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)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.

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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.
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.
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 traceImplement rob_state_trace(values). Return [skip,take] after each house for the non-adjacent maximum-sum recurrence.
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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.
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]Implement min_coin_count(coins,amount). Return the minimum number of reusable coins totaling amount, or -1.
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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).
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 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)Interview bounds depend on the stated representation, input model, and real Python operations.
| Operation | Average | Worst | Interview note |
|---|---|---|---|
| Dynamic Programming complete workflow | O(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. |
O(n) for the returned states for the focused Advance Dynamic Programming States implementation.
These are the mistakes most likely to survive a happy-path example and fail a boundary case.
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):
passUse 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 statesWhere you will hit this: Advance Dynamic Programming States(opens in a new tab)
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 statesUse 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 statesWhere you will hit this: Advance Dynamic Programming States(opens in a new tab)
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 statesUse 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 statesWhere you will hit this: House Robber(opens in a new tab)
Official Python documentation supports language behavior. Canonical problem pages provide additional practice context.
python-docs · checked 2026-07-27
python-docs · checked 2026-07-27