How should I practice Max Consecutive ON Servers?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Amazon online assessment problems in FastPrep.

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FastPrep

Problem · String

Max Consecutive ON Servers

● HardAmazonFULLTIMEOA

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Amazon operates a network of n machines. Some are currently inactive, while others are active. The engineers managing the infrastructure have access to a specific procedure that allows toggling the states of machines. In one use of this operation, they can select a continuous segment of machines and invert their statuses—turning active to inactive, and inactive to active. Due to certain restrictions, this procedure can be used at most k times.

You are given a binary string machine_status, of length n, where '1' indicates an active machine and '0' represents an inactive one, along with an integer max_flips that denotes the maximum number of operations allowed. Determine the highest possible number of adjacent active machines that can be achieved after at most max_flips operations.

Function Description

Complete the function getMaxConsecutiveON in the editor.

getMaxConsecutiveON has the following parameters:

String machine_status: states of the servers, '1' represents ON state, '0' represents OFF
int max_flips: the maximum number of operations that can be performed

Returns

int: The greatest number of back-to-back active machines achievable using no more than max_flips operations

🌷༊·˚Credit to ˚꒰აBarry ᯓᡣ𐭩໒꒱˚

Examples

01 · Example 1

machine_status = "00010"
max_flips = 1
return = 4

The optimal approach is to perform one flip on the range of indices [0..2]. After flipping: "11110" This results in a sequence of 4 consecutive active machines.

02 · Example 2

machine_status = "1001"
max_flips = 2
return = 4

You can flip indices [1..2] to obtain "1111". Only one operation is needed to achieve a maximum of 4 continuous active machines. There’s no advantage in using the second operation.

03 · Example 3

machine_status = "11101010110011"
max_flips = 2
return = 8

Flip indices [7..9]: "11101011000011" Then flip [8..11]: "11101011111111" This strategy yields a block of 8 active machines in a row.

Constraints

1 ≤ n ≤ 2 * 10^5
1 ≤ max_flips ≤ 2 * 10^5
machine_status contains only 0s and 1s