Find all the same numbers in the array

Question

I have an array with numbers in the range of 0 - 100. I need to find all the same numbers and add 1 to them.

my code worked well with arrays like [100, 2, 1, 1, 0]

const findAndChangeDuplicates = (arr: any) => {
    for (let i = arr.length - 1; i >= 0; i--) {
        if (arr[i + 1] === arr[i] && arr[i] <= 5) {
            arr[i] += 1;
        } else if (arr[i - 1] === arr[i] && arr[i] >= 5) {
            arr[i] -= 1;

            findAndChangeDuplicates(arr);
        }
    }

    return arr;
};

but when I came across this [100, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0]

my code let me down.

Expected Result: [100, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Have any ideas?

Answer 1

An approach by using at least one loop from the end to adjust the values and if necessary another loop from the beginning to set the largest value to 100.

Both loops feature a value variable v. In the first loop, it starts with the last value of the array and increments its value and check is the item is smaller than this value.

If smaller, then the value is assigned, otherwise the actual value is taken for the next item.

if necessary, the other loop works in opposite direction and with a start value of 100 and checks if the item is greater than wanted and takes the smaller value, or the value is taken from the item.

The result is an array which has a gereatest value of 100 at start and goes until zero or greater to the end of the array.

function update(array) {
    var i = array.length,
        v = array[--i];

    while (i--) if (array[i] < ++v) array[i] = v; else v = array[i];
    if (array[0] > 100) {
        v = 100;
        for (i = 0; i < array.length; i++) {
            if (array[i] > v) array[i] = v; else v = array[i];
            v--;
        }
    }
    return array;
}

console.log(update([100, 2, 1, 1, 0]));
console.log(update( [100, 100, 99, 86, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0]))

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

The following assumes you want them ordered from highest to lowest, if not this might ba as well as useless to you.

The idea is to first create an Object to keep track of how many of each number exist. We then map each value by first checking whether it's unique and if not increasing it until we can't find any value inside the Object anymore. This will not neatly order the numbers by itself so we will have to sort afterwards.

let arr1 = [100, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 0, 0],
    arr2 = [100, 2, 1, 1, 0];

const f = (arr) => arr.reduce((a,c) => (a[c] = (a[c] || 0) + 1, a),{}),
      g = (arr, obj) => arr.map(v => {
        if (obj[v] > 1) {
          let i = 1;
          obj[v] = obj[v] - 1;
          while (obj[v + i]) {
            i++;
          }
          obj[v + i] = (obj[v + i] || 0) + 1;
          return v + i;
        } else {
          return v;
        }
      }).sort((a,b) => +b - +a);

console.log(g(arr1, f(arr1)))
console.log(g(arr2, f(arr2)))