Find the smallest and second smallest elements in an array
Last Updated : 11 Mar, 2025
Given an array arr[] of size N, find the smallest and second smallest element in an array.
Examples:
Input: arr[] = {12, 13, 1, 10, 34, 1}
Output: 1 10
Explanation: The smallest element is 1 and second smallest element is 10.
Input: arr[] = {111, 13, 25, 9, 34, 1}
Output: 1 9
Explanation: The smallest element is 1 and second smallest element is 9.
Approach 1:
A Simple Solution is to sort the array in increasing order. The first two elements in the sorted array would be the two smallest elements. In this approach, if the smallest element is present more than one time then we will have to use a loop for printing the unique smallest and second smallest elements.
Below is the implementation of the above approach:
C++ // C++ simple approach to print smallest // and second smallest element. #include <bits/stdc++.h> using namespace std; int main() { int arr[] = { 111, 13, 25, 9, 34, 1}; int n = sizeof(arr) / sizeof(arr[0]); // sorting the array using // in-built sort function sort(arr, arr + n); // printing the desired element cout << "smallest element is " << arr[0] << endl; cout << "second smallest element is " << arr[1]; return 0; } // this code is contributed by Machhaliya Muhammad /*package whatever //do not write package name here */ import java.io.*; import java.util.*; class GFG { // Java simple approach to print smallest // and second smallest element. // Driver Code public static void main(String args[]) { int arr[] = { 111, 13, 25, 9, 34, 1 }; int n = arr.length; // sorting the array using // in-built sort function Arrays.sort(arr); // printing the desired element System.out.println("smallest element is " + arr[0]); System.out.println("second smallest element is " + arr[1]); } } // This code is contributed by shinjanpatra # Python3 simple approach to print smallest # and second smallest element. # driver code arr = [111, 13, 25, 9, 34, 1] n = len(arr) # sorting the array using # in-built sort function arr.sort() # printing the desired element print("smallest element is "+str(arr[0])) print("second smallest element is "+str(arr[1])) # This code is contributed by shinjanpatra // C# simple approach to print smallest // and second smallest element. using System; public class GFG { // Driver Code static public void Main() { int[] arr = { 111, 13, 25, 9, 34, 1}; int n = arr.Length; // sorting the array using // in-built sort function Array.Sort(arr); // printing the desired element Console.WriteLine("smallest element is " + arr[0]); Console.WriteLine("second smallest element is " + arr[1]); } } // This code is contributed by kothavvsaakash // JavaScript simple approach to print smallest // and second smallest element. // driver code let arr = [111, 13, 25, 9, 34, 1]; let n = arr.length; // sorting the array using // in-built sort function arr.sort((a, b) => a - b); // printing the desired element console.log("smallest element is "+arr[0]); console.log("second smallest element is "+arr[1]); Outputsmallest element is 1 second smallest element is 9
Time complexity: O(N * logN)
Auxiliary space: O(1)
Finding the smallest and second smallest elements by traversing the array twice (Two-pass):
A Better Solution is to scan the array twice. In the first traversal find the minimum element. Let this element be x. In the second traversal, find the smallest element greater than x.
Using this method, we can overcome the problem of Method 1 which occurs when the smallest element is present in an array more than one time.
The above solution requires two traversals of the input array.
C++ // C++ program to find smallest and // second smallest element in array #include <bits/stdc++.h> using namespace std; int main() { int arr[] = { 111, 13, 25, 9, 34, 1 }; int n = sizeof(arr) / sizeof(arr[0]); int smallest = INT_MAX; // traversing the array to find // smallest element. for (int i = 0; i < n; i++) { if (arr[i] < smallest) { smallest = arr[i]; } } cout << "smallest element is: " << smallest << endl; int second_smallest = INT_MAX; // traversing the array to find second smallest element for (int i = 0; i < n; i++) { if (arr[i] < second_smallest && arr[i] > smallest) { second_smallest = arr[i]; } } cout << "second smallest element is: " << second_smallest << endl; return 0; } // This code is contributed by Machhaliya Muhamma // Java program to find smallest and // second smallest element in array import java.io.*; class GFG { public static void main(String args[]) { int arr[] = { 111, 13, 25, 9, 34, 1 }; int n = arr.length; int smallest = Integer.MAX_VALUE; // traversing the array to find // smallest element. for (int i = 0; i < n; i++) { if (arr[i] < smallest) { smallest = arr[i]; } } System.out.println("smallest element is: " + smallest); int second_smallest = Integer.MAX_VALUE; // traversing the array to find second smallest // element for (int i = 0; i < n; i++) { if (arr[i] < second_smallest && arr[i] > smallest) { second_smallest = arr[i]; } } System.out.println("second smallest element is: " + second_smallest); } } // This code is contributed by Lovely Jain # python program to find smallest and second smallest element in array # import the module import sys arr = [111, 13, 25, 9, 34, 1] n = len(arr) smallest = sys.maxint # traversing the array to find smallest element. for i in range(n): if(arr[i] < smallest): smallest = arr[i] print('smallest element is: ' + str(smallest)) second_smallest = sys.maxint # traversing the array to find second smallest element for i in range(n): if(arr[i] < second_smallest and arr[i] > smallest): second_smallest = arr[i] print('second smallest element is: ' + str(second_smallest)) # This code is contributed by lokeshmvs21. // C# program to find smallest and // second smallest element in array using System; public class GFG { static public void Main () { int[] arr = { 111, 13, 25, 9, 34, 1 }; int n = arr.Length; int smallest = Int32.MaxValue; // traversing the array to find // smallest element. for (int i = 0; i < n; i++) { if (arr[i] < smallest) { smallest = arr[i]; } } Console.WriteLine("smallest element is: " + smallest); int second_smallest = Int32.MaxValue; // traversing the array to find second smallest // element for (int i = 0; i < n; i++) { if (arr[i] < second_smallest && arr[i] > smallest) { second_smallest = arr[i]; } } Console.WriteLine("second smallest element is: " + second_smallest); } } // This code is contributed by kothavvsaakash // JavaScript program to find smallest and second smallest element in array let arr = [111, 13, 25, 9, 34, 1]; let n = arr.length; // Initialize smallest with Infinity let smallest = Infinity; // Traverse the array to find the smallest element for (let i = 0; i < n; i++) { if (arr[i] < smallest) { smallest = arr[i]; } } console.log("Smallest element is: " + smallest); // Initialize second smallest with Infinity let secondSmallest = Infinity; // Traverse the array to find second smallest element for (let i = 0; i < n; i++) { if (arr[i] < secondSmallest && arr[i] > smallest) { secondSmallest = arr[i]; } } console.log("Second smallest element is: " + secondSmallest); Outputsmallest element is: 1 second smallest element is: 9
Time complexity: O(N)
Auxiliary space: O(1)
Finding the smallest and second smallest elements by traversing the array twice (One-pass):
Efficient Solution can find the minimum two elements in one traversal. Below is the complete algorithm.
Algorithm:
1. Initialize both first and second smallest as INT_MAX
first = second = INT_MAX
2. Loop through all the elements.
- If the current element is smaller than first, then update first and second.
- Else if the current element is smaller than second then update second.
Below is the implementation of the above approach:
C++ // C++ program to find smallest and // second smallest elements #include <bits/stdc++.h> using namespace std; /* For INT_MAX */ void print2Smallest(int arr[], int arr_size) { int i, first, second; /* There should be atleast two elements */ if (arr_size < 2) { cout << " Invalid Input "; return; } first = second = INT_MAX; for (i = 0; i < arr_size; i++) { /* If current element is smaller than first then update both first and second */ if (arr[i] < first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] < second && arr[i] != first) second = arr[i]; } if (second == INT_MAX) cout << "There is no second smallest element\n"; else cout << "The smallest element is " << first << " and second " "Smallest element is " << second << endl; } /* Driver code */ int main() { int arr[] = { 111, 13, 25, 9, 34, 1 }; int n = sizeof(arr) / sizeof(arr[0]); print2Smallest(arr, n); return 0; } // This is code is contributed by rathbhupendra // C program to find smallest and second smallest elements #include <limits.h> /* For INT_MAX */ #include <stdio.h> void print2Smallest(int arr[], int arr_size) { int i, first, second; /* There should be atleast two elements */ if (arr_size < 2) { printf(" Invalid Input "); return; } first = second = INT_MAX; for (i = 0; i < arr_size; i++) { /* If current element is smaller than first then update both first and second */ if (arr[i] < first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] < second && arr[i] != first) second = arr[i]; } if (second == INT_MAX) printf("There is no second smallest element\n"); else printf("The smallest element is %d and second " "Smallest element is %d\n", first, second); } /* Driver program to test above function */ int main() { int arr[] = { 111, 13, 25, 9, 34, 1 }; int n = sizeof(arr) / sizeof(arr[0]); print2Smallest(arr, n); return 0; } // Java program to find smallest and second smallest // elements import java.io.*; class SecondSmallest { /* Function to print first smallest and second smallest elements */ static void print2Smallest(int arr[]) { int first, second, arr_size = arr.length; /* There should be atleast two elements */ if (arr_size < 2) { System.out.println(" Invalid Input "); return; } first = second = Integer.MAX_VALUE; for (int i = 0; i < arr_size; i++) { /* If current element is smaller than first then update both first and second */ if (arr[i] < first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] < second && arr[i] != first) second = arr[i]; } if (second == Integer.MAX_VALUE) System.out.println("There is no second" + "smallest element"); else System.out.println("The smallest element is " + first + " and second Smallest" + " element is " + second); } /* Driver program to test above functions */ public static void main(String[] args) { int arr[] = { 111, 13, 25, 9, 34, 1 }; print2Smallest(arr); } } /*This code is contributed by Devesh Agrawal*/ # Python program to find smallest and second smallest elements import math def print2Smallest(arr): # There should be atleast two elements arr_size = len(arr) if arr_size < 2: print("Invalid Input") return first = second = math.inf for i in range(0, arr_size): # If current element is smaller than first then # update both first and second if arr[i] < first: second = first first = arr[i] # If arr[i] is in between first and second then # update second elif (arr[i] < second and arr[i] != first): second = arr[i] if (second == math.inf): print("No second smallest element") else: print('The smallest element is', first, 'and', ' second smallest element is', second) # Driver function to test above function arr = [111, 13, 25, 9, 34, 1] print2Smallest(arr) # This code is contributed by Devesh Agrawal // C# program to find smallest // and second smallest elements using System; class GFG { /* Function to print first smallest and second smallest elements */ static void print2Smallest(int[] arr) { int first, second, arr_size = arr.Length; /* There should be atleast two elements */ if (arr_size < 2) { Console.Write(" Invalid Input "); return; } first = second = int.MaxValue; for (int i = 0; i < arr_size; i++) { /* If current element is smaller than first then update both first and second */ if (arr[i] < first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] < second && arr[i] != first) second = arr[i]; } if (second == int.MaxValue) Console.Write("There is no second" + "smallest element"); else Console.Write("The smallest element is " + first + " and second Smallest" + " element is " + second); } /* Driver program to test above functions */ public static void Main() { int[] arr = { 111, 13, 25, 9, 34, 1 }; print2Smallest(arr); } } // This code is contributed by Sam007 // Javascript program to find smallest and // second smallest elements function print2Smallest( arr, arr_size) { let i, first, second; /* There should be atleast two elements */ if (arr_size < 2) { console.log(" Invalid Input "); return; } first=Number.MAX_VALUE ; second=Number.MAX_VALUE ; for (i = 0; i < arr_size ; i ++) { /* If current element is smaller than first then update both first and second */ if (arr[i] < first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] < second && arr[i] != first) second = arr[i]; } if (second == Number.MAX_VALUE ) console.log("There is no second smallest element\n"); else console.log("The smallest element is " + first + " and second "+ "Smallest element is " + second +'\n'); } // Driver program let arr = [ 111, 13, 25, 9, 34, 1 ]; let n = arr.length; print2Smallest(arr, n); OutputThe smallest element is 1 and second Smallest element is 9
The same approach can be used to find the largest and second-largest elements in an array.
Time Complexity: O(N)
Auxiliary Space: O(1)
Related Article: Minimum and Second minimum elements using minimum comparisons
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