Formatted question description: https://leetcode.ca/all/767.html

# 767. Reorganize String (Medium)

Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.

If possible, output any possible result.  If not possible, return the empty string.

Example 1:

Input: S = "aab"
Output: "aba"


Example 2:

Input: S = "aaab"
Output: ""


Note:

• S will consist of lowercase letters and have length in range [1, 500].

Companies:

Related Topics:
String, Heap, Greedy, Sort

Similar Questions:

## Solution 1.

// OJ: https://leetcode.com/problems/reorganize-string/

// Time: O(NA) where A is the size of the alphabet
// Space: O(A)
class Solution {
public:
string reorganizeString(string S) {
int cnts = {};
for (char c : S) cnts[c - 'a']++;
string ans(S.size(), '\0');
for (int i = 0; i < S.size(); ++i) {
int maxIndex = -1;
for (int j = 0; j < 26; ++j) {
if (!cnts[j] || (i > 0 && ans[i - 1] == j + 'a')) continue;
if (maxIndex == -1 || cnts[j] > cnts[maxIndex]) maxIndex = j;
}
if (maxIndex == -1) return "";
cnts[maxIndex]--;
ans[i] = maxIndex + 'a';
}
return ans;
}
};


## Solution 2. Interleaving Placement

// OJ: https://leetcode.com/problems/reorganize-string/

// Time: O(AlogA + N) where A is the size of the alphabet
// Space: O(A)
// Ref: https://leetcode.com/problems/reorganize-string/solution/
class Solution {
public:
string reorganizeString(string S) {
int N = S.size(), cnt = {}, j = 1;
for (char c : S) cnt[c - 'a'] += 100;
for (int i = 0; i < 26; ++i) cnt[i] += i;
sort(begin(cnt), end(cnt));
string ans(N, ' ');
for (int n : cnt) {
int ct = n / 100, ch = n % 100;
if (ct == 0) continue;
if (ct > (N + 1) / 2) return "";
while (ct--) {
ans[j] = ch + 'a';
j = (j + 2) % N;
if (j == 1) j = 0;
}
}
return ans;
}
};


## Solution 3. Greedy + Heap

// OJ: https://leetcode.com/problems/reorganize-string/

// Time: O(A + NlogA) where A is the size of the alphabet
// Space: O(A)
class Solution {
public:
string reorganizeString(string S) {
int cnt = {}, prev = -1;
for (char c : S) cnt[c - 'a']++;
auto cmp = [&](int a, int b) { return cnt[a] < cnt[b]; };
priority_queue<int, vector<int>, decltype(cmp)> q(cmp);
for (int i = 0; i < 26; ++i) if (cnt[i]) q.push(i);
string ans;
while (q.size()) {
int c = q.top();
q.pop();
ans.push_back('a' + c);
if (prev != -1) q.push(prev);
if (--cnt[c]) prev = c;
else prev = -1;
}
if (prev != -1) return "";
return ans;
}
};


Java

class Solution {
public String reorganizeString(String S) {
if (S == null || S.length() < 2)
return S;
if (S.length() == 2)
return S.charAt(0) == S.charAt(1) ? "" : S;
int[] counts = new int;
int maxCount = 0;
int length = S.length();
for (int i = 0; i < length; i++) {
char c = S.charAt(i);
counts[c - 'a']++;
maxCount = Math.max(maxCount, counts[c - 'a']);
}
if (maxCount > (length + 1) / 2)
return "";
char[] reorganizeArray = new char[length];
int evenIndex = 0, oddIndex = 1;
int maxPossible = length / 2 + 1;
for (int i = 0; i < 26; i++) {
while (counts[i] > 0 && counts[i] < maxPossible && oddIndex < length) {
reorganizeArray[oddIndex] = (char) ('a' + i);
counts[i]--;
oddIndex += 2;
}
while (counts[i] > 0) {
reorganizeArray[evenIndex] = (char) ('a' + i);
counts[i]--;
evenIndex += 2;
}
}
return new String(reorganizeArray);
}
}