1037EasyArray

The Biggest Later Rise

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Trains the technique from

LeetCode 2016Maximum Difference Between Increasing Elements

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A log holds the readings readings.

Return the largest amount by which some reading exceeds an earlier reading, or -1 when no reading in the log exceeds any reading before it.

Examples

Example 1

Input: readings = [8, 2, 9]
Output: 7

The 9 at the end exceeds the 2 before it by seven, which beats the 9 measured against the 8 at the front.

Example 2

Input: readings = [5, 4, 3]
Output: -1

Every reading is below the one before it, so nothing anywhere exceeds an earlier reading.

Example 3

Input: readings = [3, 8, 1, 9, 2]
Output: 8

The 9 measured against the 1 before it gives eight. The 8 near the front measured against the 3 gives only five, and nothing later beats eight.

Constraints

2 <= readings.length <= 1000
1 <= readings[i] <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def maximum_difference(readings: list[int]) -> int:

Java

public int maximumDifference(int[] readings)

The intended approach

Time O(n) over the log's length.Space O(1).

For a reading to be exceeded by as much as possible, the earlier reading it is measured against should be the smallest one seen up to that point. Anything larger would only narrow the gap. So walk the log once carrying the smallest reading met so far, and at each step measure the reading in hand against it, keeping the largest positive gap. A gap of nothing or less means the reading does not exceed the smallest earlier one, so it contributes nothing. Only after measuring is the smallest updated, which is what stops a reading being measured against itself. Finishing with no positive gap means no reading ever exceeded an earlier one, and that is the case answering `-1`.

Pythonentry point: maximum_difference

def maximum_difference(readings):
    low = readings[0]
    best = -1
    for at in range(1, len(readings)):
        value = readings[at]
        if value > low and value - low > best:
            best = value - low
        if value < low:
            low = value
    return best

The obvious version

Take every pair of positions with one before the other, and where the later reading exceeds the earlier one, measure the gap and keep the largest. Costs the log's length squared rather than one pass with a running smallest.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_maximum_difference(readings):
    n = len(readings)
    best = -1
    for i in range(n):
        for j in range(i + 1, n):
            if readings[j] > readings[i] and readings[j] - readings[i] > best:
                best = readings[j] - readings[i]
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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