https://leetcode.com/problems/minimum-knight-moves/

minimumKnightMoves.js

// Straight-up BFS. Very slow but accepted...
var minKnightMoves = function(x, y) {
    
    const queue = [[0, 0, 0]]; // x, y, distance
    const visited = new Map(); // `x:y`: distance
    while(queue.length){
        const [X, Y, dist] = queue.shift();
        if(visited.has(`${X}:${Y}`)) continue;
        visited.set(`${X}:${Y}`, dist);
        if(X === x && Y === y) return dist;
        
        [
            [X + 2, Y + 1],
            [X + 2, Y - 1],
            [X - 2, Y + 1],
            [X - 2, Y - 1],
            [X + 1, Y + 2],
            [X + 1, Y - 2],
            [X - 1, Y + 2],
            [X - 1, Y - 2],
        ].forEach(([x1, y1]) => {
           if(!visited.has(`${x1}:${y1}`)){
               queue.push([x1, y1, dist + 1]);
           }
        });
    }  
};

// Seriously Optimized after getting some inspiration...
var minKnightMoves = function(x, y) {

    x = Math.abs(x); // since the board is symetrical, make everything positive
    y = Math.abs(y); // to make life easier for ourselves
    const s = [ // if our knight is < 5 squares away, we just cache the answer.
        [0, 3, 2, 3, 2], // s[x][y] = the number of moves the knight must make
        [3, 2, 1, 2, 3], // to get from s[x][y] to 0,0.
        [2, 1, 4, 3, 2],
        [3, 2, 3, 2, 3],
        [2, 3, 2, 3, 4]
    ];
    
    let result = 0; // count moves
    
    // while the knight is > 4 squares away, be greedy.
    while(x > 4 || y > 4){ // if its further away in the x axis
        if(x > y){ // prioritise moving x
            if(x > 0) x-=2; // move two squares along the x
            else x+=2; // and one along the Y
            y = y > 0 ? (y-1) : (y+1);
        }else{ // prioritise moving y
            if(y > 0) y-=2;
            else y+=2;
            x = x > 0 ? (x-1) : (x+1);
        }
        result++;
    }
    // Woohoo! Our knight is now < 5 squares from the end
    return result + s[x][y]; // return the cached answer from this point
    
};