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17.04. Missing Number

Description

An array contains all the integers from 0 to n, except for one number which is missing.  Write code to find the missing integer. Can you do it in O(n) time?

Note: This problem is slightly different from the original one the book.

Example 1: 

Input: [3,0,1]

Output: 2

 

Example 2: 

Input: [9,6,4,2,3,5,7,0,1]

Output: 8

Solutions

Solution 1

Thinking

One number is missing from \(0\ldots n\). A boolean mark array works but uses linear extra space.

After sorting, index should equal value; the first mismatch is the missing number, or \(n\) if none.

sort then scan is the shortest correct solution, in \(O(n\log n)\).

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class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        nums.sort()
        for i, x in enumerate(nums):
            if i != x:
                return i
        return len(nums)

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class Solution {
    public int missingNumber(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (i != nums[i]) {
                return i;
            }
        }
        return n;
    }
}

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class Solution {
public:
    int missingNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            if (i != nums[i]) {
                return i;
            }
        }
        return n;
    }
};

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func missingNumber(nums []int) int {
    sort.Ints(nums)
    for i, x := range nums {
        if i != x {
            return i
        }
    }
    return len(nums)
}

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impl Solution {
    pub fn missing_number(mut nums: Vec<i32>) -> i32 {
        nums.sort();
        let n = nums.len() as i32;
        for i in 0..n {
            if i != nums[i as usize] {
                return i;
            }
        }
        n
    }
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        if (i != nums[i]) {
            return i;
        }
    }
    return n;
};

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class Solution {
    func missingNumber(_ nums: [Int]) -> Int {
        let nums = nums.sorted()
        for (i, x) in nums.enumerated() {
            if i != x {
                return i
            }
        }
        return nums.count
    }
}

Solution 2

Thinking

The log factor of sorting is unnecessary: the closed sum is \(n(n+1)/2\), minus the array sum.

One accumulation and constant extra space, with no comparisons.

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class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        return sum(range(len(nums) + 1)) - sum(nums)

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class Solution {
    public int missingNumber(int[] nums) {
        int n = nums.length;
        int ans = n;
        for (int i = 0; i < n; ++i) {
            ans += i - nums[i];
        }
        return ans;
    }
}

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

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func missingNumber(nums []int) (ans int) {
    ans = len(nums)
    for i, x := range nums {
        ans += i - x
    }
    return
}

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impl Solution {
    pub fn missing_number(nums: Vec<i32>) -> i32 {
        let n = nums.len() as i32;
        let mut sum = 0;
        let mut max = 0;
        for num in nums {
            sum += num;
            max = max.max(num);
        }
        if max == n {
            ((1 + max) * max) / 2 - sum
        } else {
            n
        }
    }
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
    const n = nums.length;
    let ans = n;
    for (let i = 0; i < n; ++i) {
        ans += i - nums[i];
    }
    return ans;
};

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class Solution {
    func missingNumber(_ nums: [Int]) -> Int {
        let n = nums.count
        return n * (n + 1) / 2 - nums.reduce(0, +)
    }
}

Solution 3

Thinking

The sum may overflow in a narrow integer type.

XOR of \(0\ldots n\) with the array cancels pairs and leaves the missing value, without carry or a table.

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class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        ans = 0
        for i, x in enumerate(nums, 1):
            ans ^= i ^ x
        return ans

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class Solution {
    public int missingNumber(int[] nums) {
        int ans = 0;
        for (int i = 1; i <= nums.length; ++i) {
            ans ^= i ^ nums[i - 1];
        }
        return ans;
    }
}

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class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int ans = 0;
        for (int i = 1; i <= nums.size(); ++i) {
            ans ^= i ^ nums[i - 1];
        }
        return ans;
    }
};

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func missingNumber(nums []int) (ans int) {
    for i, x := range nums {
        ans ^= (i + 1) ^ x
    }
    return
}

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impl Solution {
    pub fn missing_number(nums: Vec<i32>) -> i32 {
        let mut res = 0;
        let n = nums.len();
        for i in 0..n {
            res ^= nums[i] ^ ((i + 1) as i32);
        }
        res
    }
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
    let ans = 0;
    for (let i = 1; i <= nums.length; ++i) {
        ans ^= i ^ nums[i - 1];
    }
    return ans;
};

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