reverse paris Archives - Huahua's Tech Road

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)

C++

// Author: Huahua

class Solution {

public:

string minInteger(string num, int k) {

const int n = num.size();

if (k >= n * (n - 1) / 2) {

sort(begin(num), end(num));

return num;

}

int s = 0;

while (k > 0 && s < n) {

auto bit = begin(num);

auto it = min_element(bit + s, bit + min(s + k + 1, n));

k -= distance(bit + s, it);

rotate(bit + (s++), it, next(it));

}

return num;

}

};

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)

C++

// Author: Huahua

class Fenwick {

public:

explicit Fenwick(int n) : sums_(n + 1) {}

void update(int i, int delta) {

++i;

while (i < sums_.size()) {

sums_[i] += delta;

i += i & -i;

}

int query(int i) {

++i;

int ans = 0;

while (i > 0) {

ans += sums_[i];

i -= i & -i;

}

return ans;

}

private:

vector<int> sums_;

};

class Solution {

public:

string minInteger(string num, int k) {

const int n = num.length();

vector<queue<int>> pos(10);

for (int i = 0; i < n; ++i)

pos[num[i] - '0'].push(i);

Fenwick tree(n);

vector<int> used(n);

string ans;

while (k > 0 & ans.length() < n) {

for (int d = 0; d < 10; ++d) {

if (pos[d].empty()) continue;

const int i = pos[d].front();

const int cost = i - tree.query(i - 1);

if (cost > k) continue;

k -= cost;

ans += ('0' + d);

tree.update(i, 1);

used[i] = true;

pos[d].pop();

break;

}

};

for (int i = 0; i < n; ++i)

if (!used[i]) ans += num[i];

return ans;

}

};

Java

// Author: Huahua

class Solution {

class Fenwick {

private int[] sums;

public Fenwick(int n) {

this.sums = new int[n + 1];

}

public void update(int i, int delta) {

++i;

while (i < sums.length) {

this.sums[i] += delta;

i += i & -i;

}

public int query(int i) {

int ans = 0;

++i;

while (i > 0) {

ans += this.sums[i];

i -= i & -i;

}

return ans;

}

public String minInteger(String num, int k) {

int n = num.length();

var used = new boolean[n];

var pos = new ArrayList<ArrayDeque<Integer>>();

for (int i = 0; i <= 9; ++i)

pos.add(new ArrayDeque<Integer>());

for (int i = 0; i < num.length(); ++i)

pos.get(num.charAt(i) - '0').offer(i);

var tree = new Fenwick(n);

var sb = new StringBuilder();

while (k > 0 && sb.length() < n) {

for (int d = 0; d <= 9; ++d) {

Integer i = pos.get(d).peek();

if (i == null) continue;

int cost = i - tree.query(i - 1);

if (cost > k) continue;

sb.append((char)(d + '0'));

k -= cost;

pos.get(d).removeFirst();

tree.update(i, 1);

used[i] = true;

break;

}

for (int i = 0; i < n; ++i)

if (!used[i]) sb.append(num.charAt(i));

return sb.toString();

}

Python3

# Author: Huahua

class Fenwick:

def __init__(self, n):

self.sums = [0] * (n + 1)

def query(self, i):

ans = 0

i += 1

while i > 0:

ans += self.sums[i];

i -= i & -i

return ans

def update(self, i, delta):

i += 1

while i < len(self.sums):

self.sums[i] += delta

i += i & -i

class Solution:

def minInteger(self, num: str, k: int) -> str:

n = len(num)

used = [False] * n

pos = [deque() for _ in range(10)]

for i, c in enumerate(num):

pos[ord(c) - ord("0")].append(i)

tree = Fenwick(n)

ans = []

while k > 0 and len(ans) < n:

for d in range(10):

if not pos[d]: continue

i = pos[d][0]

cost = i - tree.query(i - 1)

if cost > k: continue

k -= cost

ans.append(chr(d + ord("0")))

tree.update(i, 1)

used[i] = True

pos[d].popleft()

break

for i in range(n):

if not used[i]: ans.append(num[i])

return "".join(ans)

Posts tagged as “reverse paris”

花花酱 LeetCode 1505. Minimum Possible Integer After at Most K Adjacent Swaps On Digits

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

C++

Solution 2: Binary Indexed Tree / Fenwick Tree

C++

Java

Python3