C++ Exercises: Check if the number of 3's is greater than the number of 5's
C++ Basic Algorithm: Exercise-105 with Solution
Write a C++ program to check if the number of 3's is greater than the number of 5's.
Sample Solution:
C++ Code :
#include <iostream>
using namespace std;
static bool test(int nums[], int arr_length)
{
int no_3 = 0, no_5 = 0;
for (int i = 0; i < arr_length; i++)
{
if (nums[i] == 3) no_3++;
if (nums[i] == 5) no_5++;
}
return no_3 > no_5;
}
int main()
{
int nums1[] = {1, 5, 6, 9, 3, 3};
int arr_length = sizeof(nums1) / sizeof(nums1[0]);
cout << test(nums1, arr_length) << endl;
int nums2[] = {1, 5, 5, 5, 10, 17};
arr_length = sizeof(nums2) / sizeof(nums2[0]);
cout << test(nums2, arr_length) << endl;
int nums3[] = {1, 3, 3, 5, 5, 5};
arr_length = sizeof(nums3) / sizeof(nums3[0]);
cout << test(nums3, arr_length) << endl;
return 0;
}
Sample Output:
1 0 0
Pictorial Presentation:

Flowchart:

C++ Code Editor:
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Previous: Write a C++ program to check if the sum of all 5' in the array exactly 15 in a given array of integers.
Next: Write a C++ program to check if a given array of integers contains a 3 or a 5.
What is the difficulty level of this exercise?
C++ Programming: Tips of the Day
Advantage of switch over if-else statement
Use switch.
In the worst case the compiler will generate the same code as a if-else chain, so you don't lose anything. If in doubt put the most common cases first into the switch statement.
In the best case the optimizer may find a better way to generate the code. Common things a compiler does is to build a binary decision tree (saves compares and jumps in the average case) or simply build a jump-table (works without compares at all).
Ref: https://bit.ly/3G0uBqS
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