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1405. Longest Happy String

SourceWeekly Contest 183 Q3DifficultyMediumRating1820

Description

A string s is called happy if it satisfies the following conditions:

  • s only contains the letters 'a', 'b', and 'c'.
  • s does not contain any of "aaa", "bbb", or "ccc" as a substring.
  • s contains at most a occurrences of the letter 'a'.
  • s contains at most b occurrences of the letter 'b'.
  • s contains at most c occurrences 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 <= 100
  • a + b + c > 0

Solutions

Solution 1: Greedy + Priority Queue

Thinking

\(a+b+c\le 300\) would allow search, but the optimum follows a local rule: always spend the letter that remains most often, without three identical characters in a row.

If the last two characters are already that letter, take the second-most instead. A max-heap by remaining count implements this: pop, append if legal, otherwise pop the next, then push leftovers back.

Stop when no letter can be appended, which yields the longest happy string.

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).

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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)

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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();
    }
}

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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;
    }
};

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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)
}

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function longestDiverseString(a: number, b: number, c: number): string {
    const pq = new PriorityQueue<number[]>((a, b) => b[1] - a[1]);

    if (a > 0) pq.enqueue(['a'.charCodeAt(0), a]);
    if (b > 0) pq.enqueue(['b'.charCodeAt(0), b]);
    if (c > 0) pq.enqueue(['c'.charCodeAt(0), c]);

    const sb: number[] = [];

    while (!pq.isEmpty()) {
        const cur = pq.dequeue();
        const n = sb.length;

        if (n >= 2 && sb[n - 1] === cur[0] && sb[n - 2] === cur[0]) {
            if (pq.isEmpty()) break;

            const next = pq.dequeue();
            sb.push(next[0]);

            if (next[1] > 1) {
                next[1]--;
                pq.enqueue(next);
            }
            pq.enqueue(cur);
        } else {
            sb.push(cur[0]);
            if (cur[1] > 1) {
                cur[1]--;
                pq.enqueue(cur);
            }
        }
    }

    return String.fromCharCode(...sb);
}

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