Problem

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.

Examples

Example 1:

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

Example 2:

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

Solution

We can use usual methods like hashmap, etc, but here we can use numbers as indices.

Method 1 - Using Numbers as Indices

As numbers are between [1,n], we can use them as index [0, n-1]. When we see the number, we will make nums[num-1] as negative as a visited array.

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public List<Integer> findDisappearedNumbers(int[] nums) {
	List<Integer> ans = new ArrayList<>();

	for (int i = 0; i < nums.length; i++) {
		int idx = Math.abs(nums[i]) - 1;
		if (nums[idx] > 0) {
			nums[idx] = -nums[idx];
		}
	}
	
	for (int i = 0; i < nums.length; i++) {
		if (nums[i] > 0) {
			result.add(i + 1);
		}
	}
	return result;
}

Complexity

  • ⏰ Time complexity: O(n)
  • 🧺 Space complexity: O(1)

Con of this approach is array modification.