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Pattern1 Practice3 Interview

Tabulation

Order bottom-up subproblems so every dependency is available before a state is computed. Learn its decision rule, maintained state, Python cost model, and transfer from a focused drill to a full Interview Problem.

Recognize it when

Consider Tabulation when the prompt's constraints and required operations match this shape: Order bottom-up subproblems so every dependency is available before a state is computed.

Pybit demonstrates the Tabulation decision pattern in a professional coding interview workspace.
On this page · Lay Out States in Dependency Order

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Tabulation Code Labs

Lay Out States in Dependency Order

Tabulation evaluates a finite state graph bottom-up. Every table cell must be visited only after the cells named by its recurrence are already final.

Seed Reachable Base States

Base entries encode complete smallest subproblems. Initialize only states that are actually reachable; accidental zeroes can masquerade as valid optima or counts.

Trace Bottom-Up Table Fills

Reference
def fibonacci_table(n):
    table=[0]*(n+1)
    if n>=1: table[1]=1
    trace=[table.copy()]
    for index in range(2,n+1):
        table[index]=table[index-1]+table[index-2]; trace.append(table.copy())
    return trace
Practice

Implement fibonacci_table(n). Return the table after each index 2 through n is filled, including the initial base table.

Public tests

  • Fill only after dependencies

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Transition Forward or Pull Backward

A pull recurrence computes one state from predecessors. A push transition sends one finalized state’s contribution to successors. Choose the direction that makes bounds and duplicate counting easiest to prove.

Count Stair Ways

Reference
def count_stair_ways(n):
    if n<2: return 1
    previous,current=1,1
    for _ in range(2,n+1): previous,current=current,previous+current
    return current
Practice

Implement count_stair_ways(n). Return ways to reach step n using moves of one or two, with one way to remain at step zero.

Public tests

  • Compress dead table history

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Compress Only Dead History

Replace a table with rolling variables only when no future transition needs overwritten states and reconstruction is not required. Compression changes space, not the recurrence or evaluation order.

Explain It in an Interview

Say: “This cell means…, these are the base states, and I fill in this order because every dependency is earlier.” Derive time as states times transitions and distinguish unreachable sentinels from valid zero values.

Python Version Note

Python 3.11+

The examples use the standard Python syntax and containers supported by the browser Judge; the Tabulation invariant is independent of a minor Python release.

Verify in Python docs(opens in a new tab)

Complexity & Invariants

Interview bounds depend on the stated representation, input model, and real Python operations.

OperationAverageWorstInterview note
Tabulation decision loopO(amount * len(coins))O(amount * len(coins))Tabulate the fewest coins for every amount from zero upward and translate the unresolved sentinel to -1.

Space

O(amount) for the focused Tabulate Minimum Coin Count implementation.

Assumptions

  • For each amount, every valid final coin points to a smaller already-correct state. Taking the minimum over all final coins therefore yields the optimal count, while an unchanged sentinel means no composition exists.
  • The bound includes real Python operations; no repeated slicing, linear list membership, or hidden sorting is treated as free.

Invariants Worth Saying Aloud

  1. State the precise Tabulation invariant before coding and preserve it after every update, traversal step, or recursive return.
  2. Every processed state already has all required predecessors.
  3. The table value has one precise meaning for every index.
  4. Choose a base state and impossible sentinel. Preserve this property after every transition.
  5. Fill states in dependency order from zero through amount. Preserve this property after every transition.

When To Use Or Avoid Tabulation

Use It When

  • Use Tabulation when the prompt exposes the ordering, monotonicity, recurrence, or constraint that makes this decision safe.
  • Use it when every processed state already has all required predecessors.

Choose Another Tool When

  • Avoid Tabulation when its decision rule cannot eliminate work or preserve the required state.
  • Avoid forcing the label onto a prompt whose simpler direct scan already meets the constraints.

Common Pitfalls

These are the mistakes most likely to survive a happy-path example and fail a boundary case.

Moving state without a proof

Applying the pattern without proving its ordering, monotonicity, or window invariant can silently miss valid candidates.

Prevent it: Write the decision rule beside the loop and verify it against Tabulate Minimum Coin Count before optimizing.

Avoid

def minimum_coin_count(coins, amount):
    pass

Use instead

def minimum_coin_count(coins, amount):
    dp = [amount + 1] * (amount + 1)
    dp[0] = 0
    for current in range(1, amount + 1):
        for coin in coins:
            if coin <= current:
                dp[current] = min(dp[current], dp[current - coin] + 1)
    return -1 if dp[amount] > amount else dp[amount]

Where you will hit this: Tabulate Minimum Coin Count(opens in a new tab)

Breaking the maintained state

Uses a greedy largest-coin choice that fails when the locally largest coin blocks the optimum.

Prevent it: Use the public tests and preserve this state: Every processed state already has all required predecessors.

Avoid

def minimum_coin_count(coins,amount):
    used=0
    for coin in sorted(coins,reverse=True):
        used += amount//coin; amount%=coin
    return used if amount==0 else -1

Use instead

def minimum_coin_count(coins, amount):
    dp = [amount + 1] * (amount + 1)
    dp[0] = 0
    for current in range(1, amount + 1):
        for coin in coins:
            if coin <= current:
                dp[current] = min(dp[current], dp[current - coin] + 1)
    return -1 if dp[amount] > amount else dp[amount]

Where you will hit this: Tabulate Minimum Coin Count(opens in a new tab)

Hiding Python work in the hot path

Python slicing, copying, membership in lists, or repeated sorting inside the loop can invalidate the intended complexity.

Prevent it: Count every slice, copy, sort, membership check, and container update before claiming O(amount * len(coins)).

Avoid

def minimum_coin_count(coins,amount):
    used=0
    for coin in sorted(coins,reverse=True):
        used += amount//coin; amount%=coin
    return used if amount==0 else -1

Use instead

def minimum_coin_count(coins, amount):
    dp = [amount + 1] * (amount + 1)
    dp[0] = 0
    for current in range(1, amount + 1):
        for coin in coins:
            if coin <= current:
                dp[current] = min(dp[current], dp[current - coin] + 1)
    return -1 if dp[amount] > amount else dp[amount]

Where you will hit this: House Robber(opens in a new tab)

Reviewed References

Official Python documentation supports language behavior. Canonical problem pages provide additional practice context.