Given a string num representing the digits of a very large integer and an integer k.

You are allowed to swap any two adjacent digits of the integer at most k times.

Return the minimum integer you can obtain also as a string.

Example 1:

Input: num = "4321", k = 4
Output: "1342"
Explanation: The steps to obtain the minimum integer from 4321 with 4 adjacent swaps are shown.


Example 2:

Input: num = "100", k = 1
Output: "010"
Explanation: It's ok for the output to have leading zeros, but the input is guaranteed not to have any leading zeros.


Example 3:

Input: num = "36789", k = 1000
Output: "36789"
Explanation: We can keep the number without any swaps.


Example 4:

Input: num = "22", k = 22
Output: "22"


Example 5:

Input: num = "9438957234785635408", k = 23
Output: "0345989723478563548"


Constraints:

• 1 <= num.length <= 30000
• num contains digits only and doesn’t have leading zeros.
• 1 <= k <= 10^9

## Solution: Greedy + Recursion(Update: TLE after 7/6/2020)

Move the smallest number to the left and recursion on the right substring with length equals to n -= 1.

4321 k = 4 => 1 + solve(432, 4-3) = 1 + solve(432, 1) = 1 + 3 + solve(42, 0) = 1 + 3 + 42 = 1342.

Time complexity: O(n^2)
Space complexity: O(1)

## Solution 2: Binary Indexed Tree / Fenwick Tree

Moving elements in a string is a very expensive operation, basically O(n) per op. Actually, we don’t need to move the elements physically, instead we track how many elements before i has been moved to the “front”. Thus we know the cost to move the i-th element to the “front”, which is i – elements_moved_before_i or prefix_sum(0~i-1) if we mark moved element as 1.

We know BIT / Fenwick Tree is good for dynamic prefix sum computation which helps to reduce the time complexity to O(nlogn).

Time complexity: O(nlogn)
Space complexity: O(n)

## Python3

If you like my articles / videos, donations are welcome.

Buy anything from Amazon to support our website

Paypal
Venmo
huahualeetcode

## Be First to Comment

Mission News Theme by Compete Themes.