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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:

C++ Basic Algorithm Exercises: Check if the number of 3's is greater than the number of 5's.

Flowchart:

Flowchart: Check if the number of 3's is greater than the number of 5's.

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.

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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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