Burst Balloons
A hard Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Given an array of integers where each element represents a balloon with a number on it, burst all balloons to maximize the total coins. Bursting balloon i gives coins equal to nums[left] * nums[i] * nums[right], where left and right are adjacent indices after previous bursts.
Example 1
- Input
- nums = [4,2,6]
- Output
- 78
- Why
- Optimal order: burst 2 first (1*2*6=12), then 4 (1*4*6=24), then 6 (1*6*1=6). Total 12+24+6=78.
Example 2
- Input
- nums = [1,5]
- Output
- 10
- Why
- With virtual padding [1,1,5,1], burst balloon 1 first (coins=1*1*5=5), then balloon 5 (coins=1*5*1=5). Total = 10.
Example 3
- Input
- nums = [1]
- Output
- 1
- Why
- With padding [1,1,1], burst the only balloon: coins = 1*1*1 = 1.
Constraints
- n == nums.length
- 1 <= n <= 300
- 0 <= nums[i] <= 100
How to think about it
Updated 2026-09-29Choosing which balloon to burst first creates awkward dependencies because remaining balloons become newly adjacent. Reverse perspective: choose which balloon to burst LAST in an open interval (left, right). That final balloon is guaranteed to multiply against the fixed boundary balloons left and right.
Approaches, worst first
Backtracking burst orders
time O(n!) · space O(n)
Try bursting every available balloon first, remove it, and recursively solve the remaining shortened array. Explores n! permutations.
Interval dynamic programmingWrite this one
time O(n^3) · space O(n^2)
Pad nums with 1 at both ends to form array of size n + 2. Iterate interval length from 1 to n. For each boundary (left, right), iterate the candidate last balloon k between left and right: dp[left][right] = max(dp[left][right], dp[left][k] + dp[k][right] + nums[left]*nums[k]*nums[right]).
Where people lose marks · 3
- Iterating interval lengths in arbitrary order. Sub-intervals must be solved strictly before larger intervals, requiring the outer loop to grow window length.
- Missing boundary padding. Forgetting the virtual 1s at left = 0 and right = n + 1 causes out-of-bounds reads on edge balloons.
- Confusing open and closed intervals: the interval (left, right) excludes the boundaries themselves, so balloons to burst lie strictly at k where left < k < right.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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