Permutation of Array such that sum of adjacent elements are not divisible by 3

Find elements of an array which are divisible by N using STL in C++

Last Updated : 18 Dec, 2019

Given an array and an integer N, find elements which are divisible by N, using STL in C++

Examples:

Input: a[] = {1, 2, 3, 4, 5, 10}, N = 2
Output: 3
Explanation:
As 2, 4, and 10 are divisible by 2
Therefore the output is 3

Input:a[] = {4, 3, 5, 9, 11}, N = 5
Output: 1
Explanation:
As only 5 is divisible by 5
Therefore the output is 3

Approach: This can be achieved using count_if() method in C++

Syntax:

count_if(lower_bound, upper_bound, function)

where function takes the element of given sequence one by one as a parameter and returns a boolean value on the basis of condition specified in that function.

In this case, the function will be:

bool isDiv(int i)
{
    if (i % N == 0)
        return true;
    else
        return false;
}

Below is the implementation of the above approach:

// C++ simple program to 
// find elements which are 
// divisible by N 
  
#include <bits/stdc++.h> 
using namespace std; 
  
int N; 
  
// Function to check 
// if the element is divisible by N 
bool isDiv(int i) 
{ 
    if (i % N == 0) 
        return true; 
    else
        return false; 
} 
  
// Driver code 
int main() 
{ 
    int a[] = { 1, 2, 6, 3, 4, 5 }; 
    N = 2; 
  
    int n = sizeof(a) / sizeof(a[0]); 
  
    int count = count_if(a, a + n, isDiv); 
  
    cout << "Elements divisible by "
         << N << ": " << count; 
  
    return 0; 
} 

Output:

Elements divisible by 2: 3

Permutation of Array such that sum of adjacent elements are not divisible by 3

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