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448. Find All Numbers Disappeared in an Array

DifficultyEasy

Description

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 <= 105
  • 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.

Solutions

Solution 1

Thinking

Values lie in \([1,n]\); missing numbers are a set difference. Store seen values in a hash set and scan \(1..n\). Linear time, \(O(n)\) extra space.

The follow-up wants constant extra memory; this solution establishes correctness first.

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class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        s = set(nums)
        return [x for x in range(1, len(nums) + 1) if x not in s]

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class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        int n = nums.length;
        boolean[] s = new boolean[n + 1];
        for (int x : nums) {
            s[x] = true;
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            if (!s[i]) {
                ans.add(i);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int n = nums.size();
        bool s[n + 1];
        memset(s, false, sizeof(s));
        for (int& x : nums) {
            s[x] = true;
        }
        vector<int> ans;
        for (int i = 1; i <= n; ++i) {
            if (!s[i]) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

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func findDisappearedNumbers(nums []int) (ans []int) {
    n := len(nums)
    s := make([]bool, n+1)
    for _, x := range nums {
        s[x] = true
    }
    for i := 1; i <= n; i++ {
        if !s[i] {
            ans = append(ans, i)
        }
    }
    return
}

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function findDisappearedNumbers(nums: number[]): number[] {
    const n = nums.length;
    const s: boolean[] = new Array(n + 1).fill(false);
    for (const x of nums) {
        s[x] = true;
    }
    const ans: number[] = [];
    for (let i = 1; i <= n; ++i) {
        if (!s[i]) {
            ans.push(i);
        }
    }
    return ans;
}

Solution 2

Thinking

Solution 1 uses a set. Value \(x\) can negate index \(|x|-1\) to mark “seen”; a still-positive slot \(i\) means \(i+1\) is missing. Absolute value before indexing avoids treating a mark as a subscript.

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class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        for x in nums:
            i = abs(x) - 1
            if nums[i] > 0:
                nums[i] *= -1
        return [i + 1 for i in range(len(nums)) if nums[i] > 0]

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class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        int n = nums.length;
        for (int x : nums) {
            int i = Math.abs(x) - 1;
            if (nums[i] > 0) {
                nums[i] *= -1;
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (nums[i] > 0) {
                ans.add(i + 1);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int n = nums.size();
        for (int& x : nums) {
            int i = abs(x) - 1;
            if (nums[i] > 0) {
                nums[i] = -nums[i];
            }
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            if (nums[i] > 0) {
                ans.push_back(i + 1);
            }
        }
        return ans;
    }
};

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func findDisappearedNumbers(nums []int) (ans []int) {
    n := len(nums)
    for _, x := range nums {
        i := abs(x) - 1
        if nums[i] > 0 {
            nums[i] = -nums[i]
        }
    }
    for i := 0; i < n; i++ {
        if nums[i] > 0 {
            ans = append(ans, i+1)
        }
    }
    return
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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function findDisappearedNumbers(nums: number[]): number[] {
    const n = nums.length;
    for (const x of nums) {
        const i = Math.abs(x) - 1;
        if (nums[i] > 0) {
            nums[i] *= -1;
        }
    }
    const ans: number[] = [];
    for (let i = 0; i < n; ++i) {
        if (nums[i] > 0) {
            ans.push(i + 1);
        }
    }
    return ans;
}

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