Print all sequences of given length

Given two integers k and n, write a function that prints all the sequences of length k composed of numbers 1,2..n. You need to print these sequences in sorted order.

Examples:

Input: k = 2, n = 3

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

Output:
1 1 1
1 1 2
1 1 3
1 1 4
1 2 1
.....
.....
4 3 4
4 4 1
4 4 2
4 4 3
4 4 4

Method 1 (Simple and Iterative):
The simple idea to print all sequences in sorted order is to start from {1 1 … 1} and keep incrementing the sequence while the sequence doesn’t become {n n … n}. Following is the detailed process.

1) Create an output array arr[] of size k. Initialize the array as {1, 1…1}.
2) Print the array arr[].
3) Update the array arr[] so that it becomes immediate successor (to be printed) of itself. For example, immediate successor of {1, 1, 1} is {1, 1, 2}, immediate successor of {1, 4, 4} is {2, 1, 1} and immediate successor of {4, 4, 3} is {4 4 4}.
4) Repeat steps 2 and 3 while there is a successor array. In other words, while the output array arr[] doesn’t become {n, n .. n}

Now let us talk about how to modify the array so that it contains immediate successor. By definition, the immediate successor should have the same first p terms and larger (p+l)th term. In the original array, (p+1)th term (or arr[p]) is smaller than n and all terms after arr[p] i.e., arr[p+1], arr[p+2], … arr[k-1] are n.
To find the immediate successor, we find the point p in the previously printed array. To find point p, we start from rightmost side and keep moving till we find a number arr[p] such that arr[p] is smaller than n (or not n). Once we find such a point, we increment it arr[p] by 1 and make all the elements after arr[p] as 1 i.e., we do arr[p+1] = 1, arr[p+2] = 1 .. arr[k-1] = 1. Following are the detailed steps to get immediate successor of arr[]

1) Start from the rightmost term arr[k-1] and move toward left. Find the first element arr[p] that is not same as n.
2) Increment arr[p] by 1
3) Starting from arr[p+1] to arr[k-1], set the value of all terms as 1.

 // CPP program of above approach #include    /* A utility function that prints a given arr[] of length size*/ void printArray(int arr[], int size) {     for(int i = 0; i < size; i++)         printf("%d ", arr[i]);     printf("\n");     return; }    /* This function returns 0 if there are no more sequences to be printed, otherwise    modifies arr[] so that arr[] contains next sequence to be printed */ int getSuccessor(int arr[], int k, int n) {     /* start from the rightmost side and find the first number less than n */     int p = k - 1;     while (arr[p] == n)         p--;        /* If all numbers are n in the array then there is no successor, return 0 */     if (p < 0)         return 0;        /* Update arr[] so that it contains successor */     arr[p] = arr[p] + 1;     for(int i = p + 1; i < k; i++)         arr[i] = 1;        return 1; }    /* The main function that prints all sequences from 1, 1, ..1 to n, n, ..n */ void printSequences(int n, int k) {     int *arr = new int[k];        /* Initialize the current sequence as the first sequence to be printed */     for(int i = 0; i < k; i++)         arr[i] = 1;        /* The loop breaks when there are no more successors to be printed */     while(1)     {         /* Print the current sequence */         printArray(arr, k);            /* Update arr[] so that it contains next sequence to be printed. And if            there are no more sequences then break the loop */         if(getSuccessor(arr, k, n) == 0)           break;     }        delete(arr); // free dynamically allocated array     return; }    /* Driver Program to test above functions */ int main() {     int n = 3;     int k = 2;     printSequences(n, k);     return 0; }

Output:

1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3

Time Complexity: There are total n^k sequences. Printing a sequence and finding its successor take O(k) time. So time complexity of above implementation is O(k*n^k).

Method 2 (Tricky and Recursive)
The recursive function printSequencesRecur generates and prints all sequences of length k. The idea is to use one more parameter index. The function printSequencesRecur keeps all the terms in arr[] sane till index, update the value at index and recursively calls itself for more terms after index.

C++

 // C++ program of above approach #include    /* A utility function that prints a given arr[] of length size*/ void printArray(int arr[], int size) {     for(int i = 0; i < size; i++)         printf("%d ", arr[i]);     printf("\n");     return; }    /* The core function that recursively generates and prints all sequences of    length k */ void printSequencesRecur(int arr[], int n, int k, int index) {    int i;    if (k == 0)    {      printArray(arr, index);    }    if (k > 0)    {       for(i = 1; i<=n; ++i)       {         arr[index] = i;         printSequencesRecur(arr, n, k-1, index+1);       }    } }       /* A function that uses printSequencesRecur() to prints all sequences     from 1, 1, ..1 to n, n, ..n */ void printSequences(int n, int k) {     int *arr = new int[k];     printSequencesRecur(arr, n, k, 0);        delete(arr); // free dynamically allocated array     return; }    /* Driver Program to test above functions */ int main() {     int n = 3;     int k = 2;     printSequences(n, k);     return 0; }

Java

 // Java program of above approach    class GfG {    /* A utility function that prints a given arr[] of length size*/ static void printArray(int arr[], int size)  {      for(int i = 0; i < size; i++)          System.out.print(arr[i] + " ");  System.out.println();     return;  }     /* The core function that recursively generates and prints all sequences of  length k */ static void printSequencesRecur(int arr[], int n, int k, int index)  {  int i;  if (k == 0)  {      printArray(arr, index);  }  if (k > 0)  {      for(i = 1; i<=n; ++i)      {          arr[index] = i;          printSequencesRecur(arr, n, k-1, index+1);      }  }  }        /* A function that uses printSequencesRecur() to prints all sequences  from 1, 1, ..1 to n, n, ..n */ static void printSequences(int n, int k)  {      int arr[] = new int[k];      printSequencesRecur(arr, n, k, 0);     return ; }     /* Driver Program to test above functions */ public static void main(String[] args)  {      int n = 3;      int k = 2;      printSequences(n, k);  }  }

Python3

 # Python3 program of above approach    # A utility function that prints a  # given arr[] of length size def printArray(arr, size):        for i in range(size):          print(arr[i], end = " ");      print("");     return;     # The core function that recursively  # generates and prints all sequences  # of length k  def printSequencesRecur(arr, n, k, index):     if (k == 0):         printArray(arr, index);         if (k > 0):         for i in range(1, n + 1):              arr[index] = i;              printSequencesRecur(arr, n, k - 1,                                     index + 1);     # A function that uses printSequencesRecur() to  # prints all sequences from 1, 1, ..1 to n, n, ..n def printSequences(n, k):      arr =  * n;      printSequencesRecur(arr, n, k, 0);         return;    # Driver Code n = 3;  k = 2;  printSequences(n, k);     # This code is contributed mits

C#

 // C# program of above approach using System;    class GFG  {    /* A utility function that prints a given arr[] of length size*/ static void printArray(int []arr, int size)  {      for(int i = 0; i < size; i++)          Console.Write(arr[i] + " ");      Console.WriteLine();     return;  }     /* The core function that recursively generates  and prints all sequences of length k */ static void printSequencesRecur(int []arr, int n,                                  int k, int index)  {      int i;      if (k == 0)      {          printArray(arr, index);      }      if (k > 0)      {          for(i = 1; i<=n; ++i)          {              arr[index] = i;              printSequencesRecur(arr, n, k - 1, index + 1);          }      }  }        /* A function that uses printSequencesRecur() to  prints all sequences from 1, 1, ..1 to n, n, ..n */ static void printSequences(int n, int k)  {      int[] arr = new int[k];      printSequencesRecur(arr, n, k, 0);         return; }     // Driver Code public static void Main()  {      int n = 3;      int k = 2;      printSequences(n, k);  }  }     // This code is contributed // by Akanksha Rai

PHP

 0)      {          for(\$i = 1; \$i <= \$n; ++\$i)          {              \$arr[\$index] = \$i;              printSequencesRecur(\$arr, \$n,                                  \$k - 1, \$index + 1);          }      }  }        /* A function that uses printSequencesRecur() to  prints all sequences from 1, 1, ..1 to n, n, ..n */ function printSequences(\$n, \$k)  {      \$arr = array();      printSequencesRecur(\$arr, \$n, \$k, 0);         return; }     // Driver Code \$n = 3;  \$k = 2;  printSequences(\$n, \$k);     // This code is contributed // by Akanksha Rai ?>

Output:

1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3

Time Complexity: There are n^k sequences and printing a sequence takes O(k) time. So time complexity is O(k*n^k).

Thanks to mopurizwarriors for suggesting above method. As suggested by alphayoung, we can avoid use of arrays for small sequences that can fit in an integer. Following is the implementation for same.

C

 // C program of above approach /* The core function that generates and prints all sequences of length k */ void printSeqRecur(int num, int pos, int k, int n) {     if (pos == k) {         printf("%d \n", num);         return;     }     for (int i = 1; i <= n; i++) {         printSeqRecur(num * 10 + i, pos + 1, k, n);     } }    /* A function that uses printSequencesRecur() to prints all sequences    from 1, 1, ..1 to n, n, ..n */ void printSequences(int k, int n) {     printSeqRecur(0, 0, k, n); }

Java

 // Java program of above approach    /* The core function that generates and prints all sequences of length k */ static void printSeqRecur(int num, int pos, int k, int n)  {      if (pos == k) {          System.out.print(num + " ");          return;      }      for (int i = 1; i <= n; i++) {          printSeqRecur(num * 10 + i, pos + 1, k, n);      }  }     /* A function that uses printSequencesRecur() to prints all sequences  from 1, 1, ..1 to n, n, ..n */ static void printSequences(int k, int n)  {      printSeqRecur(0, 0, k, n);  }

Python3

 # Python program of above approach # We have used number instead of  # arrays to prevent linear time  # required to print each string def printSeqRecur ( num , n, k ):     if n == 0: # if total digits become equal                 # to n, print the number and return         print(num )         return                for _ in range(1, k + 1):         printSeqRecur (num * 10 + _, n - 1, k)     # Driver Code if __name__ == "__main__":     k = 3 # length of k-ary string     n = 2 # string can take values           # from 1,2,3...n     printSeqRecur(0, n, k)    # This code is contributed  # by shivam purohit

C#

 // C# program of above approach    /* The core function that generates  and prints all sequences of length k */ static void printSeqRecur(int num, int pos,                             int k, int n)  {      if (pos == k)      {          Console.Write(num + " ");          return;      }      for (int i = 1; i <= n; i++)      {          printSeqRecur(num * 10 + i, pos + 1, k, n);      }  }     /* A function that uses printSequencesRecur() to prints all sequences from 1, 1, ..1 to n, n, ..n */ static void printSequences(int k, int n)  {      printSeqRecur(0, 0, k, n);  }    // This code is contributed by Code_Mech

PHP



References:
Algorithms and Programming: Problems and Solutions by Alexander Shen

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