645. Set Mismatch - Explanation

Prerequisites

Before attempting this problem, you should be comfortable with:

Arrays - Iterating through and manipulating array elements
Hash Maps / Frequency Counting - Tracking how many times each element appears
Sorting - Understanding how sorted order reveals duplicates and gaps
Math (Sum Formulas) - Using arithmetic sum formulas to detect missing or duplicate values
Bit Manipulation (XOR) - Understanding XOR properties for finding unique elements

1. Brute Force

Intuition

The array should contain each number from 1 to n exactly once, but one number is duplicated and one is missing. The most direct approach is to count how many times each number from 1 to n appears in the array. A number appearing twice is the duplicate, and a number appearing zero times is the missing one.

Algorithm

Initialize a result array to store [duplicate, missing].
For each number i from 1 to n:
- Count how many times i appears in the array.
- If the count is 0, i is the missing number.
- If the count is 2, i is the duplicate number.
Return the result.

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        res = [0, 0]
        n = len(nums)

        for i in range(1, n + 1):
            cnt = 0
            for num in nums:
                if num == i:
                    cnt += 1

            if cnt == 0:
                res[1] = i
            elif cnt == 2:
                res[0] = i

        return res

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int[] res = new int[2];
        int n = nums.length;

        for (int i = 1; i <= n; i++) {
            int cnt = 0;
            for (int num : nums) {
                if (num == i) {
                    cnt++;
                }
            }

            if (cnt == 0) {
                res[1] = i;
            } else if (cnt == 2) {
                res[0] = i;
            }
        }

        return res;
    }
}

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        vector<int> res(2, 0);
        int n = nums.size();

        for (int i = 1; i <= n; i++) {
            int cnt = 0;
            for (int num : nums) {
                if (num == i) {
                    cnt++;
                }
            }

            if (cnt == 0) {
                res[1] = i;
            } else if (cnt == 2) {
                res[0] = i;
            }
        }

        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    findErrorNums(nums) {
        const res = [0, 0];
        const n = nums.length;

        for (let i = 1; i <= n; i++) {
            let cnt = 0;
            for (const num of nums) {
                if (num === i) {
                    cnt++;
                }
            }

            if (cnt === 0) {
                res[1] = i;
            } else if (cnt === 2) {
                res[0] = i;
            }
        }

        return res;
    }
}

public class Solution {
    public int[] FindErrorNums(int[] nums) {
        int[] res = new int[2];
        int n = nums.Length;

        for (int i = 1; i <= n; i++) {
            int cnt = 0;
            foreach (int num in nums) {
                if (num == i) {
                    cnt++;
                }
            }

            if (cnt == 0) {
                res[1] = i;
            } else if (cnt == 2) {
                res[0] = i;
            }
        }

        return res;
    }
}

func findErrorNums(nums []int) []int {
    res := []int{0, 0}
    n := len(nums)

    for i := 1; i <= n; i++ {
        cnt := 0
        for _, num := range nums {
            if num == i {
                cnt++
            }
        }

        if cnt == 0 {
            res[1] = i
        } else if cnt == 2 {
            res[0] = i
        }
    }

    return res
}

class Solution {
    fun findErrorNums(nums: IntArray): IntArray {
        val res = intArrayOf(0, 0)
        val n = nums.size

        for (i in 1..n) {
            var cnt = 0
            for (num in nums) {
                if (num == i) {
                    cnt++
                }
            }

            if (cnt == 0) {
                res[1] = i
            } else if (cnt == 2) {
                res[0] = i
            }
        }

        return res
    }
}

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var res = [0, 0]
        let n = nums.count

        for i in 1...n {
            var cnt = 0
            for num in nums {
                if num == i {
                    cnt += 1
                }
            }

            if cnt == 0 {
                res[1] = i
            } else if cnt == 2 {
                res[0] = i
            }
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$

2. Sorting

Intuition

After sorting, consecutive elements should differ by exactly 1. If two adjacent elements are equal, we found the duplicate. If two adjacent elements differ by 2, the missing number lies between them. We also need to handle the edge case where the missing number is n (the last element after sorting is not n).

Algorithm

Sort the array.
Initialize the result with missing = 1 (handles the case where 1 is missing).
Iterate through adjacent pairs:
- If two elements are equal, record the duplicate.
- If two elements differ by 2, the missing number is in between.
If the last element is not n, then n is the missing number.
Return [duplicate, missing].

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        res = [0, 1]
        nums.sort()

        for i in range(1, len(nums)):
            if nums[i] == nums[i - 1]:
                res[0] = nums[i]
            elif nums[i] - nums[i - 1] == 2:
                res[1] = nums[i] - 1

        if nums[-1] != len(nums):
            res[1] = len(nums)
        return res

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int[] res = {0, 1};
        Arrays.sort(nums);

        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1]) {
                res[0] = nums[i];
            } else if (nums[i] - nums[i - 1] == 2) {
                res[1] = nums[i] - 1;
            }
        }

        if (nums[nums.length - 1] != nums.length) {
            res[1] = nums.length;
        }
        return res;
    }
}

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        vector<int> res = {0, 1};
        sort(nums.begin(), nums.end());

        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] == nums[i - 1]) {
                res[0] = nums[i];
            } else if (nums[i] - nums[i - 1] == 2) {
                res[1] = nums[i] - 1;
            }
        }

        if (nums.back() != nums.size()) {
            res[1] = nums.size();
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    findErrorNums(nums) {
        const res = [0, 1];
        nums.sort((a, b) => a - b);

        for (let i = 1; i < nums.length; i++) {
            if (nums[i] === nums[i - 1]) {
                res[0] = nums[i];
            } else if (nums[i] - nums[i - 1] === 2) {
                res[1] = nums[i] - 1;
            }
        }

        if (nums[nums.length - 1] !== nums.length) {
            res[1] = nums.length;
        }
        return res;
    }
}

public class Solution {
    public int[] FindErrorNums(int[] nums) {
        int[] res = {0, 1};
        Array.Sort(nums);

        for (int i = 1; i < nums.Length; i++) {
            if (nums[i] == nums[i - 1]) {
                res[0] = nums[i];
            } else if (nums[i] - nums[i - 1] == 2) {
                res[1] = nums[i] - 1;
            }
        }

        if (nums[nums.Length - 1] != nums.Length) {
            res[1] = nums.Length;
        }
        return res;
    }
}

class Solution {
    fun findErrorNums(nums: IntArray): IntArray {
        val res = intArrayOf(0, 1)
        nums.sort()

        for (i in 1 until nums.size) {
            if (nums[i] == nums[i - 1]) {
                res[0] = nums[i]
            } else if (nums[i] - nums[i - 1] == 2) {
                res[1] = nums[i] - 1
            }
        }

        if (nums[nums.size - 1] != nums.size) {
            res[1] = nums.size
        }
        return res
    }
}

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var res = [0, 1]
        var nums = nums.sorted()

        for i in 1..<nums.count {
            if nums[i] == nums[i - 1] {
                res[0] = nums[i]
            } else if nums[i] - nums[i - 1] == 2 {
                res[1] = nums[i] - 1
            }
        }

        if nums[nums.count - 1] != nums.count {
            res[1] = nums.count
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

3. Frequency Count (Hash Table)

Intuition

Using extra space, we can count occurrences in a single pass and then check each number's frequency. A frequency array of size n+1 lets us directly index by number value. The number with count 2 is the duplicate, and the number with count 0 is the missing one.

Algorithm

Create a count array of size n+1 initialized to 0.
Iterate through the input array and increment count for each number.
Iterate through numbers 1 to n:
- If count[i] is 0, i is the missing number.
- If count[i] is 2, i is the duplicate number.
Return [duplicate, missing].

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        res = [0, 0] # [duplicate, missing]
        count = Counter(nums)

        for i in range(1, len(nums) + 1):
            if count[i] == 0:
                res[1] = i
            if count[i] == 2:
                res[0] = i

        return res

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int n = nums.length;
        int[] count = new int[n + 1];
        int[] res = new int[2];

        for (int num : nums) {
            count[num]++;
        }

        for (int i = 1; i <= n; i++) {
            if (count[i] == 0) {
                res[1] = i;
            }
            if (count[i] == 2) {
                res[0] = i;
            }
        }

        return res;
    }
}

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        int n = nums.size();
        vector<int> count(n + 1, 0);
        vector<int> res(2, 0);

        for (int num : nums) {
            count[num]++;
        }

        for (int i = 1; i <= n; i++) {
            if (count[i] == 0) {
                res[1] = i;
            }
            if (count[i] == 2) {
                res[0] = i;
            }
        }

        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    findErrorNums(nums) {
        const n = nums.length;
        const count = Array(n + 1).fill(0);
        const res = [0, 0];

        for (const num of nums) {
            count[num]++;
        }

        for (let i = 1; i <= n; i++) {
            if (count[i] === 0) {
                res[1] = i;
            }
            if (count[i] === 2) {
                res[0] = i;
            }
        }

        return res;
    }
}

public class Solution {
    public int[] FindErrorNums(int[] nums) {
        int n = nums.Length;
        int[] count = new int[n + 1];
        int[] res = new int[2];

        foreach (int num in nums) {
            count[num]++;
        }

        for (int i = 1; i <= n; i++) {
            if (count[i] == 0) {
                res[1] = i;
            }
            if (count[i] == 2) {
                res[0] = i;
            }
        }

        return res;
    }
}

func findErrorNums(nums []int) []int {
    n := len(nums)
    count := make([]int, n+1)
    res := []int{0, 0}

    for _, num := range nums {
        count[num]++
    }

    for i := 1; i <= n; i++ {
        if count[i] == 0 {
            res[1] = i
        }
        if count[i] == 2 {
            res[0] = i
        }
    }

    return res
}

class Solution {
    fun findErrorNums(nums: IntArray): IntArray {
        val n = nums.size
        val count = IntArray(n + 1)
        val res = IntArray(2)

        for (num in nums) {
            count[num]++
        }

        for (i in 1..n) {
            if (count[i] == 0) {
                res[1] = i
            }
            if (count[i] == 2) {
                res[0] = i
            }
        }

        return res
    }
}

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        let n = nums.count
        var count = [Int](repeating: 0, count: n + 1)
        var res = [0, 0]

        for num in nums {
            count[num] += 1
        }

        for i in 1...n {
            if count[i] == 0 {
                res[1] = i
            }
            if count[i] == 2 {
                res[0] = i
            }
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

4. Negative Marking

Intuition

We can use the input array itself as a hash table by marking visited indices. For each number, we negate the value at the corresponding index. If we try to negate an already negative value, we found the duplicate. After processing, any positive value indicates its index corresponds to the missing number.

Algorithm

Iterate through the array:
- For each value, take its absolute value to get the index.
- If the value at that index is already negative, this is the duplicate.
- Otherwise, negate the value at that index.
Iterate through the array again:
- Find the index with a positive value that is not the duplicate.
- This index + 1 is the missing number.
Return [duplicate, missing].

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        res = [0, 0]

        for num in nums:
            num = abs(num)
            nums[num - 1] *= -1
            if nums[num - 1] > 0:
                res[0] = num

        for i, num in enumerate(nums):
            if num > 0 and i + 1 != res[0]:
                res[1] = i + 1
                return res

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int[] res = new int[2];

        for (int num : nums) {
            int absNum = Math.abs(num);
            if (nums[absNum - 1] < 0) {
                res[0] = absNum;
            } else {
                nums[absNum - 1] *= -1;
            }
        }

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0 && i + 1 != res[0]) {
                res[1] = i + 1;
                return res;
            }
        }

        return res;
    }
}

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        vector<int> res(2);

        for (int num : nums) {
            int absNum = abs(num);
            if (nums[absNum - 1] < 0) {
                res[0] = absNum;
            } else {
                nums[absNum - 1] *= -1;
            }
        }

        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0 && i + 1 != res[0]) {
                res[1] = i + 1;
                return res;
            }
        }

        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    findErrorNums(nums) {
        const res = [0, 0];

        for (let num of nums) {
            const absNum = Math.abs(num);
            if (nums[absNum - 1] < 0) {
                res[0] = absNum;
            } else {
                nums[absNum - 1] *= -1;
            }
        }

        for (let i = 0; i < nums.length; i++) {
            if (nums[i] > 0 && i + 1 !== res[0]) {
                res[1] = i + 1;
                return res;
            }
        }

        return res;
    }
}

public class Solution {
    public int[] FindErrorNums(int[] nums) {
        int[] res = new int[2];

        foreach (int num in nums) {
            int absNum = Math.Abs(num);
            if (nums[absNum - 1] < 0) {
                res[0] = absNum;
            } else {
                nums[absNum - 1] *= -1;
            }
        }

        for (int i = 0; i < nums.Length; i++) {
            if (nums[i] > 0 && i + 1 != res[0]) {
                res[1] = i + 1;
                return res;
            }
        }

        return res;
    }
}

func findErrorNums(nums []int) []int {
    res := []int{0, 0}

    for _, num := range nums {
        absNum := num
        if absNum < 0 {
            absNum = -absNum
        }
        if nums[absNum-1] < 0 {
            res[0] = absNum
        } else {
            nums[absNum-1] *= -1
        }
    }

    for i := 0; i < len(nums); i++ {
        if nums[i] > 0 && i+1 != res[0] {
            res[1] = i + 1
            return res
        }
    }

    return res
}

class Solution {
    fun findErrorNums(nums: IntArray): IntArray {
        val res = intArrayOf(0, 0)

        for (num in nums) {
            val absNum = Math.abs(num)
            if (nums[absNum - 1] < 0) {
                res[0] = absNum
            } else {
                nums[absNum - 1] *= -1
            }
        }

        for (i in nums.indices) {
            if (nums[i] > 0 && i + 1 != res[0]) {
                res[1] = i + 1
                return res
            }
        }

        return res
    }
}

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        var nums = nums
        var res = [0, 0]

        for num in nums {
            let absNum = abs(num)
            if nums[absNum - 1] < 0 {
                res[0] = absNum
            } else {
                nums[absNum - 1] *= -1
            }
        }

        for i in 0..<nums.count {
            if nums[i] > 0 && i + 1 != res[0] {
                res[1] = i + 1
                return res
            }
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

5. Math

Intuition

Using sum formulas, we can set up equations to solve for the duplicate and missing numbers. Let the duplicate be d and the missing be m. The difference between actual sum and expected sum gives us d - m. The difference between actual sum of squares and expected sum of squares gives us d^2 - m^2 = (d+m)(d-m). With these two equations, we can solve for both values.

Algorithm

Compute x = sum(nums) - sum(1 to n), which equals duplicate - missing.
Compute y = sum(nums^2) - sum(1^2 to n^2), which equals duplicate^2 - missing^2.
Since y = (duplicate + missing) * x, we get duplicate + missing = y / x.
Solve for missing: missing = (y/x - x) / 2 and duplicate = missing + x.
Return [duplicate, missing].

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        N = len(nums)
        x = 0 # duplicate - missing
        y = 0 # duplicate^2 - missing^2

        for i in range(1, N + 1):
            x += nums[i - 1] - i
            y += nums[i - 1]**2 - i**2

        missing = (y - x**2) // (2 * x)
        duplicate = missing + x
        return [duplicate, missing]

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int N = nums.length;
        int x = 0; // duplicate - missing
        int y = 0; // duplicate^2 - missing^2

        for (int i = 1; i <= N; i++) {
            x += nums[i - 1] - i;
            y += nums[i - 1] * nums[i - 1] - i * i;
        }

        int missing = (y - x * x) / (2 * x);
        int duplicate = missing + x;
        return new int[]{duplicate, missing};
    }
}

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        int N = nums.size();
        long long x = 0; // duplicate - missing
        long long y = 0; // duplicate^2 - missing^2

        for (int i = 1; i <= N; i++) {
            x += nums[i - 1] - i;
            y += (long long)nums[i - 1] * nums[i - 1] - (long long)i * i;
        }

        int missing = (y - x * x) / (2 * x);
        int duplicate = missing + x;
        return {duplicate, missing};
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    findErrorNums(nums) {
        const N = nums.length;
        let x = 0; // duplicate - missing
        let y = 0; // duplicate^2 - missing^2

        for (let i = 1; i <= N; i++) {
            x += nums[i - 1] - i;
            y += nums[i - 1] ** 2 - i ** 2;
        }

        const missing = (y - x ** 2) / (2 * x);
        const duplicate = missing + x;
        return [duplicate, missing];
    }
}

public class Solution {
    public int[] FindErrorNums(int[] nums) {
        int N = nums.Length;
        long x = 0; // duplicate - missing
        long y = 0; // duplicate^2 - missing^2

        for (int i = 1; i <= N; i++) {
            x += nums[i - 1] - i;
            y += (long)nums[i - 1] * nums[i - 1] - (long)i * i;
        }

        int missing = (int)((y - x * x) / (2 * x));
        int duplicate = (int)(missing + x);
        return new int[] { duplicate, missing };
    }
}

func findErrorNums(nums []int) []int {
    N := len(nums)
    x := 0 // duplicate - missing
    y := 0 // duplicate^2 - missing^2

    for i := 1; i <= N; i++ {
        x += nums[i-1] - i
        y += nums[i-1]*nums[i-1] - i*i
    }

    missing := (y - x*x) / (2 * x)
    duplicate := missing + x
    return []int{duplicate, missing}
}

class Solution {
    fun findErrorNums(nums: IntArray): IntArray {
        val N = nums.size
        var x = 0L // duplicate - missing
        var y = 0L // duplicate^2 - missing^2

        for (i in 1..N) {
            x += nums[i - 1] - i
            y += nums[i - 1].toLong() * nums[i - 1] - i.toLong() * i
        }

        val missing = ((y - x * x) / (2 * x)).toInt()
        val duplicate = missing + x.toInt()
        return intArrayOf(duplicate, missing)
    }
}

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        let N = nums.count
        var x = 0 // duplicate - missing
        var y = 0 // duplicate^2 - missing^2

        for i in 1...N {
            x += nums[i - 1] - i
            y += nums[i - 1] * nums[i - 1] - i * i
        }

        let missing = (y - x * x) / (2 * x)
        let duplicate = missing + x
        return [duplicate, missing]
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

6. Bitwise XOR

Intuition

XOR has the property that a ^ a = 0. If we XOR all numbers in the array with all numbers from 1 to n, pairs cancel out, leaving us with duplicate ^ missing. To separate them, we find a bit where they differ (using the rightmost set bit), then partition all numbers into two groups based on that bit. XORing within each group isolates the duplicate and missing values.

Algorithm

XOR all array elements with all numbers 1 to n. This gives duplicate ^ missing.
Find the rightmost set bit in the XOR result (this bit differs between duplicate and missing).
Partition all numbers (from array and from 1 to n) into two groups based on this bit.
XOR numbers within each group separately to get two candidates x and y.
Check which candidate appears in the array to identify the duplicate.
Return [duplicate, missing].

class Solution:
    def findErrorNums(self, nums: List[int]) -> List[int]:
        N = len(nums)
        # a ^ a = 0
        # xorr = (1 ^ 2 ^ ... N) ^ (nums[0] ^ nums[1] ^ ... nums[N - 1])
        # xorr = missing ^ duplicate
        xorr = 0
        for i in range(1, N + 1):
            xorr ^= i
            xorr ^= nums[i - 1]

        # bit that is set in only one number among (duplicate, missing),
        # will be set in (duplicate ^ missing)
        # take rightMost set bit for simplicity
        rightMostBit = xorr & ~(xorr - 1)

        # divide numbers (from nums, from [1, N]) into two sets w.r.t the rightMostBit
        # xorr the numbers of these sets independently
        x = y = 0
        for i in range(1, N + 1):
            if i & rightMostBit:
                x ^= i
            else:
                y ^= i

            if nums[i - 1] & rightMostBit:
                x ^= nums[i - 1]
            else:
                y ^= nums[i - 1]

        # identify the duplicate number from x and y
        for num in nums:
            if num == x:
                return [x, y]
        return [y, x]

public class Solution {
    public int[] findErrorNums(int[] nums) {
        int N = nums.length;
        // a ^ a = 0
        // xorr = (1 ^ 2 ^ ... N) ^ (nums[0] ^ nums[1] ^ ... nums[N - 1])
        // xorr = missing ^ duplicate
        int xorr = 0;
        for (int i = 1; i <= N; i++) {
            xorr ^= i;
            xorr ^= nums[i - 1];
        }

        // bit that is set in only one number among (duplicate, missing),
        // will be set in (duplicate ^ missing)
        // take rightMost set bit for simplicity
        int rightMostBit = xorr & ~(xorr - 1);

        // divide numbers (from nums, from [1, N]) into two sets w.r.t the rightMostBit
        // xorr the numbers of these sets independently
        int x = 0, y = 0;
        for (int i = 1; i <= N; i++) {
            if ((i & rightMostBit) != 0) {
                x ^= i;
            } else {
                y ^= i;
            }

            if ((nums[i - 1] & rightMostBit) != 0) {
                x ^= nums[i - 1];
            } else {
                y ^= nums[i - 1];
            }
        }

        // identify the duplicate number from x and y
        for (int num : nums) {
            if (num == x) {
                return new int[]{x, y};
            }
        }
        return new int[]{y, x};
    }
}

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        int N = nums.size();
        // a ^ a = 0
        // xorr = (1 ^ 2 ^ ... N) ^ (nums[0] ^ nums[1] ^ ... nums[N - 1])
        // xorr = missing ^ duplicate
        int xorr = 0;
        for (int i = 1; i <= N; i++) {
            xorr ^= i;
            xorr ^= nums[i - 1];
        }

        // bit that is set in only one number among (duplicate, missing),
        // will be set in (duplicate ^ missing)
        // take rightMost set bit for simplicity
        int rightMostBit = xorr & ~(xorr - 1);

        // divide numbers (from nums, from [1, N]) into two sets w.r.t the rightMostBit
        // xorr the numbers of these sets independently
        int x = 0, y = 0;
        for (int i = 1; i <= N; i++) {
            if (i & rightMostBit) {
                x ^= i;
            } else {
                y ^= i;
            }

            if (nums[i - 1] & rightMostBit) {
                x ^= nums[i - 1];
            } else {
                y ^= nums[i - 1];
            }
        }

        // identify the duplicate number from x and y
        for (int num : nums) {
            if (num == x) {
                return {x, y};
            }
        }
        return {y, x};
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {number[]}
     */
    findErrorNums(nums) {
        const N = nums.length;
        // a ^ a = 0
        // xorr = (1 ^ 2 ^ ... N) ^ (nums[0] ^ nums[1] ^ ... nums[N - 1])
        // xorr = missing ^ duplicate
        let xorr = 0;
        for (let i = 1; i <= N; i++) {
            xorr ^= i;
            xorr ^= nums[i - 1];
        }

        // bit that is set in only one number among (duplicate, missing),
        // will be set in (duplicate ^ missing)
        // take rightMost set bit for simplicity
        const rightMostBit = xorr & ~(xorr - 1);

        // divide numbers (from nums, from [1, N]) into two sets w.r.t the rightMostBit
        // xorr the numbers of these sets independently
        let x = 0,
            y = 0;
        for (let i = 1; i <= N; i++) {
            if (i & rightMostBit) {
                x ^= i;
            } else {
                y ^= i;
            }

            if (nums[i - 1] & rightMostBit) {
                x ^= nums[i - 1];
            } else {
                y ^= nums[i - 1];
            }
        }

        // identify the duplicate number from x and y
        for (let num of nums) {
            if (num === x) {
                return [x, y];
            }
        }
        return [y, x];
    }
}

public class Solution {
    public int[] FindErrorNums(int[] nums) {
        int N = nums.Length;
        int xorr = 0;
        for (int i = 1; i <= N; i++) {
            xorr ^= i;
            xorr ^= nums[i - 1];
        }

        int rightMostBit = xorr & ~(xorr - 1);

        int x = 0, y = 0;
        for (int i = 1; i <= N; i++) {
            if ((i & rightMostBit) != 0) {
                x ^= i;
            } else {
                y ^= i;
            }

            if ((nums[i - 1] & rightMostBit) != 0) {
                x ^= nums[i - 1];
            } else {
                y ^= nums[i - 1];
            }
        }

        foreach (int num in nums) {
            if (num == x) {
                return new int[] { x, y };
            }
        }
        return new int[] { y, x };
    }
}

func findErrorNums(nums []int) []int {
    N := len(nums)
    xorr := 0
    for i := 1; i <= N; i++ {
        xorr ^= i
        xorr ^= nums[i-1]
    }

    rightMostBit := xorr & ^(xorr - 1)

    x, y := 0, 0
    for i := 1; i <= N; i++ {
        if i&rightMostBit != 0 {
            x ^= i
        } else {
            y ^= i
        }

        if nums[i-1]&rightMostBit != 0 {
            x ^= nums[i-1]
        } else {
            y ^= nums[i-1]
        }
    }

    for _, num := range nums {
        if num == x {
            return []int{x, y}
        }
    }
    return []int{y, x}
}

class Solution {
    fun findErrorNums(nums: IntArray): IntArray {
        val N = nums.size
        var xorr = 0
        for (i in 1..N) {
            xorr = xorr xor i
            xorr = xorr xor nums[i - 1]
        }

        val rightMostBit = xorr and (xorr - 1).inv()

        var x = 0
        var y = 0
        for (i in 1..N) {
            if (i and rightMostBit != 0) {
                x = x xor i
            } else {
                y = y xor i
            }

            if (nums[i - 1] and rightMostBit != 0) {
                x = x xor nums[i - 1]
            } else {
                y = y xor nums[i - 1]
            }
        }

        for (num in nums) {
            if (num == x) {
                return intArrayOf(x, y)
            }
        }
        return intArrayOf(y, x)
    }
}

class Solution {
    func findErrorNums(_ nums: [Int]) -> [Int] {
        let N = nums.count
        var xorr = 0
        for i in 1...N {
            xorr ^= i
            xorr ^= nums[i - 1]
        }

        let rightMostBit = xorr & ~(xorr - 1)

        var x = 0
        var y = 0
        for i in 1...N {
            if i & rightMostBit != 0 {
                x ^= i
            } else {
                y ^= i
            }

            if nums[i - 1] & rightMostBit != 0 {
                x ^= nums[i - 1]
            } else {
                y ^= nums[i - 1]
            }
        }

        for num in nums {
            if num == x {
                return [x, y]
            }
        }
        return [y, x]
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Common Pitfalls

Confusing the Output Order

The problem asks to return [duplicate, missing] in that specific order. Accidentally swapping the order and returning [missing, duplicate] produces wrong answers even when both numbers are correctly identified.

Not Handling Edge Cases at Array Boundaries

When using the sorting approach, special handling is needed if the missing number is 1 (the first element) or n (the last element). These boundary cases require explicit checks beyond just comparing adjacent elements.