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Algorithm1 Practice4 Interview

Dynamic Programming

Solve overlapping subproblems once and combine their results under a state transition. Learn the algorithm's preconditions, state transition, correctness argument, Python cost model, and transfer from one focused drill to a full Interview Problem.

Recognize it when

Consider Dynamic Programming when the prompt's constraints and required operations match this shape: Solve overlapping subproblems once and combine their results under a state transition.

Pybit demonstrates Dynamic Programming in a professional Python interview workspace.
On this page · Name the Repeated State

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Dynamic Programming Code Labs

Name the Repeated State

Dynamic programming begins when different decision paths reach the same subproblem. Define the smallest state whose value fully answers the remaining question; extra history prevents reuse.

Write the Recurrence Before the Table

Express the answer for one state using already-defined smaller states. State what each option means—take, skip, extend, split—and why those options are exhaustive and non-overlapping where required.

Trace Take-or-Skip States

Reference
def rob_state_trace(values):
    skip=take=0; trace=[]
    for value in values:
        skip,take=max(skip,take),skip+value
        trace.append([skip,take])
    return trace
Practice

Implement rob_state_trace(values). Return [skip,take] after each house for the non-adjacent maximum-sum recurrence.

Public tests

  • Advance from prior states only

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Choose Evaluation Order and Base Cases

Base cases are valid states, not initialization tricks. Tabulation must visit dependencies before consumers; memoized recursion must decrease toward a base case. Use a sentinel that cannot be confused with a valid value.

Compute Minimum Coin Count

Reference
def min_coin_count(coins,amount):
    dp=[amount+1]*(amount+1); dp[0]=0
    for total in range(1,amount+1):
        for coin in coins:
            if coin<=total: dp[total]=min(dp[total],dp[total-coin]+1)
    return -1 if dp[amount]>amount else dp[amount]
Practice

Implement min_coin_count(coins,amount). Return the minimum number of reusable coins totaling amount, or -1.

Public tests

  • Use solved smaller amounts

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Choose Memoization or Tabulation

Memoization computes only reached states and mirrors the recurrence, but uses recursion and cache overhead. Tabulation exposes order, avoids stack depth, and often permits rolling-state compression. Both are O(number of states × transitions per state).

Explain It in an Interview

Say: “State means this. The recurrence considers these complete choices, and these base cases anchor it. I evaluate in this dependency order.” Then derive time from states times transitions and space from the table plus call stack. House Robber(opens in a new tab) is the canonical take-or-skip transfer.

Python Version Note

Python 3.11+

The code uses standard Python syntax and containers supported by the browser Judge. The Dynamic Programming proof does not depend on 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
Dynamic Programming complete workflowO(n)O(n)When choosing the current item forbids only the immediately previous item, the optimum depends on two earlier prefix states. state[i] is the optimum over exactly the first i values, whether the i-th value is skipped or selected.

Space

O(n) for the returned states for the focused Advance Dynamic Programming States implementation.

Assumptions

  • For each prefix, every valid solution either skips the newest value and uses state[i-1], or selects it and combines with state[i-2]. Taking the better exhaustive case proves the final state is optimal.
  • The bound counts Python sorting, slicing, hashing, heap, recursion, and container-copy costs rather than treating syntax as free.

Invariants Worth Saying Aloud

  1. State the precise Dynamic Programming invariant before coding and preserve it after every update, traversal step, or recursive return.
  2. Every stored state equals the answer for its documented subproblem.
  3. A transition reads only valid predecessor states and includes every legal choice.
  4. When choosing the current item forbids only the immediately previous item, the optimum depends on two earlier prefix states. Preserve this claim after every transition.
  5. state[i] is the optimum over exactly the first i values, whether the i-th value is skipped or selected. Preserve this claim after every transition.

When To Use Or Avoid Dynamic Programming

Use It When

  • Use Dynamic Programming when this precondition is stated or can be proved: The problem has reusable subproblems and a state definition whose dependencies are acyclic or memoizable.
  • Use it when this maintained state removes repeated work: Every stored state equals the answer for its documented subproblem.

Choose Another Tool When

  • Avoid Dynamic Programming when this precondition is absent: The problem has reusable subproblems and a state definition whose dependencies are acyclic or memoizable.
  • Avoid it when a simpler direct scan satisfies the constraints with less implementation risk.

Common Pitfalls

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

Applying the algorithm without its precondition

Verify algorithm preconditions such as sorted input, nonnegative weights, acyclicity, or admissible heuristics before applying it.

Prevent it: State and verify this precondition before coding: The problem has reusable subproblems and a state definition whose dependencies are acyclic or memoizable.

Avoid

def non_adjacent_prefix_states(values):
    pass

Use instead

def non_adjacent_prefix_states(values):
    states = [0]
    previous_two = 0
    previous_one = 0
    for value in values:
        current = max(previous_one, previous_two + value)
        states.append(current)
        previous_two, previous_one = previous_one, current
    return states

Breaking the state transition

Always adds the current value and therefore selects adjacent positions instead of comparing skip and take transitions.

Prevent it: Preserve this proof obligation: The recurrence covers every legal final choice, and the evaluation order makes each dependency available before use.

Avoid

def non_adjacent_prefix_states(values):
    states = [0]
    for value in values:
        states.append(states[-1] + value)
    return states

Use instead

def non_adjacent_prefix_states(values):
    states = [0]
    previous_two = 0
    previous_one = 0
    for value in values:
        current = max(previous_one, previous_two + value)
        states.append(current)
        previous_two, previous_one = previous_one, current
    return states

Hiding Python work in the claimed bound

Include visited state, boundary cases, recursion depth, and hidden copying costs in correctness and complexity analysis.

Prevent it: Count every sort, slice, copy, membership check, heap update, and recursive frame before claiming O(n).

Avoid

def non_adjacent_prefix_states(values):
    states = [0]
    for value in values:
        states.append(states[-1] + value)
    return states

Use instead

def non_adjacent_prefix_states(values):
    states = [0]
    previous_two = 0
    previous_one = 0
    for value in values:
        current = max(previous_one, previous_two + value)
        states.append(current)
        previous_two, previous_one = previous_one, current
    return states

Reviewed References

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