levenshtein distance

levenshtein.js

// https://en.wikipedia.org/wiki/Levenshtein_distance
function lv(A, B) {
    // create two work vectors of integer distances
    let v0 = new Array(A.length);
    let v1 = new Array(B.length);
    // initialize v0 (the previous row of distances)
    // this row is A[0][i]: edit distance for an empty s
    // the distance is just the number of characters to delete from t
    for(let i = 0; i <= B.length + 1; i++){
        v0[i] = i;
    }
    for(let i = 0; i < A.length + 1; i++){
        // calculate v1 (current row distances) from the previous row v0
        // first element of v1 is A[i+1][0]
        //   edit distance is delete (i+1) chars from s to match empty t
        v1[0] = i + 1
        // use formula to fill in the rest of the row
        for(let j = 0; j < B.length + 1; j++){
            // calculating costs for A[i+1][j+1]
            let deletionCost = v0[j + 1] + 1 || 0;
            let insertionCost = v1[j] + 1;
            let substitutionCost = A[i] === B[j] ? v0[j] : v0[j] + 1 || 0;
            v1[j + 1] = Math.min(deletionCost, insertionCost, substitutionCost)
        }
        // copy v1 (current row) to v0 (previous row) for next iteration
        // since data in v1 is always invalidated, a swap without copy could be more efficient
        let temp = v0;
        v0 = v1.concat([]);
        v1 = temp.concat([]);
    }
    // after the last swap, the results of v1 are now in v0
    return v0[B.length + 1];
}