LeetCode Entry

3423. Maximum Difference Between Adjacent Elements in a Circular Array

12.06.2025 easy 2025 kotlin rust

Max abs sibling diff

3423. Maximum Difference Between Adjacent Elements in a Circular Array easy blog post substack youtube

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Problem TLDR

Max abs sibling diff #easy

Intuition

There are many surprising ways to write that code, try all of them.

Approach

• kotlin’s last() makes runtime worse 12ms vs 1ms of n[n.size - 1]
• we can windowed
• we can zip
• we can zip with 0..100
• we can minimize the array reading

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(1)\)

Code

// 24ms
    fun maxAdjacentDistance(n: IntArray) =
        n.zip(n.drop(1) + n[0]).maxOf { (a, b) -> abs(a - b) }

// 22ms
    fun maxAdjacentDistance(n: IntArray) =
        (n + n[0]).asList().windowed(2).maxOf { abs(it[0] - it[1]) }

// 17ms
    fun maxAdjacentDistance(n: IntArray) =
        n.indices.maxOf { abs(n[(it + 1) % n.size] - n[it]) }

// 15ms
    fun maxAdjacentDistance(n: IntArray) =
        n.zip(intArrayOf(n.last()) + n).maxOf { (a, b) -> abs(a - b) }

// 11ms
    fun maxAdjacentDistance(n: IntArray): Int {
        var r = abs(n[0] - n.last())
        for (i in 1..<n.size) r = max(r, abs(n[i] - n[i - 1]))
        return r
    }

// 1ms
    fun maxAdjacentDistance(n: IntArray): Int {
        var r = abs(n[0] - n[n.size - 1])
        for (i in 1..<n.size) r = max(r, abs(n[i] - n[i - 1]))
        return r
    }

// 0ms
    pub fn max_adjacent_distance(n: Vec<i32>) -> i32 {
       (0..n.len()).map(|i| (n[(i + 1) % n.len()] - n[i]).abs()).max().unwrap()
    }

// 0ms
    pub fn max_adjacent_distance(n: Vec<i32>) -> i32 {
        (0..100).zip([&n[1..], &n[..1]].concat())
        .map(|(a, b)| (n[a] - b).abs()).max().unwrap()
    }

// 0ms
    int maxAdjacentDistance(vector<int>& n) {
        int r = 0, p = n[size(n) - 1];
        for (int x: n) r = max(r, abs(p - x)), p = x;
        return r;
    }