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;
        }
    };
    
  • # 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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