Appending only ever adds pairs, and the smallest step a '0' can add is the number of 1s so far: a prefix is valid when it already has k pairs, or when k minus its pairs is at least its count of 1s. Treat each prefix on its own. Appending characters can only add "10" pairs, never remove them, so a prefix that already has more than k pairs is out, and one with exactly k is in (append nothing). For a prefix with fewer than k pairs, ask which totals it can reach. Appending a '1' adds no pair by itself but makes every later '0' worth one more. Appending a '0' adds one pair for every '1' before it. So with `ones` 1s in the prefix, the smallest jump is `ones` (append one '0'), and any larger jump g is also reachable: append g - ones extra '1's first, then one '0'. The gap `k - pairs` is closable exactly when it is at least `ones`. When the prefix has no 1s yet, any gap works. The report that posts a solution recounts the pairs and 1s for every prefix, which is O(n^2); it passed all the tests. Both counts can instead be kept as you extend the prefix one character at a time: a '1' increases `ones`, a '0' adds `ones` to the pair count. That makes the whole scan O(n).