LeetCode #1130 — MEDIUM

Minimum Cost Tree From Leaf Values

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

Solve on LeetCode

The Problem

Problem Statement

Given an array arr of positive integers, consider all binary trees such that:

Each node has either 0 or 2 children;
The values of arr correspond to the values of each leaf in an in-order traversal of the tree.
The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree, respectively.

Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node. It is guaranteed this sum fits into a 32-bit integer.

A node is a leaf if and only if it has zero children.

Example 1:

Input: arr = [6,2,4]
Output: 32
Explanation: There are two possible trees shown.
The first has a non-leaf node sum 36, and the second has non-leaf node sum 32.

Example 2:

Input: arr = [4,11]
Output: 44

Constraints:

2 <= arr.length <= 40
1 <= arr[i] <= 15
It is guaranteed that the answer fits into a 32-bit signed integer (i.e., it is less than 2³¹).

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given an array arr of positive integers, consider all binary trees such that: Each node has either 0 or 2 children; The values of arr correspond to the values of each leaf in an in-order traversal of the tree. The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree, respectively. Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node. It is guaranteed this sum fits into a 32-bit integer. A node is a leaf if and only if it has zero children.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Stack · Greedy

Example 1

[6,2,4]

Example 2

[4,11]

Step 02

Core Insight

What unlocks the optimal approach

Do a DP, where dp(i, j) is the answer for the subarray arr[i]..arr[j].
For each possible way to partition the subarray i <= k < j, the answer is max(arr[i]..arr[k]) * max(arr[k+1]..arr[j]) + dp(i, k) + dp(k+1, j).

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 #1130: Minimum Cost Tree From Leaf Values
class Solution {
    private Integer[][] f;
    private int[][] g;

    public int mctFromLeafValues(int[] arr) {
        int n = arr.length;
        f = new Integer[n][n];
        g = new int[n][n];
        for (int i = n - 1; i >= 0; --i) {
            g[i][i] = arr[i];
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = Math.max(g[i][j - 1], arr[j]);
            }
        }
        return dfs(0, n - 1);
    }

    private int dfs(int i, int j) {
        if (i == j) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = 1 << 30;
        for (int k = i; k < j; k++) {
            ans = Math.min(ans, dfs(i, k) + dfs(k + 1, j) + g[i][k] * g[k + 1][j]);
        }
        return f[i][j] = ans;
    }
}

// Accepted solution for LeetCode #1130: Minimum Cost Tree From Leaf Values
func mctFromLeafValues(arr []int) int {
	n := len(arr)
	f := make([][]int, n)
	g := make([][]int, n)
	for i := range g {
		f[i] = make([]int, n)
		g[i] = make([]int, n)
		g[i][i] = arr[i]
		for j := i + 1; j < n; j++ {
			g[i][j] = max(g[i][j-1], arr[j])
		}
	}
	var dfs func(int, int) int
	dfs = func(i, j int) int {
		if i == j {
			return 0
		}
		if f[i][j] > 0 {
			return f[i][j]
		}
		f[i][j] = 1 << 30
		for k := i; k < j; k++ {
			f[i][j] = min(f[i][j], dfs(i, k)+dfs(k+1, j)+g[i][k]*g[k+1][j])
		}
		return f[i][j]
	}
	return dfs(0, n-1)
}

# Accepted solution for LeetCode #1130: Minimum Cost Tree From Leaf Values
class Solution:
    def mctFromLeafValues(self, arr: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> Tuple:
            if i == j:
                return 0, arr[i]
            s, mx = inf, -1
            for k in range(i, j):
                s1, mx1 = dfs(i, k)
                s2, mx2 = dfs(k + 1, j)
                t = s1 + s2 + mx1 * mx2
                if s > t:
                    s = t
                    mx = max(mx1, mx2)
            return s, mx

        return dfs(0, len(arr) - 1)[0]

// Accepted solution for LeetCode #1130: Minimum Cost Tree From Leaf Values
struct Solution;
use std::i32;

impl Solution {
    fn mct_from_leaf_values(arr: Vec<i32>) -> i32 {
        let mut res = 0;
        let mut stack: Vec<i32> = vec![i32::MAX];
        for x in arr {
            while x >= *stack.last().unwrap() {
                let mid = stack.pop().unwrap();
                let y = *stack.last().unwrap();
                res += mid * i32::min(y, x);
            }
            stack.push(x);
        }
        while stack.len() > 2 {
            let x = stack.pop().unwrap();
            let y = *stack.last().unwrap();
            res += x * y;
        }
        res
    }
}

#[test]
fn test() {
    let arr = vec![6, 2, 4];
    assert_eq!(Solution::mct_from_leaf_values(arr), 32);
}

// Accepted solution for LeetCode #1130: Minimum Cost Tree From Leaf Values
function mctFromLeafValues(arr: number[]): number {
    const n = arr.length;
    const f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(0));
    const g: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(0));
    for (let i = n - 1; i >= 0; --i) {
        g[i][i] = arr[i];
        for (let j = i + 1; j < n; ++j) {
            g[i][j] = Math.max(g[i][j - 1], arr[j]);
        }
    }
    const dfs = (i: number, j: number): number => {
        if (i === j) {
            return 0;
        }
        if (f[i][j] > 0) {
            return f[i][j];
        }
        let ans = 1 << 30;
        for (let k = i; k < j; ++k) {
            ans = Math.min(ans, dfs(i, k) + dfs(k + 1, j) + g[i][k] * g[k + 1][j]);
        }
        return (f[i][j] = ans);
    };
    return dfs(0, n - 1);
}

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^3)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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.