0754MediumArrayMathBit Manipulation

Differing Flag Bits Across Registers

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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.

Every unit on a shop floor publishes its state as one non-negative integer whose binary digits act as independent flags. Line up the binary forms of two states, padding the shorter one with leading zeros so both have the same width. Their disagreement is how many bit positions hold a 1 in exactly one of the two states.

You are handed registers, the current state of each unit in reporting order. Add up the disagreement of every unordered pair of two different positions in registers and return that total. Two units holding the same state still form a pair; their disagreement is 0.

A list with a single unit has no pairs at all.

Examples

Example 1

Input: registers = [6, 11, 20, 3]
Output: 17

Written to five bit positions the four states are 00110, 01011, 10100 and 00011. The six pairs disagree in 3, 2, 2, 5, 1 and 4 positions respectively, and those add up to the returned total.

Example 2

Input: registers = [9, 9, 24]
Output: 4

The first two states are identical, so that pair disagrees nowhere. State 9 is 01001 and state 24 is 11000, which disagree in 2 positions, and that pair occurs twice.

Example 3

Input: registers = [7, 7, 7, 7]
Output: 0

All four units publish the same state, so every one of the six pairs disagrees nowhere.

Constraints

1 <= registers.length <= 10^4
0 <= registers[i] <= 10^9
Under these bounds the returned total is at most 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 total_flag_disagreement(registers: list[int]) -> int:

Java

public int totalFlagDisagreement(int[] registers)

The intended approach

Time O(30 n) for n states: one sweep of the list per bit position.Space O(1) beyond the input.

Handle one bit position at a time. For a fixed position, count how many states hold a `1` there; call it `ones`. A pair disagrees at that position exactly when one member is in that group and the other is not, so the position contributes `ones * (n - ones)` to the total. Summing that over the 30 positions a value below 2^30 can use gives the answer without ever forming a pair explicitly.

Pythonentry point: total_flag_disagreement

def total_flag_disagreement(registers):
    # Each bit position contributes independently, so one sweep per position is
    # enough. Values stay below 2^30, so 30 positions cover every input.
    # Time O(30 n), space O(1).
    n = len(registers)
    total = 0
    for bit in range(30):
        mask = 1 << bit
        ones = 0
        for v in registers:
            if v & mask:
                ones += 1
        total += ones * (n - ones)
    return total

The obvious version

Walk every unordered pair of positions, exclusive-or the two states, and shift the result down one bit at a time adding up the 1 bits. Quadratic in the number of units, but it follows the definition literally.

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_total_flag_disagreement(registers):
    n = len(registers)
    total = 0
    for i in range(n):
        for j in range(i + 1, n):
            diff = registers[i] ^ registers[j]
            while diff:
                total += diff & 1
                diff >>= 1
    return total

How this problem was checked

Reference solution executed against 9 test cases — 9 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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