LeetCode 645: Set Mismatch Solution
Master LeetCode problem 645 (Set Mismatch), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
645. Set Mismatch
Problem Explanation
Explanation:
- We can solve this problem by iterating through the array
numsand 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.
:
Solution Code
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;
}
}Try It Yourself
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