Second Largest/Smallest Number in Array in C/C++, Java and Python
In an earlier tutorial, we have seen how to find the smallest or largest number from the integer array. In this tutorial, we are going to find the second smallest number in an array in a single iteration.
Input:
- integer array
Output:
- Print second smallest integer in given array.
We can easily find the second smallest number by traversing the array twice. In the first traversing, find the first smallest number. And in second traversing find the second smallest number. But this is not the optimal solution to the problem.
Finding second smallest number in array with single iteration is a bit tricky. Here is the algorithm to solve this problem.
This question was asked in the Goldman Sachs coderpad online interview.
Algorithm
- Compare the first two elements from the array.
- Initialize the
small_firstwith the smallest number. - Initialize the
small_secondwith the smallest number.
- Initialize the
- Loop over internet array from 3rd number.
- If the current number is smaller than small_first
- assign small_first to small_second
- assign a current number to small_first
- If the current number is smaller than small_first
- Print small_second.
If you understand the logic from the above algorithm, you can solve the problem in any programming language. Let’s implement this algorithm in different programming languages.
Note: By changing some conditional statements, you can also implement the program to find the second largest number in the given array. That is just another story. Let’s stick with the original problem first.
C/C++ Program
Prerequisites:
C/C++ Code:
#include <stdio.h>
int main() {
//initialize array
int arr[7] = {34, 62, 56, 13, 7, 9, 17};
int i=0;
int small_first = arr[0];
int small_second = arr[0];
//initializ first and second smallest number
if(arr[0] > arr[1]) {
small_first = arr[1];
small_second = arr[0];
}
else {
small_first = arr[0];
small_second = arr[1];
}
for(i = 2; i < 7; i++) {
//compare smallest number with current number
if(arr[i]<=small_first) {
small_second = small_first;
small_first = arr[i];
}
else if(arr[i]<small_second){
small_second = arr[i];
}
}
//print second smallest number from array
printf("Second smallest number in array is %d.", small_second);
return 0;
}Output:
Second smallest number in array is 9.
Java Program
Prerequisites:
Java Code:
public class Smallest_Largest_Element_Array {
public static void main(String[] args) {
// initialize the array
int [] arr = new int [] {67, 83, 8, 71, 29, 3, 48};
int small_first = arr[0];
int small_second = arr[0];
//initializ first and second smallest number
if(arr[0] > arr[1]) {
small_first = arr[1];
small_second = arr[0];
}
else {
small_first = arr[0];
small_second = arr[1];
}
//traverse all elements in the array
for (int i = 2; i < arr.length; i++) {
//compare smallest number with current number
if(arr[i]<=small_first) {
small_second = small_first;
small_first = arr[i];
}
else if(arr[i]<small_second){
small_second = arr[i];
}
}
//print smallest number from the array
System.out.println("Second smallest element in the array: " + small_second);
}
} Output:
Second smallest element in the array: 8
Python Program
Prerequisites:
- Python List (Python has a list that works similarly to the array from C/C++ and Java.)
- for-loop in Python
- if-else in Python
Python Code:
#initialize the Python list
int_arr = [45, 56, 83, 4, 11, 28, 36]
small_first = int_arr[0]
small_second = int_arr[0]
#initializ first and second smallest number
if int_arr[0] > int_arr[1]:
small_first = int_arr[1]
small_second = int_arr[0]
else:
small_first = int_arr[0]
small_second = int_arr[1]
#loop over all the number in the list
for num in int_arr[2:]:
#compare smallest number with current number
if num<=small_first:
small_second = small_first
small_first = num
elif num<small_second:
small_second = num
print(f"Second smallest number in the Python list: {small_second}")Output:
Second smallest number in the Python list: 11
Complexity
To find the second smallest number from the array, we are traversing the array only once. Each element in the array is visited once. So the time complexity of this algorithm is O(n). Here, n is the size of the array.
We are using extra space to store the first and second smallest number but it is constant (irrespective of the size of the array.)
What’s Next?
Write a program to find the second largest number in the array and share it with me in the comment.
