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Check if N indices of given Array can be colored by M colors using one color at most K times

  • Last Updated : 01 Feb, 2022

Find an arrangement of M colors for N indices such that no two adjacent indices have same color and each color can be used at most K times. If no such arrangement exists, output -1.

Examples:

Input: N = 6, M = 4, K = 2
Output: 1 2 3 4 1 2 
Explanation: Color 1 is assigned to the 1-st index. 
Color 2 is assigned to the 2-nd index, color 3 to the 3-rd index, 
color 4 to the 4-th index. 
Again, color 1 is assigned to 5-th index and color 2 to 6-th index.

Input: N = 20, M = 6, K = 3
Output: -1
Explanation: No such arrangement exists in which 6 colors may be assigned to at most 3 indices.

 

Approach: Observe the following points:

  • If the number of colors is only 1 and the number of indices is greater than 1, then no such assignment can exist.
  • If the total number of indices is greater than what can be colored by the colors available (M*K), then no such assignment exists.

Follow the steps below to solve the above problem:

  • If the number of colors is 1 and the number of indices is greater than 1, then output -1.
  • If the total number of colors is greater than what can be colored by the available colors (M*K), then output -1.
  • If both of the above conditions do not meet, then:
    • Initialize variable x with 1.
    • Run a loop till N and within the loop, output x. Increment x and check if x is greater than M. If x becomes greater than M, set x = 1.
    • When x is set to 1, the loop again starts printing the number of colors from 1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to find the required
// arrangement of colors to indices
void findAssignment(int N, int M, int K)
{
    // If number of colors is 1
    // and number of indices is more than 1
    if (M == 1 && N > 1) {
 
        // Output -1
        cout << -1;
    }
 
    // If number of indices is more than
    // what can be colored
    // by the available colors
    else if (N > M * K) {
 
        // Output -1
        cout << -1;
    }
    else {
 
        // Initialize x with 1
        int x = 1;
 
        // Loop to print
        // the required arrangement
        for (int i = 0; i < N; ++i) {
 
            // Output the number of colors
            cout << x << " ";
 
            // Increment the number of colors
            x++;
 
            // If x exceeds the total number
            // of available colors
            if (x > M) {
 
                // Set x to 1
                x = 1;
            }
        }
    }
}
 
// Driver Code
int main()
{
    // Given input
    int N = 6, M = 4, K = 2;
 
    // Function Call
    findAssignment(N, M, K);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the required
  // arrangement of colors to indices
  static void findAssignment(int N, int M, int K)
  {
 
    // If number of colors is 1
    // and number of indices is more than 1
    if (M == 1 && N > 1) {
 
      // Output -1
      System.out.print("-1");
    }
 
    // If number of indices is more than
    // what can be colored
    // by the available colors
    else if (N > M * K) {
 
      // Output -1
      System.out.print("-1");
    }
    else {
 
      // Initialize x with 1
      int x = 1;
 
      // Loop to print
      // the required arrangement
      for (int i = 0; i < N; ++i) {
 
        // Output the number of colors
        System.out.print(x + " ");
 
        // Increment the number of colors
        x++;
 
        // If x exceeds the total number
        // of available colors
        if (x > M) {
 
          // Set x to 1
          x = 1;
        }
      }
    }
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given input
    int N = 6, M = 4, K = 2;
 
    // Function Call
    findAssignment(N, M, K);
  }
}
 
// This code is contributed by hrithikgarg03188

Python3




# Python code for the above approach
 
# Function to find the required
# arrangement of colors to indices
def findAssignment(N, M, K):
 
    # If number of colors is 1
    # and number of indices is more than 1
    if (M == 1 and N > 1):
 
        # Output -1
        print(-1)
     
    # If number of indices is more than
    # what can be colored
    # by the available colors
    elif (N > M * K) :
 
        # Output -1
        print(-1)
     
    else:
 
        # Initialize x with 1
        x = 1;
 
        # Loop to print
        # the required arrangement
        for i in range(N):
 
            # Output the number of colors
            print(x, end= " ");
 
            # Increment the number of colors
            x += 1
 
            # If x exceeds the total number
            # of available colors
            if (x > M):
 
                # Set x to 1
                x = 1;
 
# Driver Code
 
# Given input
N = 6
M = 4
K = 2
 
# Function Call
findAssignment(N, M, K);
 
# This code is contributed by gfgking

C#




// C# program for the above approach
using System;
class GFG
{
   
// Function to find the required
// arrangement of colors to indices
static void findAssignment(int N, int M, int K)
{
   
    // If number of colors is 1
    // and number of indices is more than 1
    if (M == 1 && N > 1) {
 
        // Output -1
        Console.Write(-1);
    }
 
    // If number of indices is more than
    // what can be colored
    // by the available colors
    else if (N > M * K) {
 
        // Output -1
        Console.Write(-1);
    }
    else {
 
        // Initialize x with 1
        int x = 1;
 
        // Loop to print
        // the required arrangement
        for (int i = 0; i < N; ++i) {
 
            // Output the number of colors
            Console.Write(x + " ");
 
            // Increment the number of colors
            x++;
 
            // If x exceeds the total number
            // of available colors
            if (x > M) {
 
                // Set x to 1
                x = 1;
            }
        }
    }
}
 
// Driver Code
public static void Main()
{
   
    // Given input
    int N = 6, M = 4, K = 2;
 
    // Function Call
    findAssignment(N, M, K);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to find the required
       // arrangement of colors to indices
       function findAssignment(N, M, K)
       {
        
           // If number of colors is 1
           // and number of indices is more than 1
           if (M == 1 && N > 1) {
 
               // Output -1
               document.write(-1)
           }
 
           // If number of indices is more than
           // what can be colored
           // by the available colors
           else if (N > M * K) {
 
               // Output -1
               document.write(-1)
           }
           else {
 
               // Initialize x with 1
               let x = 1;
 
               // Loop to print
               // the required arrangement
               for (let i = 0; i < N; ++i) {
 
                   // Output the number of colors
                   document.write(x + " ");
 
                   // Increment the number of colors
                   x++;
 
                   // If x exceeds the total number
                   // of available colors
                   if (x > M) {
 
                       // Set x to 1
                       x = 1;
                   }
               }
           }
       }
 
       // Driver Code
 
       // Given input
       let N = 6, M = 4, K = 2;
 
       // Function Call
       findAssignment(N, M, K);
 
        // This code is contributed by Potta Lokesh
   </script>

 
 

Output
1 2 3 4 1 2 

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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