645. Set Mismatch - DevCodeEx

Explanation:

We can solve this problem by iterating through the array nums and keeping track of the frequency of each element.
We can identify the duplicate number by checking the frequency of each element.
The missing number can be calculated by comparing the expected sum of numbers from 1 to n with the actual sum of elements in the array.
The difference between the expected sum and the actual sum will be the missing number.
The time complexity of this solution is O(n) where n is the length of the input array nums.
The space complexity is O(n) to store the frequency of each element.

:

class Solution {
    public int[] findErrorNums(int[] nums) {
        int[] result = new int[2];
        int[] count = new int[nums.length + 1];
        
        for (int num : nums) {
            count[num]++;
            if (count[num] == 2) {
                result[0] = num;
            }
        }
        
        for (int i = 1; i < count.length; i++) {
            if (count[i] == 0) {
                result[1] = i;
                break;
            }
        }
        
        return result;
    }
}

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