Finding the first index of a sequence inside an array

Question

In this function I count the number of consecutive zeros that should be at least of length zerosMin. Is there any way I could return the first index of the start of the sequences? For example, for arr = [1,0,0,0,0,0,1,0,0,0,0] it would be [1,7]

function SmallestBlockOfZeros(arr, zerosMin) {
    let result = [];
    let counter = 1;

    for (let i = 0; i < arr.length; i++) {
        if (arr[i] == 0) {
            if (arr[i] === arr[i + 1]) {
                counter++;
            } else if (counter >= zerosMin) {
                result.push(counter);

                counter = 1;
            } else {
                counter = 1;
            }
        }
    }
    return result;
}
let arr = [1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0];
let zerosMin = 4;

console.log(SmallestBlockOfZeros(arr, zerosMin));
//Console output : [5,4]

score 1 · Accepted Answer · answered Jan 06 '21 at 01:09

You can iterate through the array, pushing an index into the result array whenever you encounter a block of 0 values that is at least zerosMin long, and then skipping all those 0 values until you find the next non-0 value or pass the point at which a zerosMin string of 0s could occur:

let arr = [1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0];
let zerosMin = 4;

let zeroStarts = [];
for (let i = 0; i <= arr.length - zerosMin; i++) {
  if (arr.slice(i, i + zerosMin).filter(Boolean).length == 0) {
    zeroStarts.push(i);
    i += zerosMin;
    while (i <= arr.length - zerosMin && arr[i] == 0) {
      i++;
    }
  }
}
console.log(zeroStarts);

Note I've used filter(Boolean) because 0 is falsey; if you were searching for a run of non-zero values you would use something like filter(v => v != 0)

score 1 · Answer 2 · answered Jan 06 '21 at 01:18

1

If you don't need to actually iterate elements, you can use a shortcut

function foo(arr, min)
{
  let el = arr.join("").match(new RegExp('0{' + min + ',}'));
  
  return [el.index, el.index + el[0].length];
}

console.log(foo([1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0], 3));

answered Jan 06 '21 at 01:18

John

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This is a good idea but your implementation doesn't exactly match what OP asks for – Nick Jan 06 '21 at 01:23

score 0 · Answer 3 · answered Jan 06 '21 at 01:08

For start of each chain of zeros:

function indexOfValue(arr, val) {
    var indexes = [], i;
    for(i = 0; i < arr.length; i++)
        if (arr[i] === val)
            indexes.push(i);
    return indexes;
}

(see How to find index of all occurrences of element in array?)

The code you posted above appears to work for length of blocks of 0s.

score 0 · Answer 4 · answered Jan 06 '21 at 05:31

While iterating the items, keep track of last non-zero index and push the index to output array when criteria matches.

const SmallestBlockOfZeros = (arr, zerosMin) => {
  let index = 0;
  const output = [];
  for (let i = 0; i <= arr.length; i++) {
    if (arr[i] !== 0 || !(i in arr)) {
      i - index - 1 >= zerosMin && output.push(index + 1);
      index = i;
    }
  }
  return output;
};

const arr = [1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0];
const zerosMin = 4;
console.log(SmallestBlockOfZeros(arr, zerosMin));

Finding the first index of a sequence inside an array

4 Answers4