Find Missing and Repeated Values - Practice Coding Problems

Find Missing and Repeated Values - Problem

Find Missing and Repeated Values

Imagine you're auditing a digital inventory system that should contain exactly one instance of each item numbered from 1 to n². However, due to a system glitch, one item appears twice while another item is completely missing.

You are given a n × n integer matrix grid where each cell should contain a unique value from 1 to n². Your task is to identify:
• a = the repeated value (appears twice)
• b = the missing value (should be present but isn't)

Goal: Return an array [a, b] where a is the repeated value and b is the missing value.

Example: In a 3×3 grid with values [1,2,3,4,5,6,7,8,9], if we have [1,2,3,4,5,5,7,8,9], then a=5 (repeated) and b=6 (missing).

Input & Output

example_1.py — Basic Case

$ Input: grid = [[1,3],[2,2]]

› Output: [2,4]

💡 Note: In this 2×2 grid, we should have numbers 1,2,3,4. Number 2 appears twice (repeated) and number 4 is missing.

example_2.py — Larger Grid

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

› Output: [9,5]

💡 Note: In this 3×3 grid, we should have numbers 1-9. Number 9 appears twice (at positions [0,0] and [1,1]), and number 5 is missing.

example_3.py — Edge Case

$ Input: grid = [[1]]

› Output: [1,1] is impossible since n≥2. Minimum case: grid = [[2,2],[2,3]] → [2,1]

💡 Note: For a 2×2 grid with values [2,2,2,3], number 2 appears three times but problem states exactly one repeat, so this violates constraints. Valid minimum: [[2,2],[1,3]] has repeated=2, missing=4.

Constraints

2 ≤ n ≤ 1000
1 ≤ grid[i][j] ≤ n²
The grid contains exactly one repeated number and exactly one missing number
All other numbers from 1 to n² appear exactly once

Visualization

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Understanding the Visualization

Start Roll Call

Begin with empty attendance sheet (hash table)

Mark Present

For each student number called, mark them present

Catch Duplicate

If someone already marked present answers again, found our duplicate!

Find Absent

Check attendance sheet to see who never answered

Report Results

Return [duplicate_student, absent_student]

Key Takeaway

🎯 Key Insight: Hash tables excel at detecting duplicates because lookups are O(1). The moment we see a number twice, we've found our answer!

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 Apple 6

The optimal solution uses a hash table to track seen numbers in a single pass through the grid. When we encounter a number already in our hash table, that's the repeated value. After processing all cells, we check which number from 1 to n² is missing from our hash table. This achieves O(n²) time and O(n²) space complexity, making it efficient for the given constraints.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n⁴)	O(1)	For each number 1 to n², count its occurrences in the grid
One-Pass Hash Table (Optimal)	O(n²)	O(n²)	Track seen numbers in hash table while iterating through grid once

Brute Force (Nested Loops) — Algorithm Steps

Flatten the 2D grid into a 1D array for easier processing
For each number from 1 to n², count its occurrences in the flattened array
The number with count 2 is the repeated value
The number with count 0 is the missing value

Visualization

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Step-by-Step Walkthrough

Flatten Grid

Convert 2D grid to 1D array: [1,3,3,2,5,6,7,8,9]

Check Number 1

Scan array, count = 1 (normal)

Check Number 2

Scan array, count = 1 (normal)

Check Number 3

Scan array, count = 2 (repeated!)

Check Number 4

Scan array, count = 0 (missing!)

Code -

solution.c — C

int* findMissingAndRepeatedValues(int** grid, int gridSize, int* gridColSize, int* returnSize) {
    int n = gridSize;
    int* result = (int*)malloc(2 * sizeof(int));
    *returnSize = 2;
    
    // Flatten the grid
    int* flat = (int*)malloc(n * n * sizeof(int));
    int idx = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            flat[idx++] = grid[i][j];
        }
    }
    
    // Check each number from 1 to n*n
    for (int num = 1; num <= n * n; num++) {
        int count = 0;
        for (int i = 0; i < n * n; i++) {
            if (flat[i] == num) {
                count++;
            }
        }
        
        if (count == 2) {
            result[0] = num;
        } else if (count == 0) {
            result[1] = num;
        }
    }
    
    free(flat);
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(n⁴)

For each of n² possible values, we scan through n² grid elements

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for counting and iteration

✓ Linear Space

12.2K Views

Medium Frequency

~15 min Avg. Time

485 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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