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Number of ways to arrange N numbers which are in a range from 1 to K under given constraints.

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  • Last Updated : 20 Jun, 2022

Given Four integers N, K, P and Q. The task is to calculate the number of ways to arrange N numbers which are in a range from 1 to K such that the first number is P, the last number is Q and no two adjacent numbers are consecutive. 
Examples: 
 

Input:  N = 4, K = 3, P = 2, Q = 3 
Output: 3
Explanation:
For N=4, K=3, P=2, Q=3,
ways are [2, 1, 2, 3], [2, 3, 1, 3], [2, 3, 2, 3]

Input:  N = 5, K = 3, P = 2, Q = 1 
Output: 5

Approach: The idea is to use Dynamic Programming to solve this problem. 

  • Let’s try to understand this by taking an example, N = 4, K = 3, P = 2, Q = 1. 
    We will observe all possible arrangements starting from P and try to find any pattern that can be useful to apply Dynamic programming. 
     
  • Below is the image showing all possible arrangements starting from P = 2. 
     

  • Let A be the array that consists of the number of nodes ending at Q at a particular level 
    A = { 0, 1, 1, 3 }
    Let B be the array that consists of the number of nodes NOT ending at Q at a particular level 
    B = {1, 1, 3, 5 }
  • On carefull observation it may be noted that:
    1. A[i] = B[i-1] 
      Reason : 
      All the favourable nodes ( ending at Q ) will only be produced by non-favourable nodes(NOT ending at Q) of the previous level. 
       
    2. B[i] = A[i-1]*(K – 1) + B[i-1]*(K – 2) 
      Reason : 
      • For A[i-1]*(K – 1), some of the non-favourable nodes are produced by favourable nodes of the previous level, multiply by (K – 1) as each favourable node will produce K-1 non-favourable nodes 
         
      • For B[i-1]*(K – 2), rest of the non-favourable nodes are produced by non-favourable nodes of the previous level, multiply by (K-2), as one produced node is favourable, so we subtract 2 from this. 

C++




// C++ program to calculate Number of
// ways to arrange N numbers under
// given constraints.
#include <bits/stdc++.h>
using namespace std;
 
class element {
public:
    // For favourable nodes
    // (ending at Q)
    int A;
     
    // For Non-favourable nodes
    // (NOT ending at Q)
    int B;
};
 
// Function to print Total number
// of ways
void NumberOfWays(int n, int k, int p,
                                int q)
{
    element* dp = new element[n];
 
    // If the First number and the
    // last number is same.
    if (p == q) {
        dp[0].A = 1;
        dp[0].B = 0;
    }
    else
    {
        dp[0].A = 0;
        dp[0].B = 1;
    }
 
    // DP approach to find current state
    // with the help of previous state.
    for (int i = 1; i < n; i++)
    {
        dp[i].A = dp[i - 1].B;
        dp[i].B = (dp[i - 1].A * (k - 1))
                 + (dp[i - 1].B * (k - 2));
    }
     
    cout << dp[n - 1].A << endl;
 
    return;
}
 
// Driver code
int main()
{
    
   int N = 5;
   int K = 3;
   int P = 2;
   int Q = 1;
     
   // Function call
   NumberOfWays(N, K, P, Q);
}
    

Java




// Java program to calculate number of 
// ways to arrange N numbers under
// given constraints.
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to print Total number
// of ways
static void NumberOfWays(int n, int k,
                         int p, int q)
{
    int[][] dp = new int[n][2];
 
    // If the First number and the
    // last number is same.
    if (p == q)
    {
        dp[0][0] = 1;
        dp[0][1] = 0;
    }
    else
    {
        dp[0][0] = 0;
        dp[0][1] = 1;
    }
 
    // DP approach to find current state
    // with the help of previous state.
    for(int i = 1; i < n; i++)
    {
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = (dp[i - 1][0] * (k - 1)) +
                   (dp[i - 1][1] * (k - 2));
    }
    System.out.println(dp[n - 1][0]);
}
 
// Driver Code
public static void main(String args[])
{
    int N = 5;
    int K = 3;
    int P = 2;
    int Q = 1;
         
    // Function call
    NumberOfWays(N, K, P, Q);
}
}
 
// This code is contributed by offbeat

C#




// C# code to implement the approach
using System;
 
public class GFG {
 
  // Function to print Total number
  // of ways
  static void NumberOfWays(int n, int k,
                           int p, int q)
  {
    int[,] dp = new int[n, 2];
 
    // If the First number and the
    // last number is same.
    if (p == q)
    {
      dp[0, 0] = 1;
      dp[0, 1] = 0;
    }
    else
    {
      dp[0, 0] = 0;
      dp[0, 1] = 1;
    }
 
    // DP approach to find current state
    // with the help of previous state.
    for(int i = 1; i < n; i++)
    {
      dp[i, 0] = dp[i - 1, 1];
      dp[i, 1] = (dp[i - 1, 0] * (k - 1)) +
        (dp[i - 1, 1] * (k - 2));
    }
    Console.WriteLine(dp[n - 1, 0]);
  }
 
  // Driver code
  public static void Main(string []args)
  {
    int N = 5;
    int K = 3;
    int P = 2;
    int Q = 1;
 
    // Function call
    NumberOfWays(N, K, P, Q);
 
  }
}
 
// This code is contributed by sanjoy_62.

Output: 

5

 

Time Complexity: O(N).


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