LeetCode #3796 — MEDIUM

Find Maximum Value in a Constrained Sequence

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

The Problem

Problem Statement

You are given an integer n, a 2D integer array restrictions, and an integer array diff of length n - 1. Your task is to construct a sequence of length n, denoted by a[0], a[1], ..., a[n - 1], such that it satisfies the following conditions:

a[0] is 0.
All elements in the sequence are non-negative.
For every index i (0 <= i <= n - 2), abs(a[i] - a[i + 1]) <= diff[i].
For each restrictions[i] = [idx, maxVal], the value at position idx in the sequence must not exceed maxVal (i.e., a[idx] <= maxVal).

Your goal is to construct a valid sequence that maximizes the largest value within the sequence while satisfying all the above conditions.

Return an integer denoting the largest value present in such an optimal sequence.

Example 1:

Input: n = 10, restrictions = [[3,1],[8,1]], diff = [2,2,3,1,4,5,1,1,2]

Output: 6

Explanation:

The sequence a = [0, 2, 4, 1, 2, 6, 2, 1, 1, 3] satisfies the given constraints (a[3] <= 1 and a[8] <= 1).
The maximum value in the sequence is 6.

Example 2:

Input: n = 8, restrictions = [[3,2]], diff = [3,5,2,4,2,3,1]

Output: 12

Explanation:

The sequence a = [0, 3, 3, 2, 6, 8, 11, 12] satisfies the given constraints (a[3] <= 2).
The maximum value in the sequence is 12.

Constraints:

2 <= n <= 10⁵
1 <= restrictions.length <= n - 1
restrictions[i].length == 2
restrictions[i] = [idx, maxVal]
1 <= idx < n
1 <= maxVal <= 10⁶
diff.length == n - 1
1 <= diff[i] <= 10
The values of restrictions[i][0] are unique.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n, a 2D integer array restrictions, and an integer array diff of length n - 1. Your task is to construct a sequence of length n, denoted by a[0], a[1], ..., a[n - 1], such that it satisfies the following conditions: a[0] is 0. All elements in the sequence are non-negative. For every index i (0 <= i <= n - 2), abs(a[i] - a[i + 1]) <= diff[i]. For each restrictions[i] = [idx, maxVal], the value at position idx in the sequence must not exceed maxVal (i.e., a[idx] <= maxVal). Your goal is to construct a valid sequence that maximizes the largest value within the sequence while satisfying all the above conditions. Return an integer denoting the largest value present in such an optimal sequence.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

10
[[3,1],[8,1]]
[2,2,3,1,4,5,1,1,2]

Example 2

8
[[3,2]]
[3,5,2,4,2,3,1]

Core Insight

What unlocks the optimal approach

The problem can be solved greedily.
Any restriction at index <code>i</code> also indirectly limits nearby positions due to the cumulative <code>diff</code> bounds.
The maximum possible value at each index is the minimum of all bounds propagated from the left and from the right.
Once all upper bounds are fixed, the answer is simply the maximum value among them.

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 #3796: Find Maximum Value in a Constrained Sequence
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
// package main
// 
// import (
// 	"math"
// 	"slices"
// )
// 
// // https://space.bilibili.com/206214
// func findMaxVal(n int, restrictions [][]int, diff []int) int {
// 	maxVal := make([]int, n)
// 	for i := range maxVal {
// 		maxVal[i] = math.MaxInt
// 	}
// 	for _, r := range restrictions {
// 		maxVal[r[0]] = r[1]
// 	}
// 
// 	a := make([]int, n)
// 	for i, d := range diff {
// 		a[i+1] = min(a[i]+d, maxVal[i+1])
// 	}
// 	for i := n - 2; i > 0; i-- {
// 		a[i] = min(a[i], a[i+1]+diff[i])
// 	}
// 	return slices.Max(a)
// }

// Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
package main

import (
	"math"
	"slices"
)

// https://space.bilibili.com/206214
func findMaxVal(n int, restrictions [][]int, diff []int) int {
	maxVal := make([]int, n)
	for i := range maxVal {
		maxVal[i] = math.MaxInt
	}
	for _, r := range restrictions {
		maxVal[r[0]] = r[1]
	}

	a := make([]int, n)
	for i, d := range diff {
		a[i+1] = min(a[i]+d, maxVal[i+1])
	}
	for i := n - 2; i > 0; i-- {
		a[i] = min(a[i], a[i+1]+diff[i])
	}
	return slices.Max(a)
}

# Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
# Time:  O(n)
# Space: O(n)

# greedy, dp
class Solution(object):
    def findMaxVal(self, n, restrictions, diff):
        """
        :type n: int
        :type restrictions: List[List[int]]
        :type diff: List[int]
        :rtype: int
        """
        dp = [float("inf")]*n
        dp[0] = 0
        for i, x in restrictions:
            dp[i] = x
        for i in xrange(n-1):
            dp[i+1] = min(dp[i+1], dp[i]+diff[i])
        for i in reversed(xrange(n-1)):
            dp[i] = min(dp[i], dp[i+1]+diff[i])
        return max(dp)

// Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
// package main
// 
// import (
// 	"math"
// 	"slices"
// )
// 
// // https://space.bilibili.com/206214
// func findMaxVal(n int, restrictions [][]int, diff []int) int {
// 	maxVal := make([]int, n)
// 	for i := range maxVal {
// 		maxVal[i] = math.MaxInt
// 	}
// 	for _, r := range restrictions {
// 		maxVal[r[0]] = r[1]
// 	}
// 
// 	a := make([]int, n)
// 	for i, d := range diff {
// 		a[i+1] = min(a[i]+d, maxVal[i+1])
// 	}
// 	for i := n - 2; i > 0; i-- {
// 		a[i] = min(a[i], a[i+1]+diff[i])
// 	}
// 	return slices.Max(a)
// }

// Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3796: Find Maximum Value in a Constrained Sequence
// package main
// 
// import (
// 	"math"
// 	"slices"
// )
// 
// // https://space.bilibili.com/206214
// func findMaxVal(n int, restrictions [][]int, diff []int) int {
// 	maxVal := make([]int, n)
// 	for i := range maxVal {
// 		maxVal[i] = math.MaxInt
// 	}
// 	for _, r := range restrictions {
// 		maxVal[r[0]] = r[1]
// 	}
// 
// 	a := make([]int, n)
// 	for i, d := range diff {
// 		a[i+1] = min(a[i]+d, maxVal[i+1])
// 	}
// 	for i := n - 2; i > 0; i-- {
// 		a[i] = min(a[i], a[i+1]+diff[i])
// 	}
// 	return slices.Max(a)
// }

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(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.