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Formatted question description: https://leetcode.ca/all/1802.html

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

Level

Medium

Description

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 <= 10^9
  • 0 <= index < n

Solution

Use binary search. Initialize low = 1 and high = maxSum. Each time calculate mid as the mean of low and high, and calculate the minimum possible sum of nums when nums[index] = mid. If the minimum possible sum is greater than maxSum, let high = mid - 1. Otherwise, let low = mid. Finally, the answer is low.

Note that all elements in nums must be positive, which should be considered during binary search.

  • class Solution {
    public int maxValue(int n, int index, int maxSum) {
        int low = 1, high = maxSum;
        while (low < high) {
            long mid = (high - low + 1) / 2 + low;
            long sumLeft = ((mid - index) + (mid - 1)) * index / 2;
            if (mid <= index)
                sumLeft = (mid - 1) * mid / 2 + index - mid + 1;
            sumLeft = Math.max(sumLeft, index);
            long sumRight = ((mid - 1) + (mid - (n - 1 - index))) * (n - index - 1) / 2;
            if (mid <= n - 1 - index)
                sumRight = (mid - 1) * mid / 2 + (n - 1 - index - mid) + 1;
            sumRight = Math.max(sumRight, n - index - 1);
            long sum = sumLeft + sumRight + mid;
            if (sum > maxSum)
                high = (int) mid - 1;
            else
                low = (int) mid;
        }
        return low;
    }
}

  • // OJ: https://leetcode.com/problems/maximum-value-at-a-given-index-in-a-bounded-array/
// 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;
    }
};

  • class Solution:
    def maxValue(self, n: int, index: int, maxSum: int) -> int:
        def sum(x, cnt):
            return (
                (x + x - cnt + 1) * cnt // 2 if x >= cnt else (x + 1) * x // 2 + cnt - x
            )

        left, right = 1, maxSum
        while left < right:
            mid = (left + right + 1) >> 1
            if sum(mid - 1, index) + sum(mid, n - index) <= maxSum:
                left = mid
            else:
                right = mid - 1
        return left

############

# 1802. Maximum Value at a Given Index in a Bounded Array
# https://leetcode.com/problems/maximum-value-at-a-given-index-in-a-bounded-array/

class Solution:
    def maxValue(self, n: int, index: int, maxSum: int) -> int:
        def good(a):
            b = max(a - index, 0)
            res = (a + b) * (a - b + 1) // 2
            b = max(a - ((n - 1) - index), 0)
            res += (a + b) * (a - b + 1) // 2
            return res - a
        
        maxSum -= n
        left, right = 0, maxSum
        
        while left < right:
            mid = (left + right + 1) // 2
            if good(mid) <= maxSum:
                left = mid
            else:
                right = mid - 1
        
        return left + 1

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