Print all sequences starting with n and consecutive difference limited to k in C++

C++Server Side ProgrammingProgramming

In this problem, we are given three variables n, s, and k and we have to print all the possible sequences that start with the number n and length s having the absolute difference between consecutive elements less than k.

Let’s take an example to understand the topic better −

Input: n = 3, s = 3 , k = 2
Output:
3 3 3
3 3 4
3 3 2
3 4 4
3 4 5
3 4 3
3 2 2
3 2 3
3 2 1

In this problem, we need to obtain the absolute difference less k. For this, we can get a sequence that has elements that are greater to obtain positive difference and lesser to obtain a negative difference.

For this, we will start with n and the make a recursive call to the elements at each consecutive position. A loop from 0 to k-1 and add it to the number to the number. Similarly going for the negative side also.

Example

#include <bits/stdc++.h>
using namespace std;
void printConsicutiveNumbers(vector& v, int n, int s, int k){
   if (s == 0) {
      for (int i = 0; i < v.size(); i++)
         cout<<v[i]<<" ";
      cout << endl;
      return;
   }
   for (int i = 0; i < k; i++) {
      v.push_back(n + i);
      printConsicutiveNumbers(v, n + i, s - 1, k);
      v.pop_back();
   }
   for (int i = 1; i < k; i++) {
      v.push_back(n - i);
      printConsicutiveNumbers(v, n - i, s - 1, k);
      v.pop_back();
   }
}
int main(){
   int n = 3, s = 3, k = 2;
   cout<<"The sequence is :\n";
   vector<int> v;
   v.push_back(n);
   printConsicutiveNumbers(v, n, s - 1, k);
   return 0;
}

Output

The sequence is −

3 3 3
3 3 4
3 3 2
3 4 4
3 4 5
3 4 3
3 2 2
3 2 3
3 2 1
raja
Published on 17-Jan-2020 10:58:13
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