Create Second Smallest and Second Largest Element in an Array.md

Author: bakisama

content/CP/tuf/Second Smallest and Second Largest Element in an Array.md

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+---
+title: Second Smallest and Second Largest Element in an Array
+---
+## Problem Statement
+
+I am given an array of integers of size `n`.  
+My task is to find:
+
+- the **second smallest** element
+- the **second largest** element
+
+If either does not exist, I should return `-1` for that value.
+
+---
+
+## Examples
+
+### `Example 1`
+
+Input: [1, 2, 4, 7, 7, 5]  
+Output: Second Smallest = 2, Second Largest = 5
+
+`### Example 2`
+
+Input: [1]  
+Output: -1 -1
+
+`### Example 3`
+
+Input: [5, 5, 5]  
+Output: -1 -1
+
+
+Complexity Analysis
+
+    Time Complexity: O(n)
+
+    Space Complexity: O(1)
+
+This is optimal since I must inspect every element at least once.
+Edge Cases to Remember
+
+    Array size < 2 → no second elements
+
+    All elements equal → no second smallest or largest
+
+    Duplicate largest/smallest values should be ignored
+
+---  
+## Intuition  
+To find the second smallest and second largest elements, I need **distinct values**.  
+A naive idea would be: 
+- sort the array 
+- take the second element from start and end  
+
+However, sorting costs **O(n log n)** time, which is unnecessary.  
+
+I can do this more efficiently using a **single traversal**.  
+
+---  
+## Key Observations  
+- The **smallest** and **largest** elements are easy to track. 
+- The **second smallest** must be:  
+	- greater than the smallest  
+	- smaller than all other candidates
+- The **second largest** must be:  
+	- smaller than the largest  
+	- greater than all other candidates 
+---  
+## Optimal Approach (Single Pass)
+
+### Strategy
+
+I maintain four variables:
+- `smallest`
+- `secondSmallest`
+- `largest`
+- `secondLargest`
+
+I iterate through the array once and update these values carefully.
+
+---
+
+## Algorithm
+
+1. Initialize:
+   - `smallest = +∞`
+   - `secondSmallest = +∞`
+   - `largest = -∞`
+   - `secondLargest = -∞`
+
+2. For each element `x` in the array:
+   - If `x < smallest`:
+     - `secondSmallest = smallest`
+     - `smallest = x`
+   - Else if `x > smallest` and `x < secondSmallest`:
+     - `secondSmallest = x`
+
+   - If `x > largest`:
+     - `secondLargest = largest`
+     - `largest = x`
+   - Else if `x < largest` and `x > secondLargest`:
+     - `secondLargest = x`
+
+1. If `secondSmallest` or `secondLargest` was never updated, return `-1`.
+
+---
+## Code (Optimal – O(n))
+
+```cpp
+#include <climits>
+#include <vector>
+using namespace std;
+
+pair<int, int> secondSmallestAndLargest(vector<int> &arr) {
+    int n = arr.size();
+    if (n < 2) return {-1, -1};
+
+    int smallest = INT_MAX, secondSmallest = INT_MAX;
+    int largest = INT_MIN, secondLargest = INT_MIN;
+
+    for (int x : arr) {
+        // For smallest
+        if (x < smallest) {
+            secondSmallest = smallest;
+            smallest = x;
+        } 
+        else if (x > smallest && x < secondSmallest) {
+            secondSmallest = x;
+        }
+
+        // For largest
+        if (x > largest) {
+            secondLargest = largest;
+            largest = x;
+        } 
+        else if (x < largest && x > secondLargest) {
+            secondLargest = x;
+        }
+    }
+
+    if (secondSmallest == INT_MAX || secondLargest == INT_MIN)
+        return {-1, -1};
+
+    return {secondSmallest, secondLargest};
+}
+```
+---
+
+## Complexity Analysis
+
+- **Time Complexity:** O(n)
+    
+- **Space Complexity:** O(1)
+    
+
+This is optimal since I must inspect every element at least once.
+
+---
+
+## Edge Cases to Remember
+
+- Array size < 2 → no second elements
+    
+- All elements equal → no second smallest or largest
+    
+- Duplicate largest/smallest values should be ignored
+    
+
+---