Python 3.11+
The examples use the standard Python syntax and containers supported by the browser Judge; the Interval DP invariant is independent of a minor Python release.
Verify in Python docs(opens in a new tab)Pattern1 Practice3 Interview
Interval DP
Solve progressively larger contiguous intervals by combining solutions around split points. Learn its decision rule, maintained state, Python cost model, and transfer from a focused drill to a full Interview Problem.
Recognize it when
Consider Interval DP when the prompt's constraints and required operations match this shape: Solve progressively larger contiguous intervals by combining solutions around split points.
On this page · Define a State over One Interval
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Interval DP Code Labs
Define a State over One Interval
Interval DP assigns an answer to a contiguous range, commonly dp[left][right]. The state contract must specify inclusive or half-open endpoints and what has already been resolved.
Grow from Shorter Intervals
Larger ranges depend on smaller subranges, so fill by increasing interval length. Base intervals of length zero or one must be complete before any larger transition.
Generate Interval DP Fill Order
Reference
def interval_order(n):
return [[left,left+length-1] for length in range(1,n+1) for left in range(n-length+1)]Practice
Implement interval_order(n). Return [left,right] inclusive intervals ordered by increasing length from 1 through n.
Public tests
- Visit dependencies before larger intervals
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Enumerate the Final Split or Choice
Many recurrences choose the final split point, last operation, or paired endpoint. Enumerate every legal choice and combine the two resulting smaller intervals without overlap.
Compute Minimum Interval Merge Cost
Reference
def minimum_merge_cost(values):
n=len(values)
if n<2:return 0
prefix=[0]
for value in values:prefix.append(prefix[-1]+value)
dp=[[0]*n for _ in range(n)]
for length in range(2,n+1):
for left in range(n-length+1):
right=left+length-1;total=prefix[right+1]-prefix[left]
dp[left][right]=min(dp[left][split]+dp[split+1][right]+total for split in range(left,right))
return dp[0][n-1]Practice
Implement minimum_merge_cost(values). Merging adjacent groups costs their total; return minimum cost to merge all values.
Public tests
- Try every final split
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Choose Interval DP or Prefix Optimization
Use interval DP when decisions recursively divide a contiguous range. Prefix sums can make each interval aggregate O(1), but do not remove the split dimension. Typical time is O(n cubed) and space O(n squared).
Explain It in an Interview
Say: “dp[left][right] means…, and the last split at k leaves these two already-solved intervals. I fill increasing lengths.” State base cases, split bounds, prefix optimization, and final target cell.
Python Version Note
Complexity & Invariants
Interview bounds depend on the stated representation, input model, and real Python operations.
| Operation | Average | Worst | Interview note |
|---|---|---|---|
| Interval DP decision loop | O(n^3) | O(n^3) | Tabulate minimum costs for increasing interval lengths and try every possible final split. |
Space
O(n^2) for the focused Minimize Adjacent Merge Cost implementation.
Assumptions
- Every full merge of an interval has one final split into two adjacent subintervals. The recurrence chooses the best solved cost for both sides plus the unavoidable final interval sum, covering all legal merge trees.
- The bound includes real Python operations; no repeated slicing, linear list membership, or hidden sorting is treated as free.
Invariants Worth Saying Aloud
- State the precise Interval DP invariant before coding and preserve it after every update, traversal step, or recursive return.
- Every child interval is solved before its parent interval.
- Each recurrence covers every possible final split exactly once.
- Define dp[left][right] over contiguous intervals. Preserve this property after every transition.
- Try every final split and add the interval sum once. Preserve this property after every transition.
When To Use Or Avoid Interval DP
Use It When
- Use Interval DP when the prompt exposes the ordering, monotonicity, recurrence, or constraint that makes this decision safe.
- Use it when every child interval is solved before its parent interval.
Choose Another Tool When
- Avoid Interval DP 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 Minimize Adjacent Merge Cost before optimizing.
Avoid
def minimum_adjacent_merge_cost(weights):
passUse instead
def minimum_adjacent_merge_cost(weights):
n = len(weights)
if n < 2:
return 0
prefix = [0]
for weight in weights:
prefix.append(prefix[-1] + weight)
dp = [[0] * n for _ in range(n)]
for length in range(2, n + 1):
for left in range(n - length + 1):
right = left + length - 1
interval_sum = prefix[right + 1] - prefix[left]
dp[left][right] = min(dp[left][split] + dp[split + 1][right] + interval_sum for split in range(left, right))
return dp[0][n - 1]Where you will hit this: Minimize Adjacent Merge Cost(opens in a new tab)
Breaking the maintained state
Greedily merges the currently smallest adjacent pair and can miss a cheaper global merge tree.
Prevent it: Use the public tests and preserve this state: Every child interval is solved before its parent interval.
Avoid
def minimum_adjacent_merge_cost(weights):
a=list(weights); total=0
while len(a)>1:
i=min(range(len(a)-1),key=lambda j:a[j]+a[j+1]); merged=a[i]+a[i+1]; total+=merged; a[i:i+2]=[merged]
return totalUse instead
def minimum_adjacent_merge_cost(weights):
n = len(weights)
if n < 2:
return 0
prefix = [0]
for weight in weights:
prefix.append(prefix[-1] + weight)
dp = [[0] * n for _ in range(n)]
for length in range(2, n + 1):
for left in range(n - length + 1):
right = left + length - 1
interval_sum = prefix[right + 1] - prefix[left]
dp[left][right] = min(dp[left][split] + dp[split + 1][right] + interval_sum for split in range(left, right))
return dp[0][n - 1]Where you will hit this: Minimize Adjacent Merge Cost(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(n^3).
Avoid
def minimum_adjacent_merge_cost(weights):
a=list(weights); total=0
while len(a)>1:
i=min(range(len(a)-1),key=lambda j:a[j]+a[j+1]); merged=a[i]+a[i+1]; total+=merged; a[i:i+2]=[merged]
return totalUse instead
def minimum_adjacent_merge_cost(weights):
n = len(weights)
if n < 2:
return 0
prefix = [0]
for weight in weights:
prefix.append(prefix[-1] + weight)
dp = [[0] * n for _ in range(n)]
for length in range(2, n + 1):
for left in range(n - length + 1):
right = left + length - 1
interval_sum = prefix[right + 1] - prefix[left]
dp[left][right] = min(dp[left][split] + dp[split + 1][right] + interval_sum for split in range(left, right))
return dp[0][n - 1]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.
- Python Compound Statements — Function Definitions(opens in a new tab)
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