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+// 1802. Maximum Value at a Given Index in a Bounded Array
+#include "leetcode.h"
+
+/*
+ * given three positive integers: 'n, index, and max_sum'. you want to construct
+ * an array 'nums' (0-indexed) that satisfies the following conditions:
+ * - nums.length == n
+ * - nums[i] is a positive integer where 0 <= i < n
+ * - abs(nums[i] - nums[i + 1]) <= 1 where 0 <= i < n - 1
+ * - the sum of all the elements of 'nums' does not exceed 'max_sum'
+ * - nums[index] is maximised
+ * return 'nums[index]' of the constructed array. note that 'abs(x)' equals 'x'
+ * if 'x >= 0' and '-x' otherwise
+ */
+
+class Solution {
+ bool check(int n, int i, int max_sum, int m) {
+ for (int d = 1; max_sum >= 0 && d < m && i + d < n; ++d)
+ max_sum -= m - d;
+ for (int d = 1; max_sum >= 0 && d < m && i - d >= 0; ++d)
+ max_sum -= m - d;
+ return max_sum >= 0;
+ }
+
+public:
+ int maxValue(int n, int index, int max_sum) {
+ int left = 0, right = max_sum;
+ while (left < right) {
+ int mid = (left + right + 1) / 2;
+ if (check(n, index, max_sum - n - mid, mid))
+ left = mid;
+ else
+ right = mid - 1;
+ }
+ return left + 1;
+ }
+};
+
+int main() {
+ Solution obj;
+ printf("%d\n", obj.maxValue(4, 2, 6)); // expect: 2
+ printf("%d\n", obj.maxValue(6, 1, 10)); // expect: 3
+}