I have an array of arrays that looks like this:
var grid = [
[0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0],
[0,0,0,0,0,1,1,0,0,0,1,0,0,0,1,0,0,0,1,1,0,0,0,0,0]
];
I want to count the number of 1s in a row until we hit a zero. And then if we hit a 1 again, count the sequential occurrences again.
So for grid[0], it should return 9.
For grid[1], it should return [3,3].
grid[2], [2,1,1,2].
As an extra layer of fun, I'm also trying to do this for "columns." I.e. How many times "1" appears in the 1st column, which would consist of grid[0][1], grid[1][1], grid[2][1] and so on. If there's a better way to organize the data to achieve this, I'm open to suggestions as in total I have 25 rows/arrays within the array.
I'm unsure if there is a way that doesn't involve looping through the data over and over. Currently I'm doing this:
var guides = [];
for ( var i = 0; i < grid.length; i++ ) {
var row = grid[i];
var chunks = [];
var count = 0;
for (var j = 0; j < row.length; j++ ) {
if ( j === 1 ) {
count++;
} else {
chunks.push(count);
count = 0;
}
}
guides.push(chunks);
}
Here's one example how you can use recursion to check those number 1 strikes. For the columns, you can transpose the original Grid and re-use the same function.
var grid = [
[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
]
// Returns the array of 1 strikes in the given arr
const occurrenceOfOne = (arr, strike = 0, final = []) => {
// Recursive calls until the array is empty
if (arr.length === 0) {
return strike === 0 ? final : final.concat([strike])
}
// If the item is 0, the strike ends.
if (arr[0] === 0) {
if (strike !== 0) {
return occurrenceOfOne(arr.slice(1), 0, final.concat([strike]))
}
}
// If the item is 1, the strike continues
if (arr[0] === 1) {
return occurrenceOfOne(arr.slice(1), strike + 1, final)
}
// Default value 0 found and strike is 0 as well.
return occurrenceOfOne(arr.slice(1), 0, final)
}
// Copied transpose function from this gist:
// https://gist.github.com/femto113/1784503
function transpose(a) {
return a[0].map((_, c) => a.map((r) => r[c]))
}
console.log(occurrenceOfOne(grid[0]))
console.log(occurrenceOfOne(grid[1]))
console.log(occurrenceOfOne(grid[2]))
const transposedGrid = transpose(grid)
console.log(occurrenceOfOne(transposedGrid[6]))
console.log(occurrenceOfOne(transposedGrid[7]))