0835EasyArrayMath

Smallest Folded Meter Reading

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Trains the technique from

LeetCode 3300Minimum Element After Replacement With Digit Sum

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 technician folds each meter reading before filing it. Folding a reading replaces it with the sum of its digits, so a reading of 4021 is filed as 4 + 0 + 2 + 1, which is 7.

readings holds the readings as they were taken. Every reading is folded once. Return the smallest of the filed values.

Examples

Example 1

Input: readings = [4021, 88, 705]
Output: 7

Folding gives 4 + 0 + 2 + 1 = 7 for the first reading, 8 + 8 = 16 for the second and 7 + 0 + 5 = 12 for the third. The smallest filed value is 7.

Example 2

Input: readings = [10, 9]
Output: 1

Folding gives 1 + 0 = 1 and 9. The smallest filed value is 1, taken from the larger of the two readings.

Example 3

Input: readings = [8888, 1000]
Output: 1

Folding gives 8 + 8 + 8 + 8 = 32 and 1 + 0 + 0 + 0 = 1.

Constraints

1 <= readings.length <= 100
1 <= readings[i] <= 10^4

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 smallest_after_folding(readings: list[int]) -> int:

Java

public int smallestAfterFolding(int[] readings)

The intended approach

Time O(n * d) for n readings of at most d digits.Space O(1): only the running smallest is kept.

Fold one reading at a time with arithmetic only: the remainder on division by ten is the last digit and integer division by ten drops it, so repeating that until nothing is left accumulates every digit. Keep the smallest folded value seen, which needs no storage for the rest.

Pythonentry point: smallest_after_folding

def smallest_after_folding(readings):
    best = None
    for value in readings:
        folded = 0
        left = value
        while left > 0:
            folded += left % 10
            left //= 10
        if best is None or folded < best:
            best = folded
    return best

The obvious version

Write each reading out as text, add up its characters, collect every folded value in a list, sort the list and take its first entry. Pays for a sort the running minimum does not need.

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_smallest_after_folding(readings):
    folded = []
    for value in readings:
        folded.append(sum(int(mark) for mark in str(value)))
    folded.sort()
    return folded[0]

How this problem was checked

Reference solution executed against 11 test cases — 11 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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