我想从字符串中删除 K 个字符,以使每个字符的出现最少。
例如:
字符串:abcdefghijkllllll
K:5
答案:12(abcdeghijkl)
字符串:ababac
K:4
答案:3(aac)
字符串:aaaab
K:4
答案:1(b)
我想删除5个字符。这些字符将是5 l
的
到目前为止,我所做的是使用地图计算每个字符的出现次数
但是我对下一步的工作感到困惑。
#include <bits/stdc++.h>
using namespace std;
string s;
int l, k;
map<char, int> m;
int main() {
getline(cin, s);
scanf("%d %d", &l, &k);
for(int i=0; i<s.length(); i++) {
m[s[i]]++;
}
for(auto &x : m) {
cout << x.second << "\n";
}
return 0;
}
预期结果是除去任何给定字符串的字符(可以排序或不排序)之后的最小字符串长度。
您可以删除字符串中的任何字符
#include <bits/stdc++.h>
using namespace std;
string s;
int l, k;
map<char, int> m;
int main() {
getline(cin, s);
cin >> l >> k;
for(int i = 0; i < s.length(); i++) {
m[s[i]]++;
}
for(auto it = m.end(); it != m.begin(); it--) {
// cout << it->second << "\n";
}
vector<pair<int, int>> pairs;
for (auto itr = m.begin(); itr != m.end(); itr++) {
pairs.push_back(*itr);
}
sort(pairs.begin(), pairs.end(), [=](pair<int, int>& a, pair<int, int>& b) { return a.second < b.second; } );
for(auto it = m.end(); it != m.begin(); it--) {
if(it->second - k >= 1) {
it->second-=k;
k -= it->second;
}
}
int sum = 0;
for(auto it = m.end(); it != m.begin(); it--) {
sum += it->second;
// cout << it->second << "\n";
}
cout << sum << "\n";
return 0;
}
当前的问题是它不能读取所有字符并将它们正确映射到地图。
答案 0 :(得分:0)
从您的描述和测试用例中我不确定您要寻找什么。您的答案是返回字符串中剩余的字符数,而更新后的函数将返回sum
变量。如果是这种情况,为什么不返回字符串的长度减去k?
您的第二个测试用例是:
String: ababac
K: 4
Answer: 3 (aac)
从“ ababac”(长度6)中删除4个字符将使其长度为2,而不是3。
可以按任何顺序删除字符吗?对于第三个测试用例,您具有:
String: aaaab
K: 4
Answer: 1(b)
给出描述:I want to remove K characters from a string such that the occurrence of each character is at a minimum.
删除3个字符将得出结果“ ab”。删除第四位可能会导致“ a”或“ b”。在这种情况下,您该怎么办?
这个问题有很多歧义,测试用例有点混乱。例如,给定“ aabbccdddd” k = 3,那么可接受的答案是什么? “ abcdddd”还是“ aabbccd”? “ abcdddd”将增加最少的字符数,而“ aabbccd”将减少最频繁出现的字符数。
我使用最大优先级队列/ max-heap(在Java中)与上面的后面的示例进行了汇总。假设您的所有输入都很好。
import java.util.*;
public class SO {
//Helper class to put in the priority queue.
class CharInt {
char c;
int count;
public CharInt(char c) {
this.c = c;
this.count = 1;
}
void increment() {
this.count++;
}
void decrement() {
this.count--;
}
}
public int minChar(String s, int k) {
Map<Character, CharInt> map = new HashMap<Character, CharInt>();
for (Character c : s.toCharArray()) {
if (map.get(c) == null) {
map.put(c, new CharInt(c));
}else {
map.get(c).increment();
}
}
//Makes a Max-Heap from a PriorityQueue object. The comparator makes sure the top of the PriorityQueue is the character with the highest count.
PriorityQueue<CharInt> maxHeap = new PriorityQueue<CharInt>(new Comparator<CharInt>() {
@Override
public int compare(CharInt o1, CharInt o2) {
return - Integer.compare(o1.count, o2.count);
}
});
//Add all values to the heap.
for (CharInt c : map.values()) {
maxHeap.add(c);
}
//Take the top value off, decrement its count, add it back to the heap. Do this k times.
while (k-- > 0) {
CharInt c = maxHeap.poll();
c.decrement();
maxHeap.add(c);
}
StringBuilder builder = new StringBuilder(); // Used to make output string. Can be left out.
int sum = 0;
//Remove every element from the heap and get its count value.
while(!maxHeap.isEmpty()) {
CharInt c = maxHeap.poll();
for (int i = 0; i < c.count; i++) {
sum += c.count;
builder.append(c.c); // Used to make output string. Can be left out.
}
}
char[] chars = builder.toString().toCharArray(); // Used to make output string. Can be left out.
Arrays.sort(chars); // Used to make output string. Can be left out.
System.out.println(chars); // Used to make output string. Can be left out.
return sum;
}
public static void main(String...bannaa) {
int s = new SO().minChar("abcdefghijkllllll", 5);
int s2 = new SO().minChar("ababac", 4);
int s3 = new SO().minChar("aaaab", 4);
int s4 = new SO().minChar("abbbccc", 4);
System.out.println(s + " " + s2 + " " + s3 + " " + s4);
}
}
输出:
abcdefghijkl
ac
a
abc
12 2 1 3