You are given three positive integers: n, index, and maxSum. You want to construct an array nums (0-indexed) that satisfies the following conditions:
nums.length == nnums[i] is a positive integer where 0 <= i < n.abs(nums[i] - nums[i+1]) <= 1 where 0 <= i < n-1.nums does not exceed maxSum.nums[index] is maximized.Return nums[index] of the constructed array.
Note that abs(x) equals x if x >= 0, and -x otherwise.
Example 1:
Input: n = 4, index = 2, maxSum = 6 Output: 2 Explanation: nums = [1,2,2,1] is one array that satisfies all the conditions. There are no arrays that satisfy all the conditions and have nums[2] == 3, so 2 is the maximum nums[2].
Example 2:
Input: n = 6, index = 1, maxSum = 10 Output: 3
Constraints:
1 <= n <= maxSum <= 1090 <= index < nstruct Solution;
impl Solution {
fn max_value(n: i32, index: i32, max_sum: i32) -> i32 {
let mut l = 1i64;
let mut r = max_sum as i64;
let n = n as usize;
let index = index as usize;
let mut res = 0;
let kl = index as i64;
let kr = (n - index - 1) as i64;
while l <= r {
let m = (l + (r - l) / 2) as i64;
let sum = helper(m, kl) + m + helper(m, kr);
if sum > max_sum as i64 {
r = m - 1;
} else {
res = res.max(m);
l = m + 1;
}
}
res as i32
}
}
fn helper(m: i64, k: i64) -> i64 {
if m > k {
(m - 1 + m - 1 - k + 1) * k / 2
} else {
(m - 1 + 1) * (m - 1) / 2 + (k - m + 1)
}
}
#[test]
fn test() {
let n = 4;
let index = 2;
let max_sum = 6;
let res = 2;
assert_eq!(Solution::max_value(n, index, max_sum), res);
let n = 6;
let index = 1;
let max_sum = 10;
let res = 3;
assert_eq!(Solution::max_value(n, index, max_sum), res);
}