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

• Difficulty Level : Hard
• Last Updated : 24 Jun, 2022

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``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

## 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

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## Javascript

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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)