i am looking for a way to get a sequence of two array items in an array, especially an array of length 4
for example,
[[1,0], [2,3], [5,4], [0,0], [3,2], [1,4], [0,5]]
... should return :
[[3,2], [2,3], [1,4], [0,5]]
3 --^ 2 -----^ 1 ----^ 0 -----^ so [3, 2, 1, 0]
for x
[[3,2], [2,3], [1,4], [0,5]]
2 -----^ 3 -----^ 4 ----^ 5 -----^ so [2, 3, 4, 5]
for y
[[x1, y1], [x2, y2], [x3, y3], [x4, y3]]
// 1 or -1 for the first index
// and 1 or -1 for the second index
[[3,2], [2,3], [1,4], [0,5]] // is a sequence
[[0,0], [1,1], [2,2], [3,3]] // is a sequence
[[4,4], [3,3], [2,2], [1,1]] // is a sequence
[[4,3], [3,3], [2,2], [1,1]] // is not a sequence
[[1,2], [2,3], [4,5], [5,6]] // is a sequence
i tried to use for loops but it's illegible and confusing, maybe too difficult and this but it's only counting the longest sequence, not returning it :
const Z = x.sort((a, b) => a - b).reduce((count, val, i) => {
return count = val 1 === x[i 1] ? 1 : 0
}, 1);
const Z2 = y.sort((a, b) => a - b).reduce((count, val, i) => {
return count = val 1 === y[i 1] ? 1 : 0
}, 1);
console.log(Z, Z2) // 4 4
CodePudding user response:
You could add the points to an object and check the order of four by using a factor of adding an offset for x
and y
.
If needed, you could add the check for horizontal or vertical points, too.
const
four = array => {
const
data = array.reduce((r, [x, y]) => ((r[x] ??= {})[y] = true, r), {}),
check = ([x, y], i, j) => {
const temp = [];
for (let k = 0; k < 4; k , x = i, y = j) {
if (data[x]?.[y]) temp.push([x, y]);
else break;
}
if (temp.length === 4) return temp;
};
let result;
array.some(p => result = check(p, 1, 1) || check(p, -1, 1));
return result;
},
data = [[1, 0], [2, 3], [5, 4], [0, 0], [3, 2], [1, 4], [0, 5]];
console.log(four(data)); // [[3, 2], [2, 3], [1, 4], [0, 5]]
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