Maximum number possible by doing at - most K swaps

Table of Contents#

1. Problem Statement
2. Approach and Algorithm
3. Example Walkthrough
4. Code Implementation
5. Complexity Analysis
6. Best Practices and Common Pitfalls
7. Conclusion
8. References

1. Problem Statement#

We are given a number, typically represented as a string of digits, and an integer K. Our task is to find the maximum number that can be obtained by performing at most K swaps on the digits of the given number.

For example, if the given number is "123" and K = 1, the possible swaps are:

• Swap 1 and 2: We get "213"
• Swap 1 and 3: We get "321"
• Swap 2 and 3: We get "132"

The maximum number among these is "321".

2. Approach and Algorithm#

We can use a back - tracking approach to solve this problem. The basic idea is to try all possible swaps at each step and keep track of the maximum number obtained so far.

Steps:#

1. Initialization: Convert the given number into a list of digits for easier manipulation.
2. Back - tracking:
   • Start with the current number.
   • For each position i in the number, try swapping it with all positions j > i.
   • After each swap, recursively call the function with K - 1 swaps remaining.
   • Keep track of the maximum number obtained during the process.
3. Termination:
   • If K becomes 0 or we have tried all possible swaps, stop the recursion.

3. Example Walkthrough#

Let's take the number "123" and K = 2.

Step 1:#

We start with the number "123".

• Swap 1 and 2: We get "213". Now we have 1 swap left.
  • From "213", we can swap 1 and 3 to get "231".
• Swap 1 and 3: We get "321". Now we have 1 swap left.
  • From "321", we can swap 2 and 1 to get "312".
• Swap 2 and 3: We get "132". Now we have 1 swap left.
  • From "132", we can swap 1 and 3 to get "312".

By comparing all the numbers obtained, we find that the maximum number is "321".

4. Code Implementation#

def find_max_number(num_str, k):
    num_list = list(num_str)
    max_num = num_str
 
    def backtrack(index, swaps):
        nonlocal max_num
        if swaps == 0:
            return
        for i in range(index, len(num_list)):
            for j in range(i + 1, len(num_list)):
                num_list[i], num_list[j] = num_list[j], num_list[i]
                new_num = ''.join(num_list)
                if new_num > max_num:
                    max_num = new_num
                backtrack(i, swaps - 1)
                num_list[i], num_list[j] = num_list[j], num_list[i]
 
    backtrack(0, k)
    return max_num
 
 
num = "123"
k = 2
print(find_max_number(num, k))

5. Complexity Analysis#

• Time Complexity: The time complexity of this algorithm is $O(n!)$ in the worst case, where $n$ is the number of digits in the given number. This is because in the worst case, we are exploring all possible permutations of the digits.
• Space Complexity: The space complexity is $O(n)$ due to the recursive call stack, where $n$ is the number of digits.

6. Best Practices and Common Pitfalls#

Best Practices#

• Early Termination: If at any point, we can determine that further swaps will not lead to a larger number, we can stop the recursion early.
• Use of Helper Functions: Breaking the code into smaller functions can make the code more modular and easier to understand.

Common Pitfalls#

• Not Backtracking Properly: After making a swap, we need to undo the swap to explore other possibilities. Failing to do so can lead to incorrect results.
• Inefficient Memory Usage: If the number of digits is large, the recursive approach can consume a significant amount of memory.

7. Conclusion#

The problem of finding the maximum number possible by doing at most K swaps is a challenging problem that can be solved using a back - tracking approach. By exploring all possible swaps and keeping track of the maximum number, we can find the solution. However, the time complexity of the algorithm is high, and for large numbers, more efficient algorithms may be required.

8. References#

• GeeksforGeeks: https://www.geeksforgeeks.org/find-maximum-number-possible-by-doing-at-most-k-swaps/
• LeetCode discussions on similar problems.