Formatted question description: https://leetcode.ca/all/1405.html

1405. Longest Happy String (Medium)

A string is called happy if it does not have any of the strings 'aaa', 'bbb' or 'ccc' as a substring.

Given three integers a, b and c, return any string s, which satisfies following conditions:

s is happy and longest possible.
s contains at most a occurrences of the letter 'a', at most b occurrences of the letter 'b' and at most c occurrences of the letter 'c'.
s will only contain 'a', 'b' and 'c' letters.

If there is no such string s return the empty string "".

Example 1:

Input: a = 1, b = 1, c = 7
Output: "ccaccbcc"
Explanation: "ccbccacc" would also be a correct answer.

Example 2:

Input: a = 2, b = 2, c = 1
Output: "aabbc"

Example 3:

Input: a = 7, b = 1, c = 0
Output: "aabaa"
Explanation: It's the only correct answer in this case.

Constraints:

0 <= a, b, c <= 100
a + b + c > 0

Related Topics:
Dynamic Programming, Greedy

Solution 1.

class Solution {
    public String longestDiverseString(int a, int b, int c) {
        PriorityQueue<LetterCount> priorityQueue = new PriorityQueue<LetterCount>();
        if (a > 0)
            priorityQueue.offer(new LetterCount('a', a));
        if (b > 0)
            priorityQueue.offer(new LetterCount('b', b));
        if (c > 0)
            priorityQueue.offer(new LetterCount('c', c));
        StringBuffer sb = new StringBuffer();
        char prevLetter = 'z';
        int prevCount = 0;
        while (!priorityQueue.isEmpty()) {
            LetterCount letterCount = priorityQueue.poll();
            char letter = letterCount.letter;
            int count = letterCount.count;
            if (letter == prevLetter) {
                if (prevCount < 2) {
                    int curCount = Math.min(count, 2 - prevCount);
                    for (int i = 0; i < curCount; i++)
                        sb.append(letter);
                    prevCount += curCount;
                    letterCount.count -= curCount;
                    if (letterCount.count > 0)
                        priorityQueue.offer(letterCount);
                } else {
                    if (priorityQueue.isEmpty())
                        break;
                    LetterCount letterCount2 = priorityQueue.poll();
                    sb.append(letterCount2.letter);
                    letterCount2.count--;
                    if (letterCount2.count > 0)
                        priorityQueue.offer(letterCount2);
                    priorityQueue.offer(letterCount);
                    prevLetter = letterCount2.letter;
                    prevCount = 1;
                }
            } else {
                int curCount = Math.min(count, 2);
                for (int i = 0; i < curCount; i++)
                    sb.append(letter);
                letterCount.count -= curCount;
                if (letterCount.count > 0)
                    priorityQueue.offer(letterCount);
                prevLetter = letter;
                prevCount = curCount;
            }
        }
        return sb.toString();
    }
}

class LetterCount implements Comparable<LetterCount> {
    char letter;
    int count;

    public LetterCount(char letter, int count) {
        this.letter = letter;
        this.count = count;
    }

    public int compareTo(LetterCount letterCount2) {
        if (this.count != letterCount2.count)
            return letterCount2.count - this.count;
        else
            return this.letter - letterCount2.letter;
    }
}

############

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();
    }
}

// OJ: https://leetcode.com/problems/longest-happy-string/
// Time: O(A+B+C)
// Space: O(1)
class Solution {
public:
    string longestDiverseString(int a, int b, int c) {
        int cnt[3] = {a, b, c}, prev = -1;
        string ans;
        while (true) {
            int maxIndex = -1;
            for (int i = 0; i < 3; ++i) {
                if (i != prev && (maxIndex == -1 || cnt[i] > cnt[maxIndex])) maxIndex = i;
            }
            if (cnt[maxIndex] == 0) break;
            int num = min(2, cnt[maxIndex]);
            if (prev != -1 && cnt[prev] > cnt[maxIndex]) num = 1;
            cnt[maxIndex] -= num;
            while (num--) ans.push_back(maxIndex + 'a');
            prev = maxIndex;
        }
        return ans;
    }
};

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:
    def longestDiverseString(self, a: int, b: int, c: int) -> str:
        d = [[a, "a"], [b, "b"], [c, "c"]]
        total = a + b + c
        res = ""
        def canAdd(res, x):
            return len(res) < 2 or res[-1] != x or res[-2] != x
        while total > 0:
            d.sort(reverse=True)
            for i, (count, char) in enumerate(d):
                if count == 0: continue
                if canAdd(res, char):
                    res += char
                    d[i][0] -= 1
                    total -= 1
                    break
            else:
                break
        return res

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 interface{}) { *a = append(*a, x.(pair)) }
func (a *hp) Pop() interface{}   { 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('');
}

1405 - Longest Happy String

1405. Longest Happy String (Medium)

Solution 1.

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1405. Longest Happy String (Medium)

Solution 1.

All Problems

All Solutions