LeetCode Entry

2784. Check if Array is Good

14.05.2026 easy 2026 kotlin rust

Is combination 1..n,n

2784. Check if Array is Good easy substack youtube

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Problem TLDR

Is combination 1..n,n

Intuition

• no-brain solution: check every number (1..n) is in array, and count(n)==2
• shortest solution: (1..n)+n==sorted()
• optimal solution: use array indices as visited storage

Approach

• n[0] can be used as extra storage for n[len-1] special case

Complexity

• Time complexity: \(O(nlogn|n)\)

• Space complexity: \(O(n|1)\)

Code

    fun isGood(n: IntArray) =
    (1..<n.size) + (n.size-1) == n.sorted()

    pub fn is_good(mut n: Vec<i32>) -> bool {
        (0..n.len()).all(|i| { let x = n[i].abs() as usize;
            !(x>=n.len()||n[x]<0&&(x<n.len()-1||n[0]<0)) && {
            if n[x] < 0 { n[0] *= -1 } else { n[x] *= -1 };1>0}
        }) && n[0]<0
    }