0454HardArrayGreedySorting

Smallest Starting Charge For A Round Of Calls

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Trains the technique from

LeetCode 1665Minimum Initial Energy to Finish Tasks

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 field engineer runs a round of service calls on one battery pack. calls[i] = [drain_i, floor_i] describes call i:

drain_i is the charge the call burns, deducted once the call is finished;
floor_i is the charge the pack must be showing at the moment the call is started, otherwise the diagnostic tool refuses to run it.

Every call must be made exactly once, and the engineer may take them in any order. The pack is not recharged during the round. Note that a call's threshold is never below its drain, so finishing a call never leaves the pack below zero.

Return the smallest charge the pack can start the round with so that some order gets through all of the calls.

Examples

Example 1

Input: calls = [[4, 9], [7, 8], [2, 6]]
Output: 14

A pack starting on 14 gets through the round in the order [4,9], [2,6], [7,8]: 14 clears the threshold 9 and drops to 10, 10 clears 6 and drops to 8, 8 clears 8 and drops to 1.

Example 2

Input: calls = [[5, 5], [3, 3]]
Output: 8

Each call burns exactly what its threshold demands, so the pack carries the full 8 into the round: it clears the first threshold, drops to 3, clears the second and finishes on 0.

Example 3

Input: calls = [[1, 10000], [6, 11], [9, 9]]
Output: 10000

One call cannot start below 10000, and a pack on 10000 runs it down to 9999, then the call needing 11 down to 9993, then the call needing 9 down to 9984.

Constraints

1 <= calls.length <= 10^5
calls[i].length == 2
1 <= drain_i <= floor_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_starting_charge(calls: list[list[int]]) -> int:

Java

public int smallestStartingCharge(int[][] calls)

The intended approach

Time O(n log n) for n calls, dominated by the sort.Space O(n) for the ordered copy of the calls.

Compare two calls that would sit next to each other in the round: which of the two should come first turns only on the slack each one leaves, that is its threshold less its drain, and the wider slack belongs earlier. Sorting the whole round by that slack from wide to narrow, then walking it while topping the pack up by exactly the shortfall whenever a threshold is out of reach, adds up to the answer.

Pythonentry point: smallest_starting_charge

def smallest_starting_charge(calls):
    # Order the calls so that the one leaving the widest gap between its
    # threshold and its drain is served first, then top the pack up only when a
    # threshold is out of reach.
    # O(n log n) time, O(n) space for the ordered copy.
    order = sorted(calls, key=lambda c: c[1] - c[0], reverse=True)
    start = 0
    charge = 0
    for drain, floor in order:
        if charge < floor:
            start += floor - charge
            charge = floor
        charge -= drain
    return start

The obvious version

Try every order of the calls. For a fixed order the charge needed is the largest of (drain already burnt + this call's threshold), so take the smallest such figure over all orders. O(n! * n).

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_starting_charge(calls):
    from itertools import permutations
    best = None
    for order in permutations(calls):
        used = 0
        need = 0
        for drain, floor in order:
            if used + floor > need:
                need = used + floor
            used += drain
        if best is None or need < best:
            best = need
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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