Learning pathOriginal reference
Lesson 5 of 8

Grow and repair a variable window

Distinguish longest-under-budget from shortest-reaching-target windows.

Understand the idea

For nonnegative values, extending a sum window cannot reduce its sum. To find the longest window with sum ≤ K, extend right, remove left values while over budget, then measure. To find the shortest window with sum ≥ S, measure each valid window before removing its left value.

Trace a small example

For [2, 1, 3, 2] and budget 4, the longest valid length is 2. For target 5, the shortest valid length is 2 from [3, 2]. The two tasks measure at different points: after repair for a maximum, before removal for a minimum.

Conditions and cost

Each pointer advances at most n times, giving O(n) total work. This sum-based rule needs nonnegative elements and, for the shortest nonempty formulation, S > 0. In [5, −4] with budget 1, discarding 5 too early would miss the valid length-2 window.

Prefix sum explorer

Query [l, r] is zero-based and inclusive. P[k] stores the total of the first k values; P[0] is zero.

Input array a

30
11
42
13
54
95

Prefix array P

00
31
42
83
94
145
236
P[5] − P[1] = 14 − (3) = 11

Preprocess once in O(n). Each range sum then takes O(1).

Guided simulation · Change the inputs to test your understanding.

Pause & predict

When should a shortest sum-at-least-target algorithm record a valid window?

Choose an answer
Try it yourself

Turn the idea into code

Find the shortest nonempty window with sum at least 7 in [2, 3, 1, 2, 4, 3].

Give me a hint

Keep shrinking while the sum remains sufficient.

Show the explained solution

The length-2 window [4, 3] reaches 7. No single element does, so 2 is optimal.

Your notes

Ready for the next idea?

Mark complete when you can explain it yourself.

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