Smallest in Array

Given an array of integers arr[]. Find the minimum element.

Examples:

Input: arr[] = [1, 2, 3, 4]
Output: 1
Explanation: The smallest element in the array is 1.
Input: arr[] = [3, 2, 1]
Output: 1
Explanation: The smallest element in the array is 1.
Input: arr[] = [5, 4, 2, 8, 8]
Output: 2
Explanation: The smallest element in the array is 2.

Try It Yourself

Table of Content

Iterative Approach - O(n) Time and O(1) Space

The idea is to traverse the array and keep track of the smallest element found so far.

Working of Approach:

Initialize mn with the first element of the array.
Traverse the array from the second element.
Compare each element with mn.
Update mn if the current element is smaller.
Return mn after completing the traversal.

C++

#include 
#include 
using namespace std;

int findMin(vector<int> &arr)
{
    // Initialize minimum with the first element
    int mn = arr[0];

    // Traverse the remaining elements
    for (int i = 1; i < arr.size(); i++)
    {
        // Update minimum if a smaller element is found
        if (arr[i] < mn)
            mn = arr[i];
    }

    return mn;
}

int main()
{
    vector<int> arr = {5, 4, 2, 8, 8};

    cout << findMin(arr);

    return 0;
}

Java

import java.util.Arrays;

public class GFG {
    // Initialize minimum with the first element
    static int findMin(int[] arr)
    {
        int mn = arr[0];

        // Traverse the remaining elements
        for (int i = 1; i < arr.length; i++) {
            // Update minimum if a smaller element is found
            if (arr[i] < mn)
                mn = arr[i];
        }

        return mn;
    }

    public static void main(String[] args)
    {
        int[] arr = { 5, 4, 2, 8, 8 };

        System.out.println(findMin(arr));
    }
}

Python

def findMin(arr):
    # Initialize minimum with the first element
    mn = arr[0]

    # Traverse the remaining elements
    for i in range(1, len(arr)):
        # Update minimum if a smaller element is found
        if arr[i] < mn:
            mn = arr[i]

    return mn

if __name__ == "__main__":
    arr = [5, 4, 2, 8, 8]

    print(findMin(arr))

using System;

public class GFG {
    // Initialize minimum with the first element
    public static int findMin(int[] arr)
    {
        int mn = arr[0];

        // Traverse the remaining elements
        for (int i = 1; i < arr.Length; i++) {
            // Update minimum if a smaller element is found
            if (arr[i] < mn)
                mn = arr[i];
        }

        return mn;
    }

    public static void Main()
    {
        int[] arr = { 5, 4, 2, 8, 8 };

        Console.WriteLine(findMin(arr));
    }
}

JavaScript

function findMin(arr)
{
    // Initialize minimum with the first element
    let mn = arr[0];

    // Traverse the remaining elements
    for (let i = 1; i < arr.length; i++) {
        // Update minimum if a smaller element is found
        if (arr[i] < mn)
            mn = arr[i];
    }

    return mn;
}

// Driver Code
const arr = [ 5, 4, 2, 8, 8 ];
console.log(findMin(arr));

Output

Recursive Approach - O(n) Time and O(n) Space

The idea is to recursively find the minimum in the remaining part of the array and compare it with the current element.

Working of Approach:

Start from the first element of the array.
Recursively find the minimum from the next index onward.
Use the last element as the base case.
Compare the current element with the recursive result.
Return the smaller of the two values.

C++

#include 
#include 
#include 
using namespace std;

// Recursive function to find the minimum element
int findMinRec(vector<int> &arr, int i)
{
    // Base case: last element
    if (i == arr.size() - 1)
        return arr[i];

    // Find minimum in the remaining array
    int mn = findMinRec(arr, i + 1);

    // Return the smaller element
    return min(arr[i], mn);
}

int findMin(vector<int> &arr)
{
    return findMinRec(arr, 0);
}

int main()
{
    vector<int> arr = {5, 4, 2, 8, 8};

    cout << findMin(arr);

    return 0;
}

Java

import java.util.Arrays;

public class GFG {
    // Recursive function to find the minimum element
    public static int findMinRec(int[] arr, int i)
    {
        // Base case: last element
        if (i == arr.length - 1)
            return arr[i];

        // Find minimum in the remaining array
        int mn = findMinRec(arr, i + 1);

        // Return the smaller element
        return Math.min(arr[i], mn);
    }

    public static int findMin(int[] arr)
    {
        return findMinRec(arr, 0);
    }

    public static void main(String[] args)
    {
        int[] arr = { 5, 4, 2, 8, 8 };

        System.out.println(findMin(arr));
    }
}

Python

def findMinRec(arr, i):
    # Base case: last element
    if i == len(arr) - 1:
        return arr[i]

    # Find minimum in the remaining array
    mn = findMinRec(arr, i + 1)

    # Return the smaller element
    return min(arr[i], mn)

def findMin(arr):
    return findMinRec(arr, 0)

if __name__ == '__main__':
    arr = [5, 4, 2, 8, 8]

    print(findMin(arr))

using System;

public class GFG {
    // Recursive function to find the minimum element
    public static int findMinRec(int[] arr, int i)
    {
        // Base case: last element
        if (i == arr.Length - 1)
            return arr[i];

        // Find minimum in the remaining array
        int mn = findMinRec(arr, i + 1);

        // Return the smaller element
        return Math.Min(arr[i], mn);
    }

    public static int findMin(int[] arr)
    {
        return findMinRec(arr, 0);
    }

    public static void Main()
    {
        int[] arr = { 5, 4, 2, 8, 8 };

        Console.WriteLine(findMin(arr));
    }
}

JavaScript

/* Recursive function to find the minimum element */
function findMinRec(arr, i)
{
    // Base case: last element
    if (i === arr.length - 1)
        return arr[i];

    // Find minimum in the remaining array
    let mn = findMinRec(arr, i + 1);

    // Return the smaller element
    return Math.min(arr[i], mn);
}

function findMin(arr) { return findMinRec(arr, 0); }

// Driver Code
let arr = [ 5, 4, 2, 8, 8 ];
console.log(findMin(arr));

Output

Library Methods - O(n) Time and O(1) Space

The idea is to use the language's inbuilt function to directly find the smallest element in the array.

Working of Approach:

Use the language's inbuilt function to find the minimum element.
Pass the array to the function.
The function checks the elements and finds the smallest value.
Store or return the obtained minimum element.
Return the minimum value as the answer.

C++

#include 
#include 
#include 
using namespace std;

int findMin(vector<int> &arr)
{
    // Find the iterator pointing to the minimum element
    auto it = min_element(arr.begin(), arr.end());

    // Return the minimum element
    return *it;
}

int main()
{
    vector<int> arr = {5, 4, 2, 8, 8};

    cout << findMin(arr);

    return 0;
}

Java

import java.util.*;

class GFG {
    static int findMin(int[] arr)
    {
        // Find the minimum element
        int min = Arrays.stream(arr).min().getAsInt();

        // Return the minimum element
        return min;
    }

    public static void main(String[] args)
    {
        int[] arr = { 5, 4, 2, 8, 8 };

        System.out.println(findMin(arr));
    }
}

Python

def findMin(arr):
    # Find the minimum element
    return min(arr)

if __name__ == '__main__':
    arr = [5, 4, 2, 8, 8]
    print(findMin(arr))

using System;
using System.Linq;

class GFG {
    static int findMin(int[] arr)
    {
        // Find the minimum element
        int min = arr.Min();

        // Return the minimum element
        return min;
    }

    public static void Main()
    {
        int[] arr = { 5, 4, 2, 8, 8 };

        Console.WriteLine(findMin(arr));
    }
}

JavaScript

function findMin(arr)
{
    // Find the minimum element
    return Math.min(...arr);
}

// Driver Code
let arr = [ 5, 4, 2, 8, 8 ];
console.log(findMin(arr));

Output

Iterative Approach - O(n) Time and O(1) Space

Recursive Approach - O(n) Time and O(n) Space

Library Methods - O(n) Time and O(1) Space

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