﻿ C : Find a 3 with a 5 somewhere in an array of integers

# C Exercises: Check a given array of integers and return true if there is a 3 with a 5 somewhere later in the given array

## C-programming basic algorithm: Exercise-62 with Solution

Write a C program to check a given array of integers and return true if there is a 3 with a 5 somewhere later in the given array.

C Code:

``````#include <stdio.h>
#include <stdlib.h>

// Function prototype for 'test'
int test(int numbers[], int arr_size);

int main(void){
int arr_size;

// Declaration and initialization of an integer array 'array1'
int array1[] = {1, 2, 3, 4};
arr_size = sizeof(array1)/sizeof(array1[0]);

// Printing the result of the 'test' function for 'array1'
printf("%d",test(array1, arr_size));

// Declaration and initialization of an integer array 'array2'
int array2[] = {3, 3, 5, 5, 5, 5};
arr_size = sizeof(array2)/sizeof(array2[0]);

// Printing the result of the 'test' function for 'array2'
printf("\n%d",test(array2, arr_size));

// Declaration and initialization of an integer array 'array3'
int array3[] = {2, 5, 5, 7, 8, 10};
arr_size = sizeof(array3)/sizeof(array3[0]);

// Printing the result of the 'test' function for 'array3'
printf("\n%d",test(array3, arr_size));
}

// Definition of the 'test' function
int test(int numbers[], int arr_size)
{
int three = 0;

// Looping through the elements of the array
for (int i = 0; i < arr_size; i++)
{
// Checking if 'three' is true and the current element is 5, then return 1
if (three && numbers[i] == 5)
return 1;

// Checking if the current element is 3, then set 'three' to true
if (numbers[i] == 3)
three = 1;
}

// If no consecutive occurrence of 3 followed by 5 were found
return 0;
}
``````

Sample Output:

```0
1
0
```

Pictorial Presentation:

Flowchart:

C Programming Code Editor:

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