Missing number in arrray from gfg is added

dsa-solutions/gfg-solutions/Easy problems/Missing-In-Array.md

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+---
+id: find-the-missing-number
+title: Find the Missing Number Problem 
+sidebar_label: Find-The-Missing-Number
+tags:
+  - Intermediate
+  - Array
+  - Mathematical
+  - GeeksforGeeks
+  - CPP
+  - Python
+  - DSA
+description: "This tutorial covers the solution to the Find the Missing Number problem from the GeeksforGeeks."
+---
+## Problem Description
+
+Given an array containing `n-1` distinct numbers taken from 1 to `n`, find the one number that is missing from the array.
+
+## Examples
+
+**Example 1:**
+
+```
+Input: array = [3, 1, 4]
+Output: 2
+Explanation: The missing number is 2.
+```
+
+**Example 2:**
+
+```
+Input: array = [5, 3, 1, 2]
+Output: 4
+Explanation: The missing number is 4.
+```
+
+## Your Task
+
+You don't need to read input or print anything. Your task is to complete the function `missingNumber()` which takes the size `n` of the range and the array `arr` as inputs and returns the missing number.
+
+Expected Time Complexity: $O(n)$
+
+Expected Auxiliary Space: $O(1)$
+
+## Constraints
+
+* `2 ≤ n ≤ 10^6`
+* `1 ≤ array[i] ≤ n`
+* All elements of the array are distinct.
+
+## Problem Explanation
+
+The problem is to find the missing number from an array of `n-1` elements which contains distinct numbers from 1 to `n`. The missing number is the one number that is not present in the array.
+
+## Code Implementation
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```py
+  class Solution:
+      def missingNumber(self, n, arr):
+          # Calculate the expected sum of the first n natural numbers
+          total_sum = n * (n + 1) // 2
+
+          # Calculate the sum of the elements in the array
+          arr_sum = sum(arr)
+
+          # The missing number is the difference between the expected sum and the array sum
+          return total_sum - arr_sum
+
+  # Example usage
+  if __name__ == "__main__":
+      solution = Solution()
+      print(solution.missingNumber(4, [3, 1, 4]))  # Expected output: 2
+      print(solution.missingNumber(5, [5, 3, 1, 2]))  # Expected output: 4
+  ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```cpp
+  #include <iostream>
+  #include <vector>
+  using namespace std;
+
+  class Solution {
+  public:
+      int missingNumber(int n, vector<int>& arr) {
+          // Calculate the expected sum of the first n natural numbers
+          int total_sum = n * (n + 1) / 2;
+
+          // Calculate the sum of the elements in the array
+          int arr_sum = 0;
+          for (int num : arr) {
+              arr_sum += num;
+          }
+
+          // The missing number is the difference between the expected sum and the array sum
+          return total_sum - arr_sum;
+      }
+  };
+
+  // Example usage
+  int main() {
+      int t;
+      cin >> t;
+      while (t--) {
+          int n;
+          cin >> n;
+
+          vector<int> arr(n - 1);
+          for (int i = 0; i < n - 1; ++i)
+              cin >> arr[i];
+          Solution obj;
+          cout << obj.missingNumber(n, arr) << "\n";
+      }
+      return 0;
+  }
+  ```
+
+  </TabItem>
+
+  <TabItem value="Javascript" label="Javascript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```javascript
+  class Solution {
+  missingNumber(n, arr) {
+    let totalSum = n * (n + 1) / 2;
+    let arrSum = arr.reduce((a, b) => a + b, 0);
+    return totalSum - arrSum;
+  }
+}
+
+let solution = new Solution();
+console.log(solution.missingNumber(4, [3, 1, 4])); // Expected output: 2
+console.log(solution.missingNumber(5, [5, 3, 1, 2])); // Expected output: 4
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Typescript" label="Typescript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```typescript
+  class Solution {
+  missingNumber(n: number, arr: number[]): number {
+    let totalSum: number = n * (n + 1) / 2;
+    let arrSum: number = arr.reduce((a, b) => a + b, 0);
+    return totalSum - arrSum;
+  }
+}
+
+let solution: Solution = new Solution();
+console.log(solution.missingNumber(4, [3, 1, 4])); // Expected output: 2
+console.log(solution.missingNumber(5, [5, 3, 1, 2])); // Expected output: 4
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Java" label="Java" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```java
+  public class Solution {
+  public int missingNumber(int n, int[] arr) {
+    int totalSum = n * (n + 1) / 2;
+    int arrSum = 0;
+    for (int num : arr) {
+      arrSum += num;
+    }
+    return totalSum - arrSum;
+  }
+}
+
+public class Main {
+  public static void main(String[] args) {
+    Solution solution = new Solution();
+    System.out.println(solution.missingNumber(4, new int[]{3, 1, 4})); // Expected output: 2
+    System.out.println(solution.missingNumber(5, new int[]{5, 3, 1, 2})); // Expected output: 4
+  }
+}
+
+  ```
+
+  </TabItem>
+</Tabs>
+
+## Example Walkthrough
+
+For the array `array = [3, 1, 4]` with `n = 4`:
+
+1. The expected sum of the first 4 natural numbers is `4 * (4 + 1) / 2 = 10`.
+2. The sum of the array elements is `3 + 1 + 4 = 8`.
+3. The missing number is `10 - 8 = 2`.
+
+For the array `array = [5, 3, 1, 2]` with `n = 5`:
+
+1. The expected sum of the first 5 natural numbers is `5 * (5 + 1) / 2 = 15`.
+2. The sum of the array elements is `5 + 3 + 1 + 2 = 11`.
+3. The missing number is `15 - 11 = 4`.
+
+## Solution Logic:
+
+1. Calculate the expected sum of the first `n` natural numbers using the formula `n * (n + 1) / 2`.
+2. Calculate the sum of the elements in the given array.
+3. The missing number is the difference between the expected sum and the sum of the array elements.
+
+## Time Complexity
+
+* The time complexity is $O(n)$, where n is the size of the input array.
+
+## Space Complexity
+
+* The auxiliary space complexity is $O(1)$ because we are not using any extra space proportional to the size of the input array.