The O(n) Club: Minimum Cost Tree From Leaf Values — Monotonic Stack Style

The O(n) Club: Minimum Cost Tree From Leaf Values — Monotonic Stack Style

⚡ TL;DR

Your goal: take an array of integers (leaves) and build a binary tree (no funny business: leaves stay in order!) where every non-leaf’s value is the product of the biggest leaf in its left and right subtrees. You want the lowest possible total of non-leaf nodes. Brute-force is for masochists. Here’s a monotonic stack Java dance for sanity:

// Java Monotonic Stack Solution
public int mctFromLeafValues(int[] arr) {
    int res = 0;
    Stack
<integer> stack = new Stack<>();
    stack.push(Integer.MAX_VALUE);
    for (int a : arr) {
        while (stack.peek() <= a) {
            int mid = stack.pop();
            res += mid * Math.min(stack.peek(), a);
        }
        stack.push(a);
    }
    while (stack.size() > 2) {
        res += stack.pop() * stack.peek();
    }
    return res;
}
</integer>

🧠 How Devs Usually Mess This Up

If you started multiplying every pair of adjacent leaves and called it a day—well, we’ve all been there. Many try every possible merge or, worse, shuffle leaf order like a caffeinated magician (interviewer alert: this is illegal!). And no, the greedy combine-adjacent-smallest trick doesn’t work—this isn’t Huffman coding or making a sandwich.

🔍 What’s Actually Going On

Think of it like building a robot: every time you wire two parts, you pay the price of their “loudest” signals multiplied together. To keep costs down, you always want to get rid of your weakest link early (before it costs you even more later). The stack algorithm only squishes the smallest leaf into its neighbor when that neighbor is bigger, sort of like voting the least useful team member off the island before the 360 review. If you ever reach for classic greedy or try to permute the leaves, you’re headed for the runtime woodshed.

🛠️ PseudoCode

• Start with a stack holding infinity (sentinel).
• For each value a in arr:
• Pop from stack while top ≤ a, each time adding popped * min(next stack value, a) to your total.
• Push a on the stack.
• When done, keep popping and merging the last two until only sentinel + one number left.
• Return your result.

• At every stage, small values are merged out before huge ones multiply them into orbit.
• The stack ensures you never lose track of what you might need to merge next.

💻 The Code

// Monotonic Stack: The Actual Java
public int mctFromLeafValues(int[] arr) {
    int res = 0;
    Stack
<integer> stack = new Stack<>();
    stack.push(Integer.MAX_VALUE);
    for (int a : arr) {
        while (stack.peek() <= a) {
            int mid = stack.pop();
            res += mid * Math.min(stack.peek(), a);
        }
        stack.push(a);
    }
    while (stack.size() > 2) {
        res += stack.pop() * stack.peek();
    }
    return res;
}
</integer>

⚠️ Pitfalls, Traps & Runtime Smacks

• Repeated numbers? No sweat. Zeros? Sure, if you enjoy existential dread.
• No shuffling or skipping. Array order must survive like a cockroach after SQL injection.
• Classic greedy (always merging smallest adjacent) fails on [6,2,4]—don’t go there.
• Stack solution is truly O(n) time. DP is O(n³)—if you run it in an interview, bring a sleeping bag.
• Stack uses O(n) space, but at least you’re not blowing the heap with recursive DP calls.

🧠 Interviewer Brain-Tattoo

“Merge out your weakest leaf early—otherwise, it’ll be multiplying your cost behind your back. Kind of like technical debt, only faster.”