LeetCode #1330 — HARD

Reverse Subarray To Maximize Array Value

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array nums. The value of this array is defined as the sum of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1.

You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once.

Find maximum possible value of the final array.

Example 1:

Input: nums = [2,3,1,5,4]
Output: 10
Explanation: By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.

Example 2:

Input: nums = [2,4,9,24,2,1,10]
Output: 68

Constraints:

2 <= nums.length <= 3 * 10⁴
-10⁵ <= nums[i] <= 10⁵
The answer is guaranteed to fit in a 32-bit integer.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. The value of this array is defined as the sum of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1. You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once. Find maximum possible value of the final array.

Baseline thinking

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

Pattern signal: Array · Math · Greedy

Example 1

[2,3,1,5,4]

Example 2

[2,4,9,24,2,1,10]

Step 02

Core Insight

What unlocks the optimal approach

What's the score after reversing a sub-array [L, R] ?
It's the score without reversing it + abs(a[R] - a[L-1]) + abs(a[L] - a[R+1]) - abs(a[L] - a[L-1]) - abs(a[R] - a[R+1])
How to maximize that formula given that abs(x - y) = max(x - y, y - x) ?
This can be written as max(max(a[R] - a[L - 1], a[L - 1] - a[R]) + max(a[R + 1] - a[L], a[L] - a[R + 1]) - value(L) - value(R + 1)) over all L < R where value(i) = abs(a[i] - a[i-1])
This can be divided into 4 cases.

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

Largest constraint values

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 #1330: Reverse Subarray To Maximize Array Value
class Solution {
    public int maxValueAfterReverse(int[] nums) {
        int n = nums.length;
        int s = 0;
        for (int i = 0; i < n - 1; ++i) {
            s += Math.abs(nums[i] - nums[i + 1]);
        }
        int ans = s;
        for (int i = 0; i < n - 1; ++i) {
            ans = Math.max(
                ans, s + Math.abs(nums[0] - nums[i + 1]) - Math.abs(nums[i] - nums[i + 1]));
            ans = Math.max(
                ans, s + Math.abs(nums[n - 1] - nums[i]) - Math.abs(nums[i] - nums[i + 1]));
        }
        int[] dirs = {1, -1, -1, 1, 1};
        final int inf = 1 << 30;
        for (int k = 0; k < 4; ++k) {
            int k1 = dirs[k], k2 = dirs[k + 1];
            int mx = -inf, mi = inf;
            for (int i = 0; i < n - 1; ++i) {
                int a = k1 * nums[i] + k2 * nums[i + 1];
                int b = Math.abs(nums[i] - nums[i + 1]);
                mx = Math.max(mx, a - b);
                mi = Math.min(mi, a + b);
            }
            ans = Math.max(ans, s + Math.max(0, mx - mi));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1330: Reverse Subarray To Maximize Array Value
func maxValueAfterReverse(nums []int) int {
	s, n := 0, len(nums)
	for i, x := range nums[:n-1] {
		y := nums[i+1]
		s += abs(x - y)
	}
	ans := s
	for i, x := range nums[:n-1] {
		y := nums[i+1]
		ans = max(ans, s+abs(nums[0]-y)-abs(x-y))
		ans = max(ans, s+abs(nums[n-1]-x)-abs(x-y))
	}
	dirs := [5]int{1, -1, -1, 1, 1}
	const inf = 1 << 30
	for k := 0; k < 4; k++ {
		k1, k2 := dirs[k], dirs[k+1]
		mx, mi := -inf, inf
		for i, x := range nums[:n-1] {
			y := nums[i+1]
			a := k1*x + k2*y
			b := abs(x - y)
			mx = max(mx, a-b)
			mi = min(mi, a+b)
		}
		ans = max(ans, s+max(mx-mi, 0))
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1330: Reverse Subarray To Maximize Array Value
class Solution:
    def maxValueAfterReverse(self, nums: List[int]) -> int:
        ans = s = sum(abs(x - y) for x, y in pairwise(nums))
        for x, y in pairwise(nums):
            ans = max(ans, s + abs(nums[0] - y) - abs(x - y))
            ans = max(ans, s + abs(nums[-1] - x) - abs(x - y))
        for k1, k2 in pairwise((1, -1, -1, 1, 1)):
            mx, mi = -inf, inf
            for x, y in pairwise(nums):
                a = k1 * x + k2 * y
                b = abs(x - y)
                mx = max(mx, a - b)
                mi = min(mi, a + b)
            ans = max(ans, s + max(mx - mi, 0))
        return ans

// Accepted solution for LeetCode #1330: Reverse Subarray To Maximize Array Value
struct Solution;

impl Solution {
    fn max_value_after_reverse(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut res = 0;
        let mut min = std::i32::MAX;
        let mut max = std::i32::MIN;
        let mut total = 0;
        for i in 1..n {
            let a = nums[i - 1];
            let b = nums[i];
            total += (a - b).abs();
            res = res.max((b - nums[0]).abs() - (a - b).abs());
            res = res.max((a - nums[n - 1]).abs() - (a - b).abs());
            min = min.min(a.max(b));
            max = max.max(a.min(b));
        }
        total + res.max((max - min) * 2)
    }
}

#[test]
fn test() {
    let nums = vec![2, 3, 1, 5, 4];
    let res = 10;
    assert_eq!(Solution::max_value_after_reverse(nums), res);
    let nums = vec![2, 4, 9, 24, 2, 1, 10];
    let res = 68;
    assert_eq!(Solution::max_value_after_reverse(nums), res);
}

// Accepted solution for LeetCode #1330: Reverse Subarray To Maximize Array Value
function maxValueAfterReverse(nums: number[]): number {
    const n = nums.length;
    let s = 0;
    for (let i = 0; i < n - 1; ++i) {
        s += Math.abs(nums[i] - nums[i + 1]);
    }
    let ans = s;
    for (let i = 0; i < n - 1; ++i) {
        const d = Math.abs(nums[i] - nums[i + 1]);
        ans = Math.max(ans, s + Math.abs(nums[0] - nums[i + 1]) - d);
        ans = Math.max(ans, s + Math.abs(nums[n - 1] - nums[i]) - d);
    }
    const dirs = [1, -1, -1, 1, 1];
    const inf = 1 << 30;
    for (let k = 0; k < 4; ++k) {
        let mx = -inf;
        let mi = inf;
        for (let i = 0; i < n - 1; ++i) {
            const a = dirs[k] * nums[i] + dirs[k + 1] * nums[i + 1];
            const b = Math.abs(nums[i] - nums[i + 1]);
            mx = Math.max(mx, a - b);
            mi = Math.min(mi, a + b);
        }
        ans = Math.max(ans, s + Math.max(0, mx - mi));
    }
    return ans;
}

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.

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.

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.