## Tuesday, May 6, 2014

### Problem

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

Example 1
```Input: k = 2, n = 3

Output:
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
```

Example 2
```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
```

### Solution

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.

```// 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;
}

return;
}

```

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.
```// 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);

return;
}
```

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

References