What is the time complexity of Minimum Possible Integer After at Most K Adjacent Swaps On Digits?

The optimal solution for Minimum Possible Integer After at Most K Adjacent Swaps On Digits runs in O(n log n) time and uses O(n) space. The time complexity is dominated by the heap operations, which take O(log n) time for each insertion and extraction, leading to O(n log n) overall.

What is the brute force solution for Minimum Possible Integer After at Most K Adjacent Swaps On Digits?

The brute force approach for Minimum Possible Integer After at Most K Adjacent Swaps On Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible configurations of the number by performing adjacent swaps up to k times. This method is straightforward but inefficient, as it explores many unnecessary permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Possible Integer After at Most K Adjacent Swaps On Digits?

Minimum Possible Integer After at Most K Adjacent Swaps On Digits uses the following concepts: String, Greedy, Binary Indexed Tree, Segment Tree. The recommended patterns to study are: Greedy Algorithm, Priority Queue, Sliding Window.

What are the interview tips for Minimum Possible Integer After at Most K Adjacent Swaps On Digits?

Always clarify the constraints and edge cases before diving into coding. Think about the greedy nature of the problem: how can you always make the best local choice? Practice explaining your thought process clearly as you code, as this demonstrates your understanding.

What are common mistakes when solving Minimum Possible Integer After at Most K Adjacent Swaps On Digits?

Failing to account for the remaining swaps after placing a digit. Not properly managing the indices when performing swaps, leading to incorrect results.

#1505

Minimum Possible Integer After at Most K Adjacent Swaps On Digits

Hard

StringGreedyBinary Indexed TreeSegment TreeGreedy AlgorithmPriority QueueSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses a greedy strategy with a priority queue (or min-heap) to efficiently find the smallest digit in the range allowed by k swaps. This reduces unnecessary checks and directly gives the best result.

⚙️

Algorithm

3 steps

1Step 1: Use a min-heap to keep track of the digits within the range of k swaps.
2Step 2: For each position, extract the smallest digit from the heap and place it in the current position.
3Step 3: Adjust the heap by adding the next digit from the string and removing the digit that has been placed.

solution.py18 lines

1import heapq
2
3def minInteger(num, k):
4    num = list(num)
5    n = len(num)
6    result = []
7    heap = []
8    for i in range(n):
9        heapq.heappush(heap, (num[i], i))
10        if len(heap) > k + 1:
11            heapq.heappop(heap)
12        if len(heap) > 0:
13            smallest, idx = heapq.heappop(heap)
14            result.append(smallest)
15            k -= (idx - i)
16            if k < 0:
17                break
18    return ''.join(result) + ''.join(num[i:] for i in range(len(result), n))

ℹ

Complexity note: The time complexity is dominated by the heap operations, which take O(log n) time for each insertion and extraction, leading to O(n log n) overall.

1The goal is to bring the smallest digits to the front, as they contribute most to minimizing the integer.
2Adjacent swaps mean that we can only move digits within a limited range, which requires careful management of positions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.