1802. Maximum Value at a Given Index in a Bounded Array

1802. Maximum Value at a Given Index in a Bounded Array (Medium)

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 1:

Input: n = 4, index = 2,  maxSum = 6
Output: 2
Explanation: The arrays [1,1,2,1] and [1,2,2,1] satisfy all the conditions. There are no other valid arrays with a larger value at the given index.

Example 2:

Input: n = 6, index = 1,  maxSum = 10
Output: 3

 

Constraints:

• 1 <= n <= maxSum <= 109
• 0 <= index < n

Related Topics:
Binary Search, Greedy

Solution 1. Binary Answer

// 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;
    }
};