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// 1802. Maximum Value at a Given Index in a Bounded Array
#include <stdio.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
*/
int maxValue(int n, int index, int max_sum) {
long long sum;
int cnl = index, cnr = n - cnl - 1;
int left = 1, right = max_sum, mid;
while (left < right) {
sum = 0;
mid = right - (right - left) / 2;
if (cnl) {
if (mid > cnl)
sum += ((long long)((mid - 1) + (mid - cnl)) * cnl / 2);
else {
sum += ((long long)((mid - 1) + 1) * (mid - 1) / 2);
sum += (cnl - (mid - 1));
}
}
if (cnr) {
if (mid > cnr)
sum += ((long long)((mid - 1) + (mid - cnr)) * cnr / 2);
else {
sum += ((long long)((mid - 1) + 1) * (mid - 1) / 2);
sum += (cnr - (mid - 1));
}
}
sum += mid;
if (sum <= max_sum)
left = mid;
else
right = mid - 1;
}
return left;
}
int main() {
printf("%d\n", maxValue(4, 2, 6)); // expect: 2
printf("%d\n", maxValue(6, 1, 10)); // expect: 3
}
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