Find Missing and Repeated Values

Problem Statement

In this task, you are presented with a square matrix grid of dimension n x n containing integers. These integers are typically arranged from 1 to n^2 where each number is supposed to appear exactly once. However, in this scenario, one of the numbers, labeled a, is repeated twice whereas another number, labeled b, is completely missing from the collection. The objective is to identify these two problematic numbers a (the repeat) and b (the missing number).

To fulfil the task, you must process the matrix and return an array ans of two numbers: the first being the repeated number a, and the second being the missing number b. These responses are derived from analysing the integrity of the sequence presented by the matrix.

Examples

Example 1

Input:

grid = [[1,3],[2,2]]

Output:

[2,4]

Explanation:

Number 2 is repeated and number 4 is missing so the answer is [2,4].

Example 2

Input:

grid = [[9,1,7],[8,9,2],[3,4,6]]

Output:

[9,5]

Explanation:

Number 9 is repeated and number 5 is missing so the answer is [9,5].

Constraints

• 2 <= n == grid.length == grid[i].length <= 50
• 1 <= grid[i][j] <= n * n
• For all x that 1 <= x <= n * n there is exactly one x that is not equal to any of the grid members.
• For all x that 1 <= x <= n * n there is exactly one x that is equal to exactly two of the grid members.
• For all x that 1 <= x <= n * n except two of them there is exatly one pair of i, j that 0 <= i, j <= n - 1 and grid[i][j] == x.

Approach and Intuition

When dealing with a matrix where numbers should ideally range in sequence from 1 to n^2, identifying a missing and a repeating number involves:

1. Creating a frequency count or a record of each number that appears in the matrix. This can be done using a hash table or an array (given the confined range of numbers).

2. As we iterate through the grid:

   • Increase the count for each number encountered.
   • Simultaneously, we can keep track of the total sum of numbers as they should ideally sum to the series sum of 1 to n^2 (which is (n^2 * (n^2 + 1))/2).

3. After populating the frequency table and calculating the observed sum:

   • The number which appears twice will obviously have a count of 2 in the frequency table.
   • The missing number will not be present in our records (count of 0).

4. To ascertain the answers efficiently:

   • Iterate through the expected range of numbers from 1 to n^2.
   • Check the frequency table to quickly spot the number with a frequency of 2 (indicating it’s repeated) and a frequency of 0 (indicating it’s missing).

5. The deviation of the observed sum from the expected sum can also confirm if our found numbers are logically consistent (deviation should equate to the repeated number minus the missing number).

By following these steps, we ensure that all edge cases (like minimum dimensions and maximum possible constraints) are handled effectively, given the defined range and conditions from the problem description.

Solutions

class Solution {
public:
    vector<int> detectRepeatAndMissing(vector<vector<int>>& matrix) {
        long long totalSum = 0, totalSqrSum = 0;
        long long dimension = matrix.size();
        long long perfectTotal = dimension * dimension;
    
        // Calculate sum and square sum from the matrix
        for (int i = 0; i < dimension; ++i) {
            for (int j = 0; j < dimension; ++j) {
                totalSum += matrix[i][j];
                totalSqrSum +=
                    static_cast<long long>(matrix[i][j]) * matrix[i][j];
            }
        }
    
        // Expected and actual sum computations
        long long expectedSum = perfectTotal * (perfectTotal + 1) / 2;
        long long expectedSqrSum = perfectTotal * (perfectTotal + 1) * (2 * perfectTotal + 1) / 6;
    
        // Calculate differences from expected values
        long long diffSum = totalSum - expectedSum;
        long long diffSqrSum = totalSqrSum - expectedSqrSum;
    
        // Solve the equations for missing and repeated values
        int repeated = (diffSqrSum / diffSum + diffSum) / 2;
        int missing = (diffSqrSum / diffSum - diffSum) / 2;
    
        return {repeated, missing};
    }
};