LeetCode #1802 — MEDIUM

Maximum Value at a Given Index in a Bounded Array

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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: 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 <= 10⁹
0 <= index < n

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Binary Search · Greedy

Example 1

4
2
6

Example 2

6
1
10

Step 02

Core Insight

What unlocks the optimal approach

What if the problem was instead determining if you could generate a valid array with nums[index] == target?
To generate the array, set nums[index] to target, nums[index-i] to target-i, and nums[index+i] to target-i. Then, this will give the minimum possible sum, so check if the sum is less than or equal to maxSum.
n is too large to actually generate the array, so you can use the formula 1 + 2 + ... + n = n * (n+1) / 2 to quickly find the sum of nums[0...index] and nums[index...n-1].
Binary search for the target. If it is possible, then move the lower bound up. Otherwise, move the upper bound down.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1802: Maximum Value at a Given Index in a Bounded Array
class Solution {
    public int maxValue(int n, int index, int maxSum) {
        int left = 1, right = maxSum;
        while (left < right) {
            int mid = (left + right + 1) >>> 1;
            if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private long sum(long x, int cnt) {
        return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
    }
}

// Accepted solution for LeetCode #1802: Maximum Value at a Given Index in a Bounded Array
func maxValue(n int, index int, maxSum int) int {
	sum := func(x, cnt int) int {
		if x >= cnt {
			return (x + x - cnt + 1) * cnt / 2
		}
		return (x+1)*x/2 + cnt - x
	}
	return sort.Search(maxSum, func(x int) bool {
		x++
		return sum(x-1, index)+sum(x, n-index) > maxSum
	})
}

# Accepted solution for LeetCode #1802: Maximum Value at a Given Index in a Bounded Array
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

// Accepted solution for LeetCode #1802: Maximum Value at a Given Index in a Bounded Array
struct 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);
}

// Accepted solution for LeetCode #1802: Maximum Value at a Given Index in a Bounded Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1802: Maximum Value at a Given Index in a Bounded Array
// class Solution {
//     public int maxValue(int n, int index, int maxSum) {
//         int left = 1, right = maxSum;
//         while (left < right) {
//             int mid = (left + right + 1) >>> 1;
//             if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
//                 left = mid;
//             } else {
//                 right = mid - 1;
//             }
//         }
//         return left;
//     }
// 
//     private long sum(long x, int cnt) {
//         return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.