findAllNumbersDisappearedInAnArray.ts

/*
Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

 

Example 1:
Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]

Example 2:
Input: nums = [1,1]
Output: [2]

Constraints:
n == nums.length
1 <= n <= 10^5
1 <= nums[i] <= n
 

Follow up: Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
*/

function findDisappearedNumbers(nums: number[]): number[] {
  const length = nums.length;
  const numberCount = new Array(length + 1).fill(0);
  
  for (let index = 0; index < length; index++) {
		numberCount[nums[index]]++;
  }
  
  const missingNumber = [];
  for (let index = 1; index <= length; index++) {
    if (numberCount[index] === 0) {
      missingNumber.push(index);
    }
  }

  return missingNumber;
}

/*
  We will use the input array as a marker
  With each nums, we will negate nums[index - 1] to mark it seen,
  then collect the indices where the values are still positive
*/
function findDisappearedNumbersV2(nums: number[]): number[] {
  for (const num of nums) {
    const index = Math.abs(num) - 1;
    nums[index] = -Math.abs(nums[index]);
  }

  const missingNumbers = [];
  for (let index = 0; index < nums.length; index++) {
    if (nums[index] > 0) {
      missingNumbers.push(index + 1);
    }
  }

  return missingNumbers;
}

console.log(findDisappearedNumbersV2([4, 3, 2, 7, 8, 2, 3, 1])); // [5,6]
console.log(findDisappearedNumbersV2([1, 1]));                    // [2]