Find the first, second and third minimum elements in an array in C++ program

In this article, we need to identify the three minimum elements in an array: minimum, second minimum, and third minimum. We will look at two different approaches to solve this problem and compare their time complexity.

Problem Statement

Given an array of n elements, we need to find the first, second, and third minimum elements in the array.

The first minimum is the minimum of the array's elements, the second is a minimum but larger than the first, and the third is a minimum but larger than the second.

Let us see the following example scenario to understand the problem better:

Scenario 1

Input: arr = [10, 8, 7, 4, 2]
Output: 
1st min = 2
2nd min = 4
3rd min = 7

Scenario 2

Input: arr = [4, 9, 1, 32, 12]
Output:
1st min = 1
2nd min = 4
3rd min = 9

Find 1^st, 2^nd, and 3^rd Minimum: Naive Approach

In this approach, we first sort the array and then print the first three elements:

Example

In the following C++ example, we print the first three minimum elements using the naive approach:

#include <bits/stdc++.h>
using namespace std;
void ThreeMinElement(int array[], int n) {
   // Sorting the array
   sort(array, array + n);

   if (n < 3) {
      cout << "Array has less than 3 elements.\n";
      return;
   }

   cout << "1st min = " << array[0] << "\n";
   cout << "2nd min = " << array[1] << "\n";
   cout << "3rd min = " << array[2] << "\n";
}
int main() {
   int array[] = {4, 9, 1, 32, 12};
   int n = sizeof(array) / sizeof(array[0]);
   ThreeMinElement(array, n);
   return 0;
}

Following is the output of the code:

1st min = 1
2nd min = 4
3rd min = 9

Find 1st, 2nd, and 3rd Minimum: Linear Approach

We traverse the array only once, and at each step we update the three minimums, without sorting the array. Let's see the algorithm/pseudo code:

• Initialize: first = INT_MAX, second = INT_MAX, third = INT_MAX
• Traverse each element: If arr[i] < first
• third = second, second = first, first = arr[i]
• Else if arr[i] < second
• third = second, second = arr[i]
• Else if arr[i] < third
• third = arr[i]
• Print first, second, and third

Example

This is another example of the C++ to print the first, second, and third minimum elements in an array:

#include <bits/stdc++.h>
using namespace std;

void threeMinimumElement(int array[], int n) {
   if (n < 3) {
      cout << "Array must have at least 3 elements.\n";
      return;
   }
   int first = INT_MAX, second = INT_MAX, third = INT_MAX; 
   for (int i = 0; i < n; i++) {
      if (array[i] < first) {
         third = second;
         second = first;
         first = array[i];
      }
      else if (array[i] < second) {
         third = second;
         second = array[i];
      }
      else if (array[i] < third) {
         third = array[i];
      }
   }
   cout << "First min = " << first << "\n";
   cout << "Second min = " << second << "\n";
   cout << "Third min = " << third << "\n";
}

int main() {
   int array[] = {10, 8, 7, 4, 2};
   int n = sizeof(array) / sizeof(array[0]);
   threeMinimumElement(array, n);
   return 0;
}

Following is the output of the code:

First min = 2
Second min = 4
Third min = 7