Find the smallest element in an array

Problem Statement: Given an array, we have to find the smallest element in the array.

Examples:

Example 1:
Input: arr[] = {2,5,1,3,0};
Output: 0
Explanation: 0 is the smallest element in the array. 

Example2: 
Input: arr[] = {8,10,5,7,9};
Output: 5
Explanation: 5 is the smallest element in the array.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution1: Sorting

Intuition: We can sort the array in ascending order, hence the smallest element will be at the 0th index.

Approach:

Sort the array in ascending order.
Print the 0th index.

DryRun:

Before sorting: arr[] = {2,5,1,3,0};

After sorting: arr[] = {0,1,2,3,5};

Hence answer : arr[0] = 0

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;
 
int sortArr(vector<int>& arr) {
    sort(arr.begin(),arr.end());
    return arr[0];
}
 
int main() {
    vector<int> arr1 = {2,5,1,3,0};
    vector<int> arr2 = {8,10,5,7,9};
   
    cout<<"The smallest element in the array is: "<<sortArr(arr1)<<endl;
    cout<<"The smallest element in the array is: "<<sortArr(arr2);
   
    return 0;
}

Output:

The smallest element in the array is: 0
The smallest element in the array is: 5

Time Complexity: O(N*log(N))

Space Complexity: O(1)

Java Code

import java.util.*;
public class tuf {
 
  public static void main(String args[]) {
 
    int arr1[] =  {2,5,1,3,0};
    System.out.println("The smallest element in array is: " + sort(arr1));
   
    int arr2[] =  {8,10,5,7,9};
    System.out.println("The smallest element in array is: " + sort(arr2));
  }
  static int sort(int arr[]) {
    Arrays.sort(arr);
    return arr[0];
  }
}

Output:

The smallest element in the array is: 0
The smallest element in the array is: 5

Time Complexity: O(N*log(N))

Space Complexity: O(1)

Solution2: Using a min variable

Intuition: We can maintain a min variable which will update whenever the current value is less than the value in the min variable.

Approach:

Create a min variable and initialize it with arr[0].
Use a for loop and compare it with other elements of the array
If any element is less than the min value, update min value with element’s value
Print the min variable.

Code:

C++ Code

#include <iostream>
#include <bits/stdc++.h>
 
using namespace std;
int SmallestElement(int arr[], int n) {
 
  int min = arr[0];
  for (int i = 0; i < n; i++) {
    if (min > arr[i]) {
      min = arr[i];
    }
  }
  return min;
}
int main() {
  int arr1[] = {2,5,1,3,0};
  int n = 5;
  int min = SmallestElement(arr1, n);
  cout << "The smallest element in the array is: " << min << endl;
 
  int arr2[] =  {8,10,5,7,9};
  n = 5;
  min = SmallestElement(arr2, n);
  cout << "The smallest element in the array is: " << min;
  return 0;
}

Output:

The smallest element in the array is: 0
The smallest element in the array is: 5

Time Complexity: O(N)

Space Complexity: O(1)

Java Code

import java.util.*;
 
public class tuf {
 
  public static void main(String args[]) {
 
    int arr1[] =  {2,5,1,3,0};
    System.out.println("The smallest element in the array is: "+SmallestElement(arr1));
 
    int arr2[] =  {8,10,5,7,9};
    System.out.println("The smallest element in the array is: "+SmallestElement(arr2));
  }
  static int SmallestElement(int arr[]) {
    int min = arr[0];
    for (int i = 0; i < arr.length; i++) {
      if (arr[i] < min) {
        min = arr[i];
      }
    }
    return min;
  }
}

Output:

The smallest element in the array is: 0
The smallest element in the array is: 5

Time Complexity: O(N)

Space Complexity: O(1)

Special thanks to Prashant Sahu for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article