Question 10 : Write java Program to Find Smallest and Largest Element in an Array

You are given an integer array containing 1 to n but one of the numbers from 1 to n in the array is missing. You need to provide an optimum solution to find the missing number. 

The number can not be repeated in the array.
For example:

int[] arr1={7,5,6,1,4,2};
Missing numner : 3

int[] arr2={5,3,1,2};
Missing numner : 4

Problem:

You are given an array of numbers. You need to find smallest and largest numbers in the array.

Solution:

• Initialize two variables largest and smallest with arr[0]
• Iterate over array
  • If the current element is greater than the largest, then assign the current element to the largest.
  • If the current element is smaller than the smallest, then assign the current element to the smallest.
• You will get the smallest and largest element in the end.

Java code to find the Smallest and Largest Element in an Array :

/*
Java program to Find Largest and Smallest Number in an Array 
*/
public class FindLargestSmallestNumberMain {
 
 public static void main(String[] args) {
 
 //array of 10 numbers
 int arr[] = new int[]{12,56,76,89,100,343,21,234};
 
 //assign first element of an array to largest and smallest
 int smallest = arr[0];
 int largest = arr[0];
 
 for(int i=1; i< arr.length; i++)
 {
 if(arr[i] > largest)
 largest = arr[i];
 else if (arr[i] < smallest)
 smallest = arr[i];
 
 }
 System.out.println("Smallest Number is : " + smallest);
 System.out.println("Largest Number is : " + largest); 
 }
}

When you run the above program, you will get the below output::

Largest Number is : 343
Smallest Number is : 12

Python code 

def find_smallest_largest(arr):
    smallest = arr[0]
    largest = arr[0]

    for num in arr[1:]:
        if num > largest:
            largest = num
        elif num < smallest:
            smallest = num

    return smallest, largest


arr = [12, 56, 76, 89, 100, 343, 21, 234]
s, l = find_smallest_largest(arr)
print("Smallest Number is:", s)
print("Largest  Number is:", l)

  
    
    
    
    
  

  

    
Approach
    
Time Complexity
    
Space Complexity
    
Best Use
  

  

    

      Single pass linear scan
    
    

      O(n)
    
    

      O(1)
    
    

      ⭐ Best & optimal
    
  

  

    

      Sorting the array
    
    

      O(n log n)
    
    

      O(n)
    
    

      Simple but inefficient
    
  

  

    

      Recursive divide & conquer
    
    

      O(n)
    
    

      O(log n)
    
    

      Not needed here