1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number
Description
You are given a string num
, representing a large integer, and an integer k
.
We call some integer wonderful if it is a permutation of the digits in num
and is greater in value than num
. There can be many wonderful integers. However, we only care about the smallest-valued ones.
- For example, when
num = "5489355142"
:- The 1st smallest wonderful integer is
"5489355214"
. - The 2nd smallest wonderful integer is
"5489355241"
. - The 3rd smallest wonderful integer is
"5489355412"
. - The 4th smallest wonderful integer is
"5489355421"
.
- The 1st smallest wonderful integer is
Return the minimum number of adjacent digit swaps that needs to be applied to num
to reach the kth
smallest wonderful integer.
The tests are generated in such a way that kth
smallest wonderful integer exists.
Example 1:
Input: num = "5489355142", k = 4 Output: 2 Explanation: The 4th smallest wonderful number is "5489355421". To get this number: - Swap index 7 with index 8: "5489355142" -> "5489355412" - Swap index 8 with index 9: "5489355412" -> "5489355421"
Example 2:
Input: num = "11112", k = 4 Output: 4 Explanation: The 4th smallest wonderful number is "21111". To get this number: - Swap index 3 with index 4: "11112" -> "11121" - Swap index 2 with index 3: "11121" -> "11211" - Swap index 1 with index 2: "11211" -> "12111" - Swap index 0 with index 1: "12111" -> "21111"
Example 3:
Input: num = "00123", k = 1 Output: 1 Explanation: The 1st smallest wonderful number is "00132". To get this number: - Swap index 3 with index 4: "00123" -> "00132"
Constraints:
2 <= num.length <= 1000
1 <= k <= 1000
num
only consists of digits.
Solutions
Solution 1: Find Next Permutation + Inversion Pairs
We can call the next_permutation
function $k$ times to get the $k$th smallest permutation $s$.
Next, we just need to calculate how many swaps are needed for $num$ to become $s$.
Let's first consider a simple situation where all the digits in $num$ are different. In this case, we can directly map the digit characters in $num$ to indices. For example, if $num$ is "54893"
and $s$ is "98345"
. We map each digit in $num$ to an index, that is:
$$ \begin{aligned} num[0] &= 5 &\rightarrow& \quad 0 \ num[1] &= 4 &\rightarrow& \quad 1 \ num[2] &= 8 &\rightarrow& \quad 2 \ num[3] &= 9 &\rightarrow& \quad 3 \ num[4] &= 3 &\rightarrow& \quad 4 \ \end{aligned} $$
Then, mapping each digit in $s$ to an index results in "32410"
. In this way, the number of swaps needed to change $num$ to $s$ is equal to the number of inversion pairs in the index array after $s$ is mapped.
If there are identical digits in $num$, we can use an array $d$ to record the indices where each digit appears, where $d[i]$ represents the list of indices where the digit $i$ appears. To minimize the number of swaps, when mapping $s$ to an index array, we only need to greedily select the index of the corresponding digit in $d$ in order.
Finally, we can directly use a double loop to calculate the number of inversion pairs, or we can optimize it with a Binary Indexed Tree.
The time complexity is $O(n \times (k + n))$, and the space complexity is $O(n)$. Where $n$ is the length of $num$.
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