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Use uniform balanced binary sequence sampling by choosing the positions of the heads. Number the $2n=2^m$ positions from zero to $2n-1$. Start with no selected positions. Repeatedly toss the fair coin $m$ times, interpret the resulting bits as a position, and select that position if it has not already been selected; otherwise discard this block of tosses and retry. Stop when $n$ distinct positions have been selected. Put heads at those positions and tails everywhere else.

After $s$ positions have been selected, every one of the remaining positions has the same probability of being the next new position. This follows either by summing a geometric series over rejected blocks or by conditioning a uniform distribution on a finite set on avoiding the selected positions. Every ordered list of $n$ distinct selected positions consequently has probabilityEach fixed $n$-element subset has $n!$ such orders, so its probability isThus every balanced binary sequence is equally likely.

When $s$ positions are selected, a block succeeds with probability $(2n-s)/(2n)$, so the geometric distribution gives expectation $2n/(2n-s)$ blocks until the next new position. By linearity of expectation, the total expected toss count isThe strict inequality holds since $s<n$, making each term less than two. A Boolean array records membership of each position; initializing it and constructing the output require $O(n)$ additional work. The scheme terminates with probability one, since every stage has positive success probability.