>,1] and [1,2,<strong>2</strong>,1] satisfy all the conditions. There are no other valid arrays with a larger value at the given index.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> n = 6, index = 1,  maxSum = 10
+<strong>Output:</strong> 3
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= maxSum &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= index &lt; n</code></li>
+</ul>
+
+**Related Topics**:  
+[Binary Search](https://leetcode.com/tag/binary-search/), [Greedy](https://leetcode.com/tag/greedy/)
+
+## Solution 1. Binary Answer
+
+```cpp
+// OJ: https://leetcode.com/problems/maximum-value-at-a-given-index-in-a-bounded-array/
+// Author: github.com/lzl124631x
+// Time: O(log(maxSum))
+// Space: O(1)
+class Solution {
+    long count(long len, long M) {
+        return M >= len ? (2 * M - len + 1) * len / 2 : ((M - 1) * M / 2 + len);
+    }
+    bool valid(long n, long i, long maxSum, long M) {
+        long left = count(i + 1, M), right = count(n - i, M);
+        return left + right - M <= maxSum;
+    }
+public:
+    int maxValue(int n, int index, int maxSum) {
+        int L = 1, R = maxSum;
+        while (L <= R) {
+            int M = (L + R) / 2;
+            if (valid(n, index, maxSum, M)) L = M + 1;
+            else R = M - 1;
+        }
+        return R;
+    }
+};
+```