Find Missing Number in Array using Sum Formula - Optimal Algorithm

Contents

ProblemExamplesSolutionAlgorithmCodeTimeSpace

Problem Statement❯

You are given an array of N-1 distinct integers in the range 1 to N. The task is to find the one number that is missing from the array.

The array has no duplicates, and only one number is missing. The numbers can be in any order.

Your goal is to identify the missing number efficiently.

Examples❯

Input Array Expected N Missing Number Description
[1, 2, 4, 5] 5 3 All numbers from 1 to 5 are present except 3Visualization
[2, 3, 1, 5] 5 4 Missing number is in the middleVisualization
[1, 2, 3, 4] 5 5 Last number is missingVisualization
[2, 3, 4, 5] 5 1 First number is missingVisualization
[1] 2 2 Only one number present, second one is missingVisualization
[] 1 1 Empty array, so the only number 1 is missingVisualization
[2] 2 1 Only number 2 is present, so 1 is missingVisualization

Solution❯

To find the missing number in an array of size N-1 where elements are from 1 to N, we can take advantage of the fact that the sum of first N natural numbers can be calculated directly using the formula:

Expected Sum = N * (N + 1) / 2

Let’s say we compute this expected sum and compare it with the sum of the actual elements present in the array. The difference between the two will tell us which number is missing.

Let’s Discuss a Few Scenarios:

1. Normal Case: Suppose the array is [1, 2, 4, 5] and N = 5. The expected sum is 15, but the actual sum is 12. The missing number is 15 - 12 = 3.

2. First Number Missing: In [2, 3, 4, 5], the missing number is 1. The formula still works: Expected = 15, Actual = 14, Missing = 1.

3. Last Number Missing: For [1, 2, 3, 4] with N = 5, the missing number is 5. Again, Expected = 15, Actual = 10, Missing = 5.

4. Only One Element in Array: If array = [1] and N = 2, then Expected = 3, Actual = 1, Missing = 2.

5. Empty Array Case: This is a valid edge case. If N = 1 and the array is empty [], then the only number that could have been in the array is 1. So, 1 is the missing number.

6. Random Order: It doesn’t matter what the order of elements is. Even if the array is [3, 1, 5, 2], the total will still work out because the sum is independent of order.

Why This Works

Since we are not iterating over every possible number from 1 to N, but instead just comparing totals, the time complexity is O(N) (to sum the array) and the space complexity is O(1). This makes the solution optimal and suitable for large inputs.

This technique also avoids needing any extra memory like hash sets or frequency arrays.

Algorithm Steps❯

  1. Given an array of size N-1, containing unique numbers between 1 to N.
  2. Calculate the sum of the first N natural numbers using the formula N*(N+1)/2.
  3. Calculate the sum of all elements in the given array.
  4. The missing number is the difference between the expected sum and the actual sum.

Code

Python
JavaScript
Java
C++
C
def find_missing_number(arr, N):
    total = N * (N + 1) // 2
    actual_sum = sum(arr)
    return total - actual_sum

# Sample Input
arr = [1, 2, 4, 5, 6]
N = 6
print("Missing Number:", find_missing_number(arr, N))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)We need to iterate through the array once to calculate the actual sum, which takes linear time.
Average CaseO(n)Regardless of the input distribution, we always sum all elements, resulting in linear time complexity.
Worst CaseO(n)In the worst case, the missing number is at the end, but we still iterate the full array, so the time remains linear.

Space Complexity

O(1)

Explanation: Only a constant number of variables are used to store sums and the final result. No extra space proportional to input size is required.