LeetCode 2965: Find Missing and Repeated Values Solution

Master LeetCode problem 2965 (Find Missing and Repeated Values), 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.

2965. Find Missing and Repeated Values

Easy

Array Hash Table Math Matrix

Problem Explanation

Explanation:

To solve this problem, we can use a set to keep track of the numbers we have seen so far. We iterate through the grid, calculating the expected sum of numbers from 1 to n^2, the actual sum of numbers in the grid, and keep track of the repeated and missing numbers.

Calculate the expected sum of numbers from 1 to n^2 using the formula n * (n + 1) / 2.
Calculate the actual sum of numbers in the grid.
Iterate through the grid:
- For each number in the grid:
  - Check if the number has been seen before:
    - If yes, mark it as the repeated number.
    - If no, add it to the set of seen numbers.
Calculate the missing number as the difference between the expected sum and the actual sum minus the repeated number.

The time complexity of this solution is O(n^2) as we iterate through the grid once. The space complexity is O(n^2) to store the set of seen numbers.

Solution Code

class Solution {
    public int[] findErrorNums(int[][] grid) {
        int n = grid.length;
        Set<Integer> seen = new HashSet<>();
        int expectedSum = n * (n + 1) / 2;
        int actualSum = 0;
        int repeated = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (seen.contains(grid[i][j])) {
                    repeated = grid[i][j];
                } else {
                    seen.add(grid[i][j]);
                    actualSum += grid[i][j];
                }
            }
        }
        
        int missing = expectedSum - actualSum + repeated;
        
        return new int[]{repeated, missing};
    }
}

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