0001EasyArrayHash Table

Ledger Reconciliation

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Trains the technique from

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.

An accountant is closing out a ledger. You are given an integer array adjustments, where each entry is a signed correction applied to an account, and an integer net.

Exactly one pair of entries sums to net. Return the indices of those two entries, in any order. An entry cannot be paired with itself.

Examples

Example 1

Input: adjustments = [7, -3, 11, 2], net = 8
Output: [1, 2]

The entry at index 1 holds -3 and the entry at index 2 holds 11, which together settle to 8.

Example 2

Input: adjustments = [-5, -5, 4], net = -10
Output: [0, 1]

Both entries hold the same amount but occupy separate positions, so pairing them is allowed.

Example 3

Input: adjustments = [6, 1], net = 7
Output: [0, 1]

With only two entries available, they must be the settling pair.

Constraints

2 <= adjustments.length <= 10^4
-10^9 <= adjustments[i] <= 10^9
-10^9 <= net <= 10^9
Exactly one valid pair exists

The values you return may be in any order.

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 reconcile_ledger(adjustments: list[int], net: int) -> list[int]:

Java

public int[] reconcileLedger(int[] adjustments, int net)

The intended approach

Time O(n)Space O(n)

Walk the ledger once, remembering the index of every amount seen so far. At each entry, look up whether its complement has already appeared.

Pythonentry point: reconcile_ledger

def reconcile_ledger(adjustments, net):
    seen = {}
    for i, value in enumerate(adjustments):
        if net - value in seen:
            return [seen[net - value], i]
        seen[value] = i
    return []

The obvious version

Check every unordered pair directly, O(n^2).

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_reconcile(adjustments, net):
    n = len(adjustments)
    for i in range(n):
        for j in range(i + 1, n):
            if adjustments[i] + adjustments[j] == net:
                return [i, j]
    return []

How this problem was checked

Reference solution executed against 8 test cases — 8 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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