Given two integers N and K where K < N, the task is to generate a permutation of integers from 1 to N such that the absolute difference of all the consecutive integers give exactly K distinct integers.
Examples:
Input: N = 3, K = 2
Output: 1 3 2
|1 – 3| = 2 and |3 – 2| = 1 which gives 2 distinct integers (2 and 1)Input: N = 5, K = 4
Output: 1 5 2 4 3
|1 – 5| = 4, |5 – 2| = 3, |2 – 4| = 2 and |4 – 3| = 1 gives 4 distinct integers i.e. 4, 3, 2 and 1
Approach: The problem can be easily solved by simple observation. At the odd indices place increasing sequence 1, 2, 3, … and at the even indices place the decreasing sequence N, N-1, N-2, … and so on.
For N = 10, a permutation with distinct integers for consecutive absolute difference can be 1 10 2 9 3 8 4 7 5 6. The consecutive absolute difference gives integers 9, 8, 7 and so on.
So, first print K integers of such a sequence then make the rest of the differences equal to 1. The code is quite self explanatory.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to generate a permutation of integers// from 1 to N such that the absolute difference of// all the two consecutive integers give K distinct integersvoid printPermutation(int N, int K){ // To store the permutation vector<int> res; int l = 1, r = N, flag = 0; for (int i = 0; i < K; i++) { if (!flag) { // For sequence 1 2 3... res.push_back(l); l++; } else { // For sequence N, N-1, N-2... res.push_back(r); r--; } // Flag is used to alternate between // the above if else statements flag ^= 1; } // Taking integers with difference 1 // If last element added was r + 1 if (!flag) { for (int i = r; i >= l; i--) res.push_back(i); } // If last element added was l - 1 else { for (int i = l; i <= r; i++) res.push_back(i); } // Print the permutation for (auto i : res) cout << i << " ";}// Driver codeint main(){ int N = 10, K = 4; printPermutation(N, K); return 0;} |
Java
// Java implementation of the above approachimport java.util.Vector;class GFG { // Function to generate a permutation // of integers from 1 to N such that the // absolute difference of all the two // consecutive integers give K distinct integers static void printPermutation(int N, int K) { // To store the permutation Vector<Integer> res = new Vector<>(); int l = 1, r = N, flag = 0; for (int i = 0; i < K; i++) { if (flag == 0) { // For sequence 1 2 3... res.add(l); l++; } else { // For sequence N, N-1, N-2... res.add(r); r--; } // Flag is used to alternate between // the above if else statements flag ^= 1; } // Taking integers with difference 1 // If last element added was r + 1 if (flag != 1) { for (int i = r; i >= l; i--) { res.add(i); } } // If last element added was l - 1 else { for (int i = l; i <= r; i++) { res.add(i); } } // Print the permutation for (Integer i : res) { System.out.print(i + " "); } } // Driver code public static void main(String[] args) { int N = 10, K = 4; printPermutation(N, K); }}// This code is contributed by// 29AjayKumar |
Python3
# Python 3 implementation of the approach# Function to generate a permutation # of integers from 1 to N such that the # absolute difference of all the two # consecutive integers give K distinct# integersdef printPermutation(N, K): # To store the permutation res = list(); l, r, flag = 1, N, 0 for i in range(K): if flag == False: # For sequence 1 2 3... res.append(l) l += 1 else: # For sequence N, N-1, N-2... res.append(r); r -= 1; # Flag is used to alternate between # the above if else statements flag = flag ^ 1; # Taking integers with difference 1 # If last element added was r + 1 if flag == False: for i in range(r, 2, -1): res.append(i) # If last element added was l - 1 else: for i in range(l, r): res.append(i) # Print the permutation for i in res: print(i, end = " ")# Driver codeN, K = 10, 4printPermutation(N, K)# This code is contributed by# Mohit Kumar 29 |
C#
// C# implementation of the above approachusing System;using System.Collections;class GFG { // Function to generate a permutation // of integers from 1 to N such that the // absolute difference of all the two // consecutive integers give K distinct integers static void printPermutation(int N, int K) { // To store the permutation ArrayList res = new ArrayList(); int l = 1, r = N, flag = 0; for (int i = 0; i < K; i++) { if (flag == 0) { // For sequence 1 2 3... res.Add(l); l++; } else { // For sequence N, N-1, N-2... res.Add(r); r--; } // Flag is used to alternate between // the above if else statements flag ^= 1; } // Taking integers with difference 1 // If last element added was r + 1 if (flag != 1) { for (int i = r; i >= l; i--) { res.Add(i); } } // If last element added was l - 1 else { for (int i = l; i <= r; i++) { res.Add(i); } } // Print the permutation foreach (int i in res) { Console.Write(i + " "); } } // Driver code public static void Main() { int N = 10, K = 4; printPermutation(N, K); }}// This code is contributed by PrinciRaj1992 |
PHP
<?php// PHP implementation of the approach // Function to generate a permutation // of integers from 1 to N such that the // absolute difference of all the two // consecutive integers give K distinct // integers function printPermutation($N, $K) { // To store the permutation $res = array(); $l = 1; $r = $N; $flag = 0; for ($i = 0; $i < $K; $i++) { if (!$flag) { // For sequence 1 2 3... array_push($res, $l); $l++; } else { // For sequence N, N-1, N-2... array_push($res, $r); $r--; } // Flag is used to alternate between // the above if else statements $flag ^= 1; } // Taking integers with difference 1 // If last element added was r + 1 if (!$flag) { for ($i = $r; $i >= $l; $i--) array_push($res, $i); } // If last element added was l - 1 else { for ($i = l; $i <= $r; $i++) array_push($res, $i); } // Print the permutation for($i = 0; $i < sizeof($res); $i++) echo $res[$i], " ";} // Driver code $N = 10;$K = 4; printPermutation($N, $K); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the approach// Function to generate a permutation of // integers from 1 to N such that the // absolute difference of all the two // consecutive integers give K distinct integersfunction printPermutation(N, K){ // To store the permutation var res = []; var l = 1, r = N, flag = 0; for(var i = 0; i < K; i++) { if (!flag) { // For sequence 1 2 3... res.push(l); l++; } else { // For sequence N, N-1, N-2... res.push(r); r--; } // Flag is used to alternate between // the above if else statements flag ^= 1; } // Taking integers with difference 1 // If last element added was r + 1 if (!flag) { for(var i = r; i >= l; i--) res.push(i); } // If last element added was l - 1 else { for(var i = l; i <= r; i++) res.push(i); } // Print the permutation for(var i = 0; i< res.length; i++) { document.write(res[i] + " "); }}// Driver codevar N = 10, K = 4;printPermutation(N, K);// This code is contributed by noob2000</script> |
Output:
1 10 2 9 8 7 6 5 4 3
Time Complexity : O(K+N)
Space Complexity : O(N)
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