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7. Coin Change (Static Option Pool) • Invariant Proof & Speed-Run

Theory Context•Backtracking & State Search

MediumLC 322

Coin Change (LeetCode 322)

Target Frequency:AmazonMeta

You're given an array coins of distinct coin denominations and a target amount, with an unlimited supply of every denomination.

Find the smallest number of coins whose values add up to exactly amount. If no combination of coins reaches that amount exactly, return -1.

Worked Examples

Example 1

Input:coins = [1,2,5], amount = 11

Output:3

Explanation: 11 = 5 + 5 + 1

Example 2

Input:coins = [2], amount = 3

Output:-1

Explanation: Coins of 2 can only make even amounts, never 3.

Example 3

Input:coins = [1], amount = 0

Output:0

Explanation: An amount of 0 needs no coins.

⚖️Formal Constraints & Bounds

1 <= coins.length <= 12
1 <= coins[i] <= 2³¹ - 1
0 <= amount <= 10⁴

Deep-Dive & Conceptual Insights

Core Insight

Choice Continuity allows the full coin denomination pool to be looped at every recursion frame. Because any remaining amount 'rem' is path-independent, caching memo[rem] collapses exponential O(C^A) search into linear O(C * A) subproblems (where C = len(coins), A = amount).

Real-World Scenario & Production Applications

Vending machines, currency dispensing, and resource optimization where identical denominations can be reused continuously.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Base Case Anchors (rem == 0 / rem < 0)

Zero coins are needed to make remaining amount 0 (return 0). An overshoot (rem < 0) is an invalid choice, represented as float('inf') so it never wins a min() comparison.

Mathematical Recurrence / Code Invariant

if rem == 0:
    return 0
if rem < 0:
    return float('inf')

Step-by-Step Execution Trace Table

Code / Blueprint

                         rem=3 (Root, d=0)
                      /          |          \
            choose 1 /  choose 2 | choose 5  \ (rem - 5 < 0)
                    /            |            \
             rem=2 (d=1)    rem=1 (d=1)     ✂️ PRUNED (rem < 0 -> inf)
            /     \              |
 choose 1  /       \ choose 2    | choose 1
          /         \            |
     rem=1 (d=2)   rem=0 ✅     rem=0 ✅ (1 coin + 1)
        |        (1 coin + 1)   (memo[1] = 1)
 choose 1| (d=3)
        |
     rem=0 ✅ (3 coins)

Step-by-Step Decision Trace

Step 1 (Root, rem=3, d=0): Allocate memo = [-1] * 4. Call dfs(3). Loop coins [1, 2, 5].
Step 2 (Choose 1, rem=2, d=1): Call dfs(2). Loop coins [1, 2, 5].
Step 3 (Choose 1, rem=1, d=2): Call dfs(1). Loop coins [1, 2, 5].
Step 4 (Choose 1, rem=0, d=3): Base Case reached (rem == 0): return 0 coins needed.
Step 5 (Compute rem=1, d=2): Coin 1 returns 0 $\rightarrow$ ans = min(inf, 0 + 1) = 1. Coins 2, 5 yield rem < 0 (✂️ PRUNED). Cache memo[1] = 1, return 1.
Step 6 (Backtrack to d=1, Choose 2, rem=0, d=2): Call dfs(2 - 2 = 0) $\rightarrow$ returns 0. ans = min(2, 0 + 1) = 1. Cache memo[2] = 1, return 1.
Step 7 (Backtrack to d=0, Choose 2, rem=1, d=1): Call dfs(3 - 2 = 1). Cache hit (memo[1] == 1): returns 1 in $O(1)$ ! ans = min(2, 1 + 1) = 2.
Step 8 (Choose 5, rem=-2, d=1): 3 - 5 = -2 < 0 $\rightarrow$ ✂️ PRUNED (rem < 0 returns inf).
Step 9 (Record Optimal Solution, d=0): Cache memo[3] = 2. Record Minimum Coins = 2 ✅. Search Complete: Minimum coins needed = 2.

Full Walkthrough9 Steps

Input:coins = [1,2,5], amount = 11Expected:3

1🌱 ROOT(Root, rem=3, d=0)

remaining = 3

Allocate memo = [-1] * 4. Call dfs(3). Loop coins [1, 2, 5].

2🟢 CHOOSE(Choose 1, rem=2, d=1)

remaining = 2

Call dfs(2). Loop coins [1, 2, 5].

3🟢 CHOOSE(Choose 1, rem=1, d=2)

remaining = 1

Call dfs(1). Loop coins [1, 2, 5].

4✅ RECORD / GOAL(Choose 1, rem=0, d=3)

remaining = 0

Base Case reached (rem == 0): return 0 coins needed.

5✂️ PRUNED(Compute rem=1, d=2)

remaining = 1

Coin 1 returns 0

\rightarrow

ans = min(inf, 0 + 1) = 1. Coins 2, 5 yield rem < 0 (✂️ PRUNED). Cache memo[1] = 1, return 1.

6↩️ BACKTRACK / POP(Backtrack to d=1, Choose 2, rem=0, d=2)

remaining = 0

Call dfs(2 - 2 = 0)

\rightarrow

returns 0. ans = min(2, 0 + 1) = 1. Cache memo[2] = 1, return 1.

7🌱 ROOT(Backtrack to d=0, Choose 2, rem=1, d=1)

remaining = 1

Call dfs(3 - 2 = 1). Cache hit (memo[1] == 1): returns 1 in

O(1)

! ans = min(2, 1 + 1) = 2.

8✂️ PRUNED(Choose 5, rem=-2, d=1)

3 - 5 = -2 < 0

\rightarrow

✂️ PRUNED (rem < 0 returns inf).

9🌱 ROOT(Record Optimal Solution, d=0)

memo[3] = 2`.

Cache memo[3] = 2. Record Minimum Coins = 2 ✅. Search Complete: Minimum coins needed = 2.

Code Shape

Archetype Code Shape

Python

1	memo = [-1] * (amount + 1)
2	def dfs(rem):
3	if rem == 0:
4	return 0
5	if rem < 0:
6	return float('inf')
7	if memo[rem] != -1:
8	return memo[rem]
9	ans = float('inf')
10	for coin in coins:
11	res = dfs(rem - coin)
12	if res != float('inf'):
13	ans = min(ans, res + 1)
14	memo[rem] = ans
15	return ans
16	best = dfs(amount)
17	return best if best != float('inf') else -1

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

Coin Change asks for the minimum number of coins needed to make up a given amount. Expressed as a decision tree, at any remaining amount $A$ , you can branch into choosing any coin $c \le A$ , transitioning to subproblem $A - c$ . However, because coin denominations can be reused in any order (e.g. paying 1 then 2 vs 2 then 1), pure backtracking explores the same remaining amounts repeatedly, triggering an astronomical $O(C^A)$ exponential blowup. By attaching a memoization cache memo[remain], identical subproblems collapse into a single $O(A \cdot C)$ calculation.

🏟️ The Analogy: A Subway Fare Machine with Memory

Imagine an automated transit machine calculating the fewest coins to refund 87 cents. If the machine dispenses a quarter (leaving 62 cents), it starts figuring out the optimal way to make 62 cents. If along a completely different calculation path it dispenses two dimes and a nickel (also leaving 62 cents), a naive machine would re-calculate the optimal 62-cent payout from scratch. A machine with memory writes on a digital notepad: "Optimal coins for 62 cents = 5 coins." The next time any coin branch leaves 62 cents, it reads the stored answer instead of recalculating it.

🔑 Why It Works

The decision tree's overlapping subproblems are parameterized entirely by the scalar remain. Memoizing the optimal count for each remaining value converts an exponential recursive tree into a DAG(Directed Acyclic Graph), guaranteeing $O(\text{amount} \cdot \text{len(coins)})$ runtime.

Omission of remainder memoization: Recursively branching without caching memo[rem] results in $O(C^A)$ exponential explosion. Caching the minimum coins for each remainder reduces runtime to $O(C \cdot A)$ .
Base case return offset: Returning 1 instead of 0 when rem == 0. Zero remaining value requires zero additional coins. Returning 1 shifts the answer by +1.
Unreachable amount sentinel bug: Returning float('inf') directly to the caller instead of converting to -1 when no combination can reach the target amount.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Find minimum coins/elements to make a target sum + unlimited coin reuse allowed + choice continuity.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: remaining tracks remaining amount. Optimal subproblem depends strictly on remaining, making memoization effective.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

Missing memoization: exploring all C^A paths without caching causes severe time-limit-exceeded (TLE).

The 60-Second Interview Pitch

Say this out loud before you type a single line

"I use memoized backtracking over remaining amount. For each coin, I recursively compute the minimum coins needed for remaining - coin. I cache results in memo to avoid re-evaluating identical subproblems. Time complexity is O(amount * len(coins)) with O(amount) recursion and memo space."

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(Amount · N)

There are Amount + 1 subproblems (0 to Amount). Each subproblem iterates through N coin denominations once. Total runtime is O(Amount · N).

SPACE COMPLEXITY

O(Amount) Auxiliary

Memoization dictionary/table of size Amount + 1 and recursion call stack of max depth Amount.

Formal Recurrence Relation

T(A, N) = (A + 1) · N = O(A · N)

Derivation Progression

Distinct Subproblem Count

A + 1 states for amounts 0 dots A

Memoization ensures each remainder amount is computed at most once.

Transitions Per State

N coin branches per state

Evaluates 1 + minCoins(rem - c) across all N coins.

Compound Time Bound

O(A · N)

Eliminates exponential branching completely.

Variable Definitions

$A$

Target amount value (0 to Amount subproblems)

$N$

Number of coin denominations available

Memory Architecture & Bounds

🟣 Call Stack

O(A) Recursion stack depth (worst case coin 1)

🔵 Auxiliary Heap

O(A) Memoization cache table

🟢 Output Space

O(1) Min coins integer

Boundary Best / Worst Cases

Best Case

$O(1)$ if amount == 0.

Average Case

$O(A · N)$ .

Worst Case

$O(A · N)$ with coin 1 available.

Decision & State-Space Search Tree

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "fewest number of coins", "unlimited supply". Top-down memoized backtracking: state depends solely on remaining amount rem.

CONSTRAINTS & BOUNDS

$coins \le 12, amount \le 10^4$ . Memoization reduces exponential tree to $O(\text{amount} \cdot |coins|)$ .

FAANG PRODUCTION TRAPS & EDGE CASES

Unmemoized exponential explosion. Return $-1$ if amount cannot be formed.

Theory Context•Backtracking & State Search

MediumLC 322

Coin Change (LeetCode 322)

Target Frequency:AmazonMeta

You're given an array coins of distinct coin denominations and a target amount, with an unlimited supply of every denomination.

Find the smallest number of coins whose values add up to exactly amount. If no combination of coins reaches that amount exactly, return -1.

Worked Examples

Example 1

Input:coins = [1,2,5], amount = 11

Output:3

Explanation: 11 = 5 + 5 + 1

Example 2

Input:coins = [2], amount = 3

Output:-1

Explanation: Coins of 2 can only make even amounts, never 3.

Example 3

Input:coins = [1], amount = 0

Output:0

Explanation: An amount of 0 needs no coins.

⚖️Formal Constraints & Bounds

1 <= coins.length <= 12
1 <= coins[i] <= 2³¹ - 1
0 <= amount <= 10⁴

Deep-Dive & Conceptual Insights

Core Insight

Real-World Scenario & Production Applications

Vending machines, currency dispensing, and resource optimization where identical denominations can be reused continuously.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Base Case Anchors (rem == 0 / rem < 0)

Zero coins are needed to make remaining amount 0 (return 0). An overshoot (rem < 0) is an invalid choice, represented as float('inf') so it never wins a min() comparison.

Mathematical Recurrence / Code Invariant

if rem == 0:
    return 0
if rem < 0:
    return float('inf')

Step-by-Step Execution Trace Table

Code / Blueprint

                         rem=3 (Root, d=0)
                      /          |          \
            choose 1 /  choose 2 | choose 5  \ (rem - 5 < 0)
                    /            |            \
             rem=2 (d=1)    rem=1 (d=1)     ✂️ PRUNED (rem < 0 -> inf)
            /     \              |
 choose 1  /       \ choose 2    | choose 1
          /         \            |
     rem=1 (d=2)   rem=0 ✅     rem=0 ✅ (1 coin + 1)
        |        (1 coin + 1)   (memo[1] = 1)
 choose 1| (d=3)
        |
     rem=0 ✅ (3 coins)

Step-by-Step Decision Trace

Step 1 (Root, rem=3, d=0): Allocate memo = [-1] * 4. Call dfs(3). Loop coins [1, 2, 5].
Step 2 (Choose 1, rem=2, d=1): Call dfs(2). Loop coins [1, 2, 5].
Step 3 (Choose 1, rem=1, d=2): Call dfs(1). Loop coins [1, 2, 5].
Step 4 (Choose 1, rem=0, d=3): Base Case reached (rem == 0): return 0 coins needed.
Step 5 (Compute rem=1, d=2): Coin 1 returns 0 $\rightarrow$ ans = min(inf, 0 + 1) = 1. Coins 2, 5 yield rem < 0 (✂️ PRUNED). Cache memo[1] = 1, return 1.
Step 6 (Backtrack to d=1, Choose 2, rem=0, d=2): Call dfs(2 - 2 = 0) $\rightarrow$ returns 0. ans = min(2, 0 + 1) = 1. Cache memo[2] = 1, return 1.
Step 7 (Backtrack to d=0, Choose 2, rem=1, d=1): Call dfs(3 - 2 = 1). Cache hit (memo[1] == 1): returns 1 in $O(1)$ ! ans = min(2, 1 + 1) = 2.
Step 8 (Choose 5, rem=-2, d=1): 3 - 5 = -2 < 0 $\rightarrow$ ✂️ PRUNED (rem < 0 returns inf).
Step 9 (Record Optimal Solution, d=0): Cache memo[3] = 2. Record Minimum Coins = 2 ✅. Search Complete: Minimum coins needed = 2.

Full Walkthrough9 Steps

Input:coins = [1,2,5], amount = 11Expected:3

1🌱 ROOT(Root, rem=3, d=0)

remaining = 3

Allocate memo = [-1] * 4. Call dfs(3). Loop coins [1, 2, 5].

2🟢 CHOOSE(Choose 1, rem=2, d=1)

remaining = 2

Call dfs(2). Loop coins [1, 2, 5].

3🟢 CHOOSE(Choose 1, rem=1, d=2)

remaining = 1

Call dfs(1). Loop coins [1, 2, 5].

4✅ RECORD / GOAL(Choose 1, rem=0, d=3)

remaining = 0

Base Case reached (rem == 0): return 0 coins needed.

5✂️ PRUNED(Compute rem=1, d=2)

remaining = 1

Coin 1 returns 0

\rightarrow

ans = min(inf, 0 + 1) = 1. Coins 2, 5 yield rem < 0 (✂️ PRUNED). Cache memo[1] = 1, return 1.

6↩️ BACKTRACK / POP(Backtrack to d=1, Choose 2, rem=0, d=2)

remaining = 0

Call dfs(2 - 2 = 0)

\rightarrow

returns 0. ans = min(2, 0 + 1) = 1. Cache memo[2] = 1, return 1.

7🌱 ROOT(Backtrack to d=0, Choose 2, rem=1, d=1)

remaining = 1

Call dfs(3 - 2 = 1). Cache hit (memo[1] == 1): returns 1 in

O(1)

! ans = min(2, 1 + 1) = 2.

8✂️ PRUNED(Choose 5, rem=-2, d=1)

3 - 5 = -2 < 0

\rightarrow

✂️ PRUNED (rem < 0 returns inf).

9🌱 ROOT(Record Optimal Solution, d=0)

memo[3] = 2`.

Cache memo[3] = 2. Record Minimum Coins = 2 ✅. Search Complete: Minimum coins needed = 2.

Code Shape

Archetype Code Shape

Python

1	memo = [-1] * (amount + 1)
2	def dfs(rem):
3	if rem == 0:
4	return 0
5	if rem < 0:
6	return float('inf')
7	if memo[rem] != -1:
8	return memo[rem]
9	ans = float('inf')
10	for coin in coins:
11	res = dfs(rem - coin)
12	if res != float('inf'):
13	ans = min(ans, res + 1)
14	memo[rem] = ans
15	return ans
16	best = dfs(amount)
17	return best if best != float('inf') else -1

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

🏟️ The Analogy: A Subway Fare Machine with Memory

🔑 Why It Works

Omission of remainder memoization: Recursively branching without caching memo[rem] results in $O(C^A)$ exponential explosion. Caching the minimum coins for each remainder reduces runtime to $O(C \cdot A)$ .
Base case return offset: Returning 1 instead of 0 when rem == 0. Zero remaining value requires zero additional coins. Returning 1 shifts the answer by +1.
Unreachable amount sentinel bug: Returning float('inf') directly to the caller instead of converting to -1 when no combination can reach the target amount.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Find minimum coins/elements to make a target sum + unlimited coin reuse allowed + choice continuity.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: remaining tracks remaining amount. Optimal subproblem depends strictly on remaining, making memoization effective.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

Missing memoization: exploring all C^A paths without caching causes severe time-limit-exceeded (TLE).

The 60-Second Interview Pitch

Say this out loud before you type a single line

"I use memoized backtracking over remaining amount. For each coin, I recursively compute the minimum coins needed for remaining - coin. I cache results in memo to avoid re-evaluating identical subproblems. Time complexity is O(amount * len(coins)) with O(amount) recursion and memo space."

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(Amount · N)

There are Amount + 1 subproblems (0 to Amount). Each subproblem iterates through N coin denominations once. Total runtime is O(Amount · N).

SPACE COMPLEXITY

O(Amount) Auxiliary

Memoization dictionary/table of size Amount + 1 and recursion call stack of max depth Amount.

Formal Recurrence Relation

T(A, N) = (A + 1) · N = O(A · N)

Derivation Progression

Distinct Subproblem Count

A + 1 states for amounts 0 dots A

Memoization ensures each remainder amount is computed at most once.

Transitions Per State

N coin branches per state

Evaluates 1 + minCoins(rem - c) across all N coins.

Compound Time Bound

O(A · N)

Eliminates exponential branching completely.

Variable Definitions

$A$

Target amount value (0 to Amount subproblems)

$N$

Number of coin denominations available

Memory Architecture & Bounds

🟣 Call Stack

O(A) Recursion stack depth (worst case coin 1)

🔵 Auxiliary Heap

O(A) Memoization cache table

🟢 Output Space

O(1) Min coins integer

Boundary Best / Worst Cases

Best Case

$O(1)$ if amount == 0.

Average Case

$O(A · N)$ .

Worst Case

$O(A · N)$ with coin 1 available.

Decision & State-Space Search Tree

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "fewest number of coins", "unlimited supply". Top-down memoized backtracking: state depends solely on remaining amount rem.

CONSTRAINTS & BOUNDS

$coins \le 12, amount \le 10^4$ . Memoization reduces exponential tree to $O(\text{amount} \cdot |coins|)$ .

FAANG PRODUCTION TRAPS & EDGE CASES

Unmemoized exponential explosion. Return $-1$ if amount cannot be formed.