🚀Day 13. Smallest Positive Missing Number 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given an integer array arr[]. Your task is to find the smallest positive number missing from the array.

Note: Positive number starts from 1. The array can have negative integers too.

🔍 Example Walkthrough:

Input: arr[] = [2, -3, 4, 1, 1, 7] Output: 3

Explanation: Smallest positive missing number is 3.

Input: arr[] = [5, 3, 2, 5, 1] Output: 4

Explanation: Smallest positive missing number is 4.

Input: arr[] = [-8, 0, -1, -4, -3] Output: 1

Explanation: Smallest positive missing number is 1.

Constraints:

• $1 <= arr.size() <= 10^5$

• $-10^6 <= arr[i] <= 10^6$

🎯 My Approach:

1. In-place Rearrangement:

   • The problem can be solved using an in-place rearrangement technique that places elements at their correct indices.

   • The idea is to rearrange the elements such that for any element arr[i], it should be placed at index arr[i] - 1.

   • After rearranging the elements, traverse the array again to find the smallest missing positive integer.

2. Steps:

   • Iterate through the array, and for each element that is within the valid range [1, n], place it in its correct position.

   • Once the array is rearranged, traverse the array to identify the smallest index i where arr[i] != i + 1. This indicates the missing number.

   • If all elements are in place, the missing number is n + 1.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), where n is the size of the array. The algorithm requires two linear scans of the array, making it efficient.

• Expected Auxiliary Space Complexity: O(1), as we use only a constant amount of additional space.

📝 Solution Code

Code (C)

int missingNumber(int arr[], int n) {
    for (int i = 0; i < n; i++) {
        while (arr[i] > 0 && arr[i] <= n && arr[i] != arr[arr[i] - 1]) {
            int temp = arr[i];
            arr[i] = arr[arr[i] - 1];
            arr[temp - 1] = temp;
        }
    }
    for (int i = 0; i < n; i++) {
        if (arr[i] != i + 1) {
            return i + 1;
        }
    }
    return n + 1;
}

Code (Cpp)

class Solution {
public:
    int missingNumber(vector<int>& arr) {
        int n = arr.size();
        for (int i = 0; i < n; i++) {
            while (arr[i] > 0 && arr[i] <= n && arr[i] != arr[arr[i] - 1]) {
                swap(arr[i], arr[arr[i] - 1]);
            }
        }
        for (int i = 0; i < n; i++) {
            if (arr[i] != i + 1) {
                return i + 1;
            }
        }
        return n + 1;
    }
};

Code (Java)

class Solution {
    public int missingNumber(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            while (arr[i] > 0 && arr[i] <= n && arr[i] != arr[arr[i] - 1]) {
                int temp = arr[i];
                arr[i] = arr[arr[i] - 1];
                arr[temp - 1] = temp;
            }
        }
        for (int i = 0; i < n; i++) {
            if (arr[i] != i + 1) {
                return i + 1;
            }
        }
        return n + 1;
    }
}

Code (Python)

class Solution:
    def missingNumber(self, arr):
        n = len(arr)
        for i in range(n):
            while arr[i] > 0 and arr[i] <= n and arr[i] != arr[arr[i] - 1]:
                arr[arr[i] - 1], arr[i] = arr[i], arr[arr[i] - 1]
        for i in range(n):
            if arr[i] != i + 1:
                return i + 1
        return n + 1

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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