| comments | difficulty | edit_url | rating | source | tags | |||
|---|---|---|---|---|---|---|---|---|
true |
Medium |
1820 |
Weekly Contest 183 Q3 |
|
A string s is called happy if it satisfies the following conditions:
sonly contains the letters'a','b', and'c'.sdoes not contain any of"aaa","bbb", or"ccc"as a substring.scontains at mostaoccurrences of the letter'a'.scontains at mostboccurrences of the letter'b'.scontains at mostcoccurrences of the letter'c'.
Given three integers a, b, and c, return the longest possible happy string. If there are multiple longest happy strings, return any of them. If there is no such string, return the empty string "".
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: a = 1, b = 1, c = 7 Output: "ccaccbcc" Explanation: "ccbccacc" would also be a correct answer.
Example 2:
Input: a = 7, b = 1, c = 0 Output: "aabaa" Explanation: It is the only correct answer in this case.
Constraints:
0 <= a, b, c <= 100a + b + c > 0
The greedy strategy is to prioritize the selection of characters with the most remaining occurrences. By using a priority queue or sorting, we ensure that the character selected each time is the one with the most remaining occurrences (to avoid having three consecutive identical characters, in some cases, we need to select the character with the second most remaining occurrences).
class Solution:
def longestDiverseString(self, a: int, b: int, c: int) -> str:
h = []
if a > 0:
heappush(h, [-a, 'a'])
if b > 0:
heappush(h, [-b, 'b'])
if c > 0:
heappush(h, [-c, 'c'])
ans = []
while len(h) > 0:
cur = heappop(h)
if len(ans) >= 2 and ans[-1] == cur[1] and ans[-2] == cur[1]:
if len(h) == 0:
break
nxt = heappop(h)
ans.append(nxt[1])
if -nxt[0] > 1:
nxt[0] += 1
heappush(h, nxt)
heappush(h, cur)
else:
ans.append(cur[1])
if -cur[0] > 1:
cur[0] += 1
heappush(h, cur)
return ''.join(ans)class Solution {
public String longestDiverseString(int a, int b, int c) {
Queue<int[]> pq = new PriorityQueue<>((x, y) -> y[1] - x[1]);
if (a > 0) {
pq.offer(new int[] {'a', a});
}
if (b > 0) {
pq.offer(new int[] {'b', b});
}
if (c > 0) {
pq.offer(new int[] {'c', c});
}
StringBuilder sb = new StringBuilder();
while (pq.size() > 0) {
int[] cur = pq.poll();
int n = sb.length();
if (n >= 2 && sb.codePointAt(n - 1) == cur[0] && sb.codePointAt(n - 2) == cur[0]) {
if (pq.size() == 0) {
break;
}
int[] next = pq.poll();
sb.append((char) next[0]);
if (next[1] > 1) {
next[1]--;
pq.offer(next);
}
pq.offer(cur);
} else {
sb.append((char) cur[0]);
if (cur[1] > 1) {
cur[1]--;
pq.offer(cur);
}
}
}
return sb.toString();
}
}class Solution {
public:
string longestDiverseString(int a, int b, int c) {
using pci = pair<char, int>;
auto cmp = [](pci x, pci y) { return x.second < y.second; };
priority_queue<pci, vector<pci>, decltype(cmp)> pq(cmp);
if (a > 0) pq.push({'a', a});
if (b > 0) pq.push({'b', b});
if (c > 0) pq.push({'c', c});
string ans;
while (!pq.empty()) {
pci cur = pq.top();
pq.pop();
int n = ans.size();
if (n >= 2 && ans[n - 1] == cur.first && ans[n - 2] == cur.first) {
if (pq.empty()) break;
pci nxt = pq.top();
pq.pop();
ans.push_back(nxt.first);
if (--nxt.second > 0) {
pq.push(nxt);
}
pq.push(cur);
} else {
ans.push_back(cur.first);
if (--cur.second > 0) {
pq.push(cur);
}
}
}
return ans;
}
};type pair struct {
c byte
num int
}
type hp []pair
func (a hp) Len() int { return len(a) }
func (a hp) Swap(i, j int) { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool { return a[i].num > a[j].num }
func (a *hp) Push(x any) { *a = append(*a, x.(pair)) }
func (a *hp) Pop() any { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }
func longestDiverseString(a int, b int, c int) string {
var h hp
if a > 0 {
heap.Push(&h, pair{'a', a})
}
if b > 0 {
heap.Push(&h, pair{'b', b})
}
if c > 0 {
heap.Push(&h, pair{'c', c})
}
var ans []byte
for len(h) > 0 {
cur := heap.Pop(&h).(pair)
if len(ans) >= 2 && ans[len(ans)-1] == cur.c && ans[len(ans)-2] == cur.c {
if len(h) == 0 {
break
}
next := heap.Pop(&h).(pair)
ans = append(ans, next.c)
if next.num > 1 {
next.num--
heap.Push(&h, next)
}
heap.Push(&h, cur)
} else {
ans = append(ans, cur.c)
if cur.num > 1 {
cur.num--
heap.Push(&h, cur)
}
}
}
return string(ans)
}function longestDiverseString(a: number, b: number, c: number): string {
let ans = [];
let store: Array<[string, number]> = [
['a', a],
['b', b],
['c', c],
];
while (true) {
store.sort((a, b) => b[1] - a[1]);
let hasNext = false;
for (let [i, [ch, ctn]] of store.entries()) {
if (ctn < 1) {
break;
}
const n = ans.length;
if (n >= 2 && ans[n - 1] == ch && ans[n - 2] == ch) {
continue;
}
hasNext = true;
ans.push(ch);
store[i][1] -= 1;
break;
}
if (!hasNext) {
break;
}
}
return ans.join('');
}function longestDiverseString(a: number, b: number, c: number): string {
let res = '';
let prev = { ch: '', c: 0 };
let last = { ch: '', c: 0 };
const pq = new MaxPriorityQueue({ priority: ({ c }) => c });
pq.enqueue({ ch: 'a', c: a });
pq.enqueue({ ch: 'b', c: b });
pq.enqueue({ ch: 'c', c });
while (pq.size()) {
const item = pq.dequeue().element;
let c = item.c < prev.c ? 1 : 2;
if (prev.c) pq.enqueue(prev);
if (last.ch !== item.ch && item.c) last = { ...item, c: 0 };
while (c-- && item.c && last.c++ < 2) {
item.c--;
res += item.ch;
}
prev = item;
}
return res;
}function longestDiverseString(a, b, c) {
let res = '';
let prev = { ch: '', c: 0 };
let last = { ch: '', c: 0 };
const pq = new MaxPriorityQueue({ priority: ({ c }) => c });
pq.enqueue({ ch: 'a', c: a });
pq.enqueue({ ch: 'b', c: b });
pq.enqueue({ ch: 'c', c });
while (pq.size()) {
const item = pq.dequeue().element;
let c = item.c < prev.c ? 1 : 2;
if (prev.c) pq.enqueue(prev);
if (last.ch !== item.ch && item.c) last = { ...item, c: 0 };
while (c-- && item.c && last.c++ < 2) {
item.c--;
res += item.ch;
}
prev = item;
}
return res;
}