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 == 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 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

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].

Solution

Great solution found here

/**
 * @param {number} n
 * @param {number} index
 * @param {number} maxSum
 * @return {number}
 */
var maxValue = function (n, index, maxSum) {
  //Set the min and max values
  let min = Math.floor(maxSum / n);
  let max = maxSum;
  //Find the sum of the left and right side of the array
  const findSum = (num, len) => {
    //If the length is less than the number, return the sum of the first n numbers
    if (len < num) return (len * (num + num - len + 1)) / 2;
    //Return the sum of the first n numbers plus the remaining numbers
    return (num * (num + 1)) / 2 + (len - num);
  };
  //Helper function to check if the sum of the left and right side of the array is greater than the maxSum
  const helper = (num) => {
    //Find the sum of the left and right side of the array
    const leftSum = findSum(num, index + 1);
    const rightSum = findSum(num, n - index);
    //Return true if the sum of the left and right side of the array is greater than the maxSum
    return maxSum >= leftSum + rightSum - num;
  };
  while (min <= max) {
    const mid = (min + max) >> 1;
    helper(mid) ? (min = mid + 1) : (max = mid - 1);
  }
  return max;
};