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perf(LCS): find the candidate to extend by binary search #89

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perf(LCS): find the candidate to extend by binary search #89

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20 changes: 16 additions & 4 deletions src/diff3.mjs
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Original file line number Diff line number Diff line change
Expand Up @@ -45,12 +45,24 @@ function LCS(buffer1, buffer2) {
let c = candidates[0];

for (const j of buffer2indices) {
let s;
for (s = r; s < candidates.length; s++) {
if ((candidates[s].buffer2index < j) && ((s === candidates.length - 1) || (candidates[s + 1].buffer2index > j))) {
break;
// Find the candidate `s` (at or after `r`) that `j` extends: the one with
// `candidates[s].buffer2index < j`, where the next one, if any, is past `j`.
// The `buffer2index` values from `r` on are strictly increasing, so the
// boundary is found by binary search instead of a linear scan.
let lo = r;
let hi = candidates.length;
while (lo < hi) {
const mid = (lo + hi) >>> 1;
if (candidates[mid].buffer2index < j) {
lo = mid + 1;
} else {
hi = mid;
}
}
let s = candidates.length;
if (lo > r && (lo === candidates.length || candidates[lo].buffer2index > j)) {
s = lo - 1;
}

if (s < candidates.length) {
const newCandidate = { buffer1index: i, buffer2index: j, chain: candidates[s] };
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