Find Missing and Repeated Values

Find Missing and Repeated Values - Problem

You are given a 0-indexed 2D integer matrix grid of size n * n with values in the range [1, n²]. Each integer appears exactly once except:

a which appears twice
b which is missing

Your task is to find the repeating and missing numbers a and b.

Return a 0-indexed integer array ans of size 2 where:

ans[0] equals to a (the repeated number)
ans[1] equals to b (the missing number)

Input & Output

Example 1 — Basic Case

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

› Output: [2,4]

💡 Note: In a 2×2 grid, numbers should be [1,2,3,4]. Number 2 appears twice (repeated), number 4 is missing.

Example 2 — Larger Grid

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

› Output: [9,5]

💡 Note: In a 3×3 grid, numbers should be [1,2,3,4,5,6,7,8,9]. Number 9 appears twice, number 5 is missing.

Example 3 — Edge Positions

$ Input: grid = [[1,2],[4,4]]

› Output: [4,3]

💡 Note: Number 4 appears twice at positions (1,0) and (1,1). Number 3 is missing from the sequence [1,2,3,4].

Constraints

1 ≤ n ≤ 100
1 ≤ grid[i][j] ≤ n²
Exactly one number appears twice
Exactly one number is missing

Visualization

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Asked in

a Amazon 25 M Microsoft 20 G Google 15 Apple 12

The key insight is to identify which number appears twice and which is missing from the expected sequence [1, n²]. The hash map approach provides the best balance of simplicity and efficiency. Time: O(n²), Space: O(n²) for counting, or use mathematical formulas for O(1) space.

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Two Pointers

⏱️ Time: N/A Space: N/A

Brute Force - Count Each Number

⏱️ Time: O(n⁴) Space: O(1)

For each possible number from 1 to n², iterate through the entire grid counting how many times it appears. A number appearing 0 times is missing, 2 times is repeated.

Hash Map - Count Frequencies

⏱️ Time: O(n²) Space: O(n²)

Iterate through the grid once to count frequency of each number. Then check which number has frequency 2 (repeated) and which number from 1 to n² is missing (frequency 0).

Mathematical Formula

⏱️ Time: O(n²) Space: O(1)

Calculate the expected sum and sum of squares for numbers 1 to n². Compare with actual sums from grid to derive equations that directly give us the repeated and missing numbers.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

// Simulate knows function based on graph
int knows(int a, int b, int graph[][100]) {
    return graph[a][b] == 1;
}

// Parse JSON-style 2D array
void parseGraph(char* input, int graph[][100], int n) {
    int i = 0, j = 0;
    char* ptr = input;
    
    // Skip initial characters until we find first digit
    while (*ptr && !isdigit(*ptr)) ptr++;
    
    while (*ptr && i < n) {
        if (isdigit(*ptr)) {
            graph[i][j] = *ptr - '0';
            j++;
            if (j == n) {
                i++;
                j = 0;
            }
        }
        ptr++;
    }
}

int solution(int n, int graph[][100]) {
    // Phase 1: Find candidate using elimination
    int candidate = 0;
    
    for (int i = 1; i < n; i++) {
        if (knows(candidate, i, graph)) {
            candidate = i; // Current candidate knows i, so candidate = i
        }
    }
    
    // Phase 2: Verify the candidate
    for (int i = 0; i < n; i++) {
        if (i != candidate) {
            // Candidate should not know anyone
            if (knows(candidate, i, graph)) {
                return -1;
            }
            // Everyone should know candidate
            if (!knows(i, candidate, graph)) {
                return -1;
            }
        }
    }
    
    return candidate;
}

int main() {
    int n;
    scanf("%d", &n);
    
    char graphInput[10000];
    scanf("%s", graphInput);
    
    int graph[100][100];
    parseGraph(graphInput, graph, n);
    
    printf("%d\n", solution(n, graph));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

✓ Linear Space

33.2K Views

Medium Frequency

~15 min Avg. Time

850 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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