Print all increasing sequences of length k from first n natural numbers

Given two positive integers n and k, print all increasing sequences of length k such that the elements in every sequence are from the first n natural numbers.

Examples:

Input: k = 2, n = 3
Output: 1 2
        1 3
        2 3 

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

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

We strongly recommend minimizing the browser and trying this yourself first.
It’s a good recursion question. The idea is to create an array of lengths k. The array stores the current sequence. For every position in the array, we check the previous element and one by one put all elements greater than the previous element. If there is no previous element (first position), we put all numbers from 1 to n. 

Below is the implementation of the above idea : 

C++

// C++ program to  print all increasing sequences of
// length 'k' such that the elements in every sequence
// are from first 'n' natural numbers.
#include<iostream>
using namespace std;

// A utility function to print contents of arr[0..k-1]
void printArr(int arr[], int k)
{
    for (int i=0; i<k; i++)
        cout << arr[i] << " ";
    cout << endl;
}

// A recursive function to print all increasing sequences
// of first n natural numbers.  Every sequence should be
// length k. The array arr[] is used to store current
// sequence.
void printSeqUtil(int n, int k, int &len, int arr[])
{
    // If length of current increasing sequence becomes k,
    // print it
    if (len == k)
    {
        printArr(arr, k);
        return;
    }

    // Decide the starting number to put at current position:
    // If length is 0, then there are no previous elements
    // in arr[].  So start putting new numbers with 1.
    // If length is not 0, then start from value of
    // previous element plus 1.
    int i = (len == 0)? 1 : arr[len-1] + 1;

    // Increase length
    len++;

    // Put all numbers (which are greater than the previous
    // element) at new position.
    while (i<=n)
    {
        arr[len-1] = i;
        printSeqUtil(n, k, len, arr);
        i++;
    }

    // This is important. The variable 'len' is shared among
    // all function calls in recursion tree. Its value must be
    // brought back before next iteration of while loop
    len--;
}

// This function prints all increasing sequences of
// first n natural numbers. The length of every sequence
// must be k.  This function mainly uses printSeqUtil()
void printSeq(int n, int k)
{
    int arr[k];  // An array to store individual sequences
    int len = 0; // Initial length of current sequence
    printSeqUtil(n, k, len, arr);
}

// Driver program to test above functions
int main()
{
    int k = 3, n = 7;
    printSeq(n, k);
    return 0;
}

Java

// Java program to print all
// increasing sequences of
// length 'k' such that the
// elements in every sequence
// are from first 'n'
// natural numbers.

class GFG {

    // A utility function to print
    // contents of arr[0..k-1]
    static void printArr(int[] arr, int k)
    {
        for (int i = 0; i < k; i++)
            System.out.print(arr[i] + " ");
        System.out.print("\n");
    }

    // A recursive function to print
    // all increasing sequences
    // of first n natural numbers.
    // Every sequence should be
    // length k. The array arr[] is
    // used to store current sequence
    static void printSeqUtil(int n, int k,
                             int len, int[] arr)
    {

        // If length of current increasing
        // sequence becomes k, print it
        if (len == k)
        {
            printArr(arr, k);
            return;
        }

        // Decide the starting number
        // to put at current position:
        // If length is 0, then there
        // are no previous elements
        // in arr[]. So start putting
        // new numbers with 1.
        // If length is not 0,
        // then start from value of
        // previous element plus 1.
        int i = (len == 0) ? 1 : arr[len - 1] + 1;

        // Increase length
        len++;

        // Put all numbers (which are
        // greater than the previous
        // element) at new position.
        while (i <= n)
        {
            arr[len - 1] = i;
            printSeqUtil(n, k, len, arr);
            i++;
        }

        // This is important. The
        // variable 'len' is shared among
        // all function calls in recursion
        // tree. Its value must be
        // brought back before next
        // iteration of while loop
        len--;
    }

    // This function prints all
    // increasing sequences of
    // first n natural numbers.
    // The length of every sequence
    // must be k. This function
    // mainly uses printSeqUtil()
    static void printSeq(int n, int k)
    {

        // An array to store
        // individual sequences
        int[] arr = new int[k];

        // Initial length of
        // current sequence
        int len = 0;
        printSeqUtil(n, k, len, arr);
    }

    // Driver Code
    static public void main (String[] args)
    {
        int k = 3, n = 7;
        printSeq(n, k);
    }
}

// This code is contributed by Smitha.

Python3

# Python3 program to print all
# increasing sequences of length
# 'k' such that the elements in
# every sequence are from first
# 'n' natural numbers.

# A utility function to
# print contents of arr[0..k-1]
def printArr(arr, k):
    for i in range(k):
        print(arr[i], end = " ");
    print();

# A recursive function to print
# all increasing sequences of
# first n natural numbers. Every
# sequence should be length k.
# The array arr[] is used to
# store current sequence.
def printSeqUtil(n, k,len1, arr):

    # If length of current
    # increasing sequence
    # becomes k, print it
    if (len1 == k):
        printArr(arr, k);
        return;

    # Decide the starting number
    # to put at current position:
    # If length is 0, then there
    # are no previous elements
    # in arr[]. So start putting
    # new numbers with 1. If length
    # is not 0, then start from value
    # of previous element plus 1.
    i = 1 if(len1 == 0) else (arr[len1 - 1] + 1);

    # Increase length
    len1 += 1;

    # Put all numbers (which are greater
    # than the previous element) at
    # new position.
    while (i <= n):
        arr[len1 - 1] = i;
        printSeqUtil(n, k, len1, arr);
        i += 1;

    # This is important. The variable
    # 'len' is shared among all function
    # calls in recursion tree. Its value
    # must be brought back before next
    # iteration of while loop
    len1 -= 1;

# This function prints all increasing
# sequences of first n natural numbers.
# The length of every sequence must be
# k. This function mainly uses printSeqUtil()
def printSeq(n, k):
        arr = [0] * k; # An array to store
                       # individual sequences
        len1 = 0; # Initial length of
                  # current sequence
        printSeqUtil(n, k, len1, arr);

# Driver Code
k = 3;
n = 7;
printSeq(n, k);

# This code is contributed by mits

C#

// C# program to print all
// increasing sequences of
// length 'k' such that the
// elements in every sequence
// are from first 'n'
// natural numbers.
using System;

class GFG {

    // A utility function to print
    // contents of arr[0..k-1]
    static void printArr(int[] arr, int k)
    {
        for (int i = 0; i < k; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine();
    }

    // A recursive function to print
    // all increasing sequences
    // of first n natural numbers.
    // Every sequence should be
    // length k. The array arr[] is
    // used to store current sequence
    static void printSeqUtil(int n, int k,
                             int len, int[] arr)
    {

        // If length of current increasing
        // sequence becomes k, print it
        if (len == k)
        {
            printArr(arr, k);
            return;
        }

        // Decide the starting number
        // to put at current position:
        // If length is 0, then there
        // are no previous elements
        // in arr[]. So start putting
        // new numbers with 1.
        // If length is not 0,
        // then start from value of
        // previous element plus 1.
        int i = (len == 0) ? 1 : arr[len - 1] + 1;

        // Increase length
        len++;

        // Put all numbers (which are
        // greater than the previous
        // element) at new position.
        while (i <= n)
        {
            arr[len - 1] = i;
            printSeqUtil(n, k, len, arr);
            i++;
        }

        // This is important. The
        // variable 'len' is shared among
        // all function calls in recursion
        // tree. Its value must be
        // brought back before next
        // iteration of while loop
        len--;
    }

    // This function prints all
    // increasing sequences of
    // first n natural numbers.
    // The length of every sequence
    // must be k. This function
    // mainly uses printSeqUtil()
    static void printSeq(int n, int k)
    {

        // An array to store
        // individual sequences
        int[] arr = new int[k];

        // Initial length of
        // current sequence
        int len = 0;
        printSeqUtil(n, k, len, arr);
    }

    // Driver Code
    static public void Main ()
    {
        int k = 3, n = 7;
        printSeq(n, k);
    }
}

// This code is contributed by Ajit.

PHP

<?php
// PHP program to print all
// increasing sequences of
// length 'k' such that the
// elements in every sequence
// are from first 'n' natural
// numbers.

// A utility function to
// print contents of arr[0..k-1]
function printArr($arr, $k)
{
    for ($i = 0; $i < $k; $i++)
        echo $arr[$i], " ";
        echo "\n";
}

// A recursive function to
// print all increasing
// sequences of first n
// natural numbers. Every
// sequence should be length
// k. The array arr[] is used
// to store current sequence.
function printSeqUtil($n, $k,
                      $len, $arr)
{
    // If length of current
    // increasing sequence
    // becomes k, print it
    if ($len == $k)
    {
        printArr($arr, $k);
        return;
    }

    // Decide the starting number
    // to put at current position:
    // If length is 0, then there
    // are no previous elements
    // in arr[]. So start putting
    // new numbers with 1. If length
    // is not 0, then start from value
    // of previous element plus 1.
    $i = ($len == 0)? 1 :
          $arr[$len - 1] + 1;

    // Increase length
    $len++;

    // Put all numbers (which are
    // greater than the previous
    // element) at new position.
    while ($i <= $n)
    {
        $arr[$len-1] = $i;
        printSeqUtil($n, $k,
                     $len, $arr);
        $i++;
    }

    // This is important. The
    // variable 'len' is shared
    // among all function calls
    // in recursion tree. Its
    // value must be brought back
    // before next iteration of
    // while loop

    $len--;
}

// This function prints all
// increasing sequences of
// first n natural numbers.
// The length of every sequence
// must be k. This function
// mainly uses printSeqUtil()
function printSeq($n, $k)
{
        $arr = array(); // An array to store
                        // individual sequences
        $len = 0; // Initial length of
                  // current sequence
        printSeqUtil($n, $k,
                     $len, $arr);
}

// Driver Code
$k = 3;
$n = 7;
printSeq($n, $k);

// This code is contributed by Ajit
?>

Javascript

<script>

// Javascript program to print all
// increasing sequences of
// length 'k' such that the
// elements in every sequence
// are from first 'n'
// natural numbers.

// A utility function to print
// contents of arr[0..k-1]
function printArr(arr, k)
{
    for(let i = 0; i < k; i++)
        document.write(arr[i] + " ");

    document.write("</br>");
}

// A recursive function to print
// all increasing sequences
// of first n natural numbers.
// Every sequence should be
// length k. The array arr[] is
// used to store current sequence
function printSeqUtil(n, k, len, arr)
{

    // If length of current increasing
    // sequence becomes k, print it
    if (len == k)
    {
        printArr(arr, k);
        return;
    }

    // Decide the starting number
    // to put at current position:
    // If length is 0, then there
    // are no previous elements
    // in arr[]. So start putting
    // new numbers with 1.
    // If length is not 0,
    // then start from value of
    // previous element plus 1.
    let i = (len == 0) ? 1 : arr[len - 1] + 1;

    // Increase length
    len++;

    // Put all numbers (which are
    // greater than the previous
    // element) at new position.
    while (i <= n)
    {
        arr[len - 1] = i;
        printSeqUtil(n, k, len, arr);
        i++;
    }

    // This is important. The
    // variable 'len' is shared among
    // all function calls in recursion
    // tree. Its value must be
    // brought back before next
    // iteration of while loop
    len--;
}

// This function prints all
// increasing sequences of
// first n natural numbers.
// The length of every sequence
// must be k. This function
// mainly uses printSeqUtil()
function printSeq(n, k)
{

    // An array to store
    // individual sequences
    let arr = new Array(k);

    // Initial length of
    // current sequence
    let len = 0;
    printSeqUtil(n, k, len, arr);
}

// Driver code
let k = 3, n = 7;
printSeq(n, k);

// This code is contributed by divyesh072019

</script>

Output:

1 2 3
1 2 4
1 2 5
1 2 6
1 2 7
1 3 4
1 3 5
1 3 6
1 3 7
1 4 5
1 4 6
1 4 7
1 5 6
1 5 7
1 6 7
2 3 4
2 3 5
2 3 6
2 3 7
2 4 5
2 4 6
2 4 7
2 5 6
2 5 7
2 6 7
3 4 5
3 4 6
3 4 7
3 5 6
3 5 7
3 6 7
4 5 6
4 5 7
4 6 7
5 6 7

Time Complexity: O(n*n!), as there are n! permutations and it requires O(n) time to print a permutation. T(n) = n * T(n-1) + O(1) when reduced will result in O(n*n!) time.

Auxiliary Space: O(n)
This article is contributed by Arjun. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above