Write an efficient C++/C program to find the smallest and second smallest element in an array.
Example:
Input: arr[] = {12, 13, 1, 10, 34, 1}
Output: The smallest element is 1 and
second Smallest element is 10
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. 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 |
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 |
Java
/*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
# 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 |
Javascript
<script> // 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(); // printing the desired element document.write("smallest element is "+arr[0],"</br>"); document.write("second smallest element is "+arr[1],"</br>"); // This code is contributed by shinjanpatra </script> |
Output
smallest element is 1 second smallest element is 9
Time complexity: O(N*logN)
Auxiliary space: O(1)
Approach 2A:
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[] = {12, 13, 1, 10, 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
// Java program to find smallest and // second smallest element in array import java.io.*; class GFG { public static void main(String args[]) { int arr[] = { 12, 13, 1, 10, 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
# python program to find smallest and second smallest element in array # import the module import sys arr = [12, 13, 1, 10, 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#
// C# program to find smallest and // second smallest element in array using System; public class GFG { static public void Main () { int[] arr = { 12, 13, 1, 10, 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
<script> // Javascript program to find smallest and // second smallest elements function solution( arr, arr_size) { let first = Number.MAX_VALUE, second = Number.MAX_VALUE; /* There should be atleast two elements */ if (arr_size < 2) { document.write(" Invalid Input "); return; } /* find the smallest element */ for (let i = 0; i < arr_size ; i ++) { if (arr[i] < first){ first = arr[i]; } } /* find the second smallest element */ for (let i = 0; i < arr_size ; i ++){ if (arr[i] < second && arr[i] > first){ second = arr[i]; } } if (second == Number.MAX_VALUE ) document.write("There is no second smallest element\n"); else document.write("The smallest element is " + first + " and second "+ "Smallest element is " + second +'\n'); } // Driver program let arr = [ 12, 13, 1, 10, 34, 1 ]; let n = arr.length; solution(arr, n); </script> |
Output
smallest element is: 1 second smallest element is: 10
Time complexity: O(N)
Auxiliary space: O(1)
Approach 2B:
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 <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[] = { 12, 13, 1, 10, 34, 1 }; int n = sizeof(arr) / sizeof(arr[0]); print2Smallest(arr, n); return 0; } |
C#
// 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 = { 12, 13, 1, 10, 34, 1 }; print2Smallest(arr); } } // This code is contributed by Sam007 |
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[] = { 12, 13, 1, 10, 34, 1 }; int n = sizeof(arr) / sizeof(arr[0]); print2Smallest(arr, n); return 0; } // This is code is contributed by rathbhupendra |
Java
// 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[] = { 12, 13, 1, 10, 34, 1 }; print2Smallest(arr); } } /*This code is contributed by Devesh Agrawal*/ |
Python3
# 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 = [12, 13, 1, 10, 34, 1] print2Smallest(arr) # This code is contributed by Devesh Agrawal |
PHP
<?php // PHP program to find smallest and // second smallest elements function print2Smallest($arr, $arr_size) { $INT_MAX = 2147483647; /* There should be atleast two elements */ if ($arr_size < 2) { echo(" 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 == $INT_MAX) echo("There is no second smallest element\n"); else echo "The smallest element is ",$first ," and second Smallest element is " , $second; } // Driver Code $arr = array(12, 13, 1, 10, 34, 1); $n = count($arr); print2Smallest($arr, $n) // This code is contributed by Smitha ?> |
Javascript
<script> // 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) { document.write(" 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 ) document.write("There is no second smallest element\n"); else document.write("The smallest element is " + first + " and second "+ "Smallest element is " + second +'\n'); } // Driver program let arr = [ 12, 13, 1, 10, 34, 1 ]; let n = arr.length; print2Smallest(arr, n); </script> |
Output
The smallest element is 1 and second Smallest element is 10
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)
Approach 3:
A N*log(N) approach using Priority_queue data structure. You can read about Priority Queue in more detail here.
C#
using System; using System.Collections.Generic; class GFG { static void Main(string[] args) { int[] arr = { 2, 5, 7, 3, 9, 10, 11, 1 }; int n = arr.Length; PriorityQueue<int> pq = new PriorityQueue<int>(); // adding elements to priority queue for (int i = 0; i < n; i++) { pq.Enqueue(arr[i]); } int t = pq.Peek(); // smallest element will be on // top of pq pq.Dequeue(); // removing first element so second // will // be on top. int w = pq.Peek(); Console.WriteLine("smallest element : " + t); Console.WriteLine("second smallest element : " + w); } } public class PriorityQueue<T> where T : IComparable<T> { private List<T> data; public PriorityQueue() { this.data = new List<T>(); } public void Enqueue(T item) { data.Add(item); int ci = data.Count - 1; while (ci > 0) { int pi = (ci - 1) / 2; if (data[ci].CompareTo(data[pi]) >= 0) break; T tmp = data[ci]; data[ci] = data[pi]; data[pi] = tmp; ci = pi; } } public T Dequeue() { // Assumes pq isn't empty int li = data.Count - 1; T frontItem = data[0]; data[0] = data[li]; data.RemoveAt(li); --li; int pi = 0; while (true) { int ci = pi * 2 + 1; if (ci > li) break; int rc = ci + 1; if (rc <= li && data[rc].CompareTo(data[ci]) < 0) ci = rc; if (data[pi].CompareTo(data[ci]) <= 0) break; T tmp = data[pi]; data[pi] = data[ci]; data[ci] = tmp; pi = ci; } return frontItem; } public T Peek() { T frontItem = data[0]; return frontItem; } public int Count { get { return data.Count; } } } |
C++
#include <bits/stdc++.h> using namespace std; int main() { int arr[] = { 2, 5, 7, 3, 9, 10, 11, 1 }; int n = sizeof(arr) / sizeof(arr[0]); priority_queue<int, vector<int>, greater<int> > pq; // min heap priority_queue // adding elements to priority queue for (int i = 0; i < n; i++) { pq.push(arr[i]); } int t = pq.top(); // smallest element will be on top of // pq as it is min heap pq.pop(); // removing first element so second will be on // top. int w = pq.top(); cout << "smallest element : " << t << endl; cout << "second smallest element : " << w << endl; return 0; } // this code is contributed by lakshyadeeppatel |
Java
import java.util.*; class GFG { public static void main(String[] args) { int arr[] = { 2, 5, 7, 3, 9, 10, 11, 1 }; int n = arr.length; PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); // adding elements to priority queue for (int i = 0; i < n; i++) { pq.add(arr[i]); } int t = pq.peek(); // smallest element will be on // top of pq pq.poll(); // removing first element so second will // be on top. int w = pq.peek(); System.out.println("smallest element : " + t); System.out.println("second smallest element : " + w); } } // This code is contributed by divyansh2212 |
Python3
from queue import PriorityQueue arr = [2, 5, 7, 3, 9, 10, 11, 1] n = len(arr) pq = PriorityQueue() # adding elements to priority queue for i in range(0, n): pq.put(arr[i]) t = pq.get() # smallest element will be on top of pq as it is min heap w = pq.get() print("smallest element :", t) print("second smallest element :", w) |
Javascript
class PriorityQueue { constructor() { this.values = []; } enqueue(val, priority) { this.values.push({ val, priority }); this.sort(); } dequeue() { return this.values.shift(); } sort() { this.values.sort((a, b) => a.priority - b.priority); } get top() { return this.values[0].val; } get secondTop() { return this.values[1].val; } } const arr = [2, 5, 7, 3, 9, 10, 11, 1]; const pq = new PriorityQueue();//min heap priority_queue //adding elements to priority queue for (let i = 0; i < arr.length; i++) { pq.enqueue(arr[i], arr[i]); } const t = pq.top;// smallest element will be on // top of pq pq.dequeue(); const w = pq.top; console.log("smallest element: ", t); console.log("second smallest element: ", w); |
Output
smallest element : 1 second smallest element : 2
Time Complexity: O(n*log(n))
In a priority queue time for adding each element is O(logn) and we are adding all the elements so the total time taken by the priority queue will be O(n*logn)
Auxiliary Space: O(n)
The extra space is used to store the elements in the priority queue.
Related Article: Minimum and Second minimum elements using minimum comparisons
Approach 4: Using list properties
Algorithm:
- Step 1: Declare a new list
- Step 2: Check length of the original list is more than 2 or not.
- Step 3: Append the smallest element from the original list into new list.
- Step 4: Count the number times smallest element appeared in the original list, then run a loop to remove all the smallest elements from the original list.
- Step 5: After removing all smallest elements, now find for 2nd smallest element using min function.
- Step 6: return new list which contains the smallest element and 2nd smallest element.
Example:
C++
#include <bits/stdc++.h> using namespace std; vector<int> print2Smallest(vector<int> arr) { vector<int> lst; // vector is declared if (unordered_set<int>(arr.begin(), arr.end()).size() == 1) { return { -1, -1 }; // checking length of array is // greater than 2 or not } lst.push_back(*min_element( arr.begin(), arr.end())); // appending smallest // element in new vector int p = *min_element(arr.begin(), arr.end()); // storing min element int c = count(arr.begin(), arr.end(), p); // counting number of smallest element arr.erase( remove(arr.begin(), arr.end(), p), arr.end()); // removing all the smallest element lst.push_back(*min_element( arr.begin(), arr.end())); // appending 2nd smallest element return lst; } // Driver function to test above function int main() { vector<int> arr = { 12, 13, 1, 10, 34, 1 }; vector<int> res = print2Smallest(arr); for (int i = 0; i < res.size(); i++) { cout << res[i] << " "; } return 0; } |
Java
import java.util.*; public class Main { public static List<Integer> print2Smallest(List<Integer> arr) { List<Integer> lst = new ArrayList<>(); // List is declared if (new HashSet<Integer>(arr).size() == 1) { // Checking length of array is greater // than 2 or not return Arrays.asList(-1, -1); } lst.add(Collections.min( arr)); // Appending smallest element in new list int p = Collections.min(arr); // Storing min element int c = Collections.frequency( arr, p); // Counting number of smallest element arr.removeAll(Collections.singleton( p)); // Removing all the smallest element lst.add(Collections.min( arr)); // Appending 2nd smallest element return lst; } // Driver function to test above function public static void main(String[] args) { List<Integer> arr = new ArrayList<>( Arrays.asList(12, 13, 1, 10, 34, 1)); List<Integer> res = print2Smallest(arr); for (int i = 0; i < res.size(); i++) { System.out.print(res.get(i) + " "); } } } |
Python3
def print2Smallest(arr): lst = [] # list is declared if len(set(arr)) == 1: # checking length of array is return [-1, -1] # greater than 2 or not lst.append(min(arr)) # appending small element in new list p = min(arr) # storing min element c = arr.count(p) # counting number of a smallest element for i in range(c): arr.remove(p) # removing all the smallest element lst.append(min(arr)) # appending 2nd smallest element return lst # Driver function to test above function arr = [12, 13, 1, 10, 34, 1] print(print2Smallest(arr)) # This code is contributed by Vaibhav Kumar |
Javascript
function print2Smallest(arr) { let lst = []; // list is declared if (new Set(arr).size === 1) { // checking length of array is greater than 2 or not return [-1, -1]; } lst.push(Math.min(...arr)); // appending small element in new list let p = Math.min(...arr); // storing min element let c = arr.filter((el) => el === p).length; // counting number of a smallest element for (let i = 0; i < c; i++) { arr.splice(arr.indexOf(p), 1); // removing all the smallest element } lst.push(Math.min(...arr)); // appending 2nd smallest element return lst; } // Driver function to test above function let arr = [12, 13, 1, 10, 34, 1]; console.log(print2Smallest(arr)); |
C#
using System; using System.Collections.Generic; using System.Linq; public class MainClass { public static List<int> Print2Smallest(List<int> arr) { List<int> lst = new List<int>(); // List is declared if (new HashSet<int>(arr).Count == 1) { // Checking length of array is greater than 2 or not return new List<int> {-1, -1}; } lst.Add(arr.Min()); // Appending smallest element in new list int p = arr.Min(); // Storing min element int c = arr.Count(x => x == p); // Counting number of smallest element arr.RemoveAll(x => x == p); // Removing all the smallest element lst.Add(arr.Min()); // Appending 2nd smallest element return lst; } // Driver function to test above function public static void Main() { List<int> arr = new List<int>{12, 13, 1, 10, 34, 1}; List<int> res = Print2Smallest(arr); foreach (int i in res) { Console.Write(i + " "); } } } |
Output
[1, 10]
Time Complexity: O(N)
Auxiliary Space: O(1)
Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.
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