C Exercises: Check if a triple is presents in an array of integers or not

C-programming basic algorithm: Exercise-16 with Solution

Write a C program to check if a triple is present in an array of integers or not. If a value appears three times in a row in an array it is called a triple.

C Code:

#include <stdio.h> // Include standard input/output library
#include <stdlib.h> // Include standard library for additional functions

// Function declaration for 'test' with an array of integers and its size as parameters
int test(int nums[], int arr_size);

int main(void){
    int arr_size; // Declare an integer variable to store the size of the array

    // Initialize an integer array 'array1' with given values
    int array1[] = {1, 1, 2, 2, 1};
    arr_size = sizeof(array1)/sizeof(array1[0]); // Calculate the size of 'array1'
    printf("%d",test(array1, arr_size)); // Call the 'test' function with 'array1' and its size, and print the result

    // Initialize another integer array 'array2' with given values
    int array2[] = {1, 1, 2, 1, 2, 3};
    arr_size = sizeof(array2)/sizeof(array2[0]); // Calculate the size of 'array2'
    printf("\n%d",test(array2, arr_size)); // Call the 'test' function with 'array2' and its size, and print the result

    // Initialize yet another integer array 'array3' with given values
    int array3[] = {1, 1, 1, 2, 2, 2, 1 };
    arr_size = sizeof(array3)/sizeof(array3[0]); // Calculate the size of 'array3'
    printf("\n%d",test(array3, arr_size)); // Call the 'test' function with 'array3' and its size, and print the result
}

// Function definition for 'test'
int test(int nums[], int arr_size)
{
    // Iterate through the elements of the array 'nums'
    for (int i = 0; i < arr_size - 1; i++)
    {
        // Check if the current element is equal to the next two consecutive elements
        if(nums[i] == nums[i + 1] && nums[i + 2] == nums[i])
            return 1; // If the condition is met, return 1 (true)
    }
    return 0; // If no match is found, return 0 (false)
}

Sample Output:

0
0
1

Explanation:

int test(int nums[], int arr_size) {
  for (int i = 0; i < arr_size - 1; i++) {
    if (nums[i] == nums[i + 1] && nums[i + 2] == nums[i])
      return 1;
  }
  return 0;
}

The above function takes an array of integers and its size as input. It checks if the array contains any sequence of three integers where the first and third integers are the same. If such a sequence is found, the function returns 1, otherwise, it returns 0.

Time complexity and space complexity:

The time complexity of the given function is O(n), where n is the size of the input array. This is because the function iterates over the array once to check for the required sequence of integers.

The space complexity of the given function is O(1), as it uses a constant amount of additional space to store the loop variable and the result variable.

Pictorial Presentation:

Flowchart:

C Programming Code Editor:

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