C  Bit Counting Sequence

Observation C.1. For a non-negative integer \(y\), if \(p(y) - p(y+1) = t \ge 0\), the rightmost \((t+1)\) bits in the binary representation of \(y\) are all \(1\). Additionally, the \((t+2)\)-th bit of \(y\) from the right is \(0\).

For example, consider \(y = 87 = (1010111)_2\). Here, \(t = p(y) - p(y+1) = 5 - 3 = 2\). As the observation states, the rightmost three bits of \(y\) are \(1\), and the fourth bit of \(y\) from the right is \(0\).

Let \(k\) be the smallest index such that \(a_k = \max(a_1, \ldots, a_n)\). Also, let \(x\) denote the solution to this problem.

When \(k \lt n\), let \(t = a_k - a_{k+1}\). It’s worth noting that \(t \ge 0\) according to the definition of \(k\). Based on the previous observation, the binary representation of \(x + k - 1\) should be like

The bits starting from the \((t+3)\)-th position from the right of \(x + k - 1\) cannot be determined solely from this observation. However, considering that \(a_k\) is the largest value among \(a_1, \ldots, a_n\), for \(s = a_k - (t+1)\),

is the smallest possible value for \(x + k - 1\). Consequently, we can compute the smallest possible value of \(x\). If any of the following conditions hold: \(s \lt 0\), \(x \lt 0\), or \(p(x + i - 1) \ne a_i\) for some \(i\), then the solution does not exist.

Similarly, when \(k = n\), the smallest possible value for \(x + n - 1\) is \(2^{a_n} - 1\). The runtime complexity is \(O(n)\).