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ReorganizeString767.java
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ReorganizeString767.java
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/**
* 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].
*/
public class ReorganizeString767 {
public String reorganizeString(String S) {
int[] hash = new int[26];
for(char c : S.toCharArray()) hash[c-'a']++;
int max = 0;
for(int i = 0; i<26; ++i){
if(hash[i] > hash[max]) max = i;
}
if(hash[max] > (S.length()+1)/2) return "";
char[] res = new char[S.length()];
int ptr = 0;
while(hash[max]-- > 0){
res[ptr] = (char)(max + 'a');
ptr += 2;
}
for(int i = 0; i<26; ++i){
while(hash[i]-- > 0){
if(ptr >= S.length()) ptr = 1;
res[ptr] = (char)(i + 'a');
ptr += 2;
}
}
return new String(res);
}
/**
* https://leetcode.com/problems/reorganize-string/solution/
*/
public String reorganizeString2(String S) {
int N = S.length();
int[] count = new int[26];
for (char c: S.toCharArray()) count[c-'a']++;
PriorityQueue<MultiChar> pq = new PriorityQueue<MultiChar>((a, b) ->
a.count == b.count ? a.letter - b.letter : b.count - a.count);
for (int i = 0; i < 26; ++i) if (count[i] > 0) {
if (count[i] > (N + 1) / 2) return "";
pq.add(new MultiChar(count[i], (char) ('a' + i)));
}
StringBuilder ans = new StringBuilder();
while (pq.size() >= 2) {
MultiChar mc1 = pq.poll();
MultiChar mc2 = pq.poll();
/*This code turns out to be superfluous, but explains what is happening
if (ans.length() == 0 || mc1.letter != ans.charAt(ans.length() - 1)) {
ans.append(mc1.letter);
ans.append(mc2.letter);
} else {
ans.append(mc2.letter);
ans.append(mc1.letter);
}*/
ans.append(mc1.letter);
ans.append(mc2.letter);
if (--mc1.count > 0) pq.add(mc1);
if (--mc2.count > 0) pq.add(mc2);
}
if (pq.size() > 0) ans.append(pq.poll().letter);
return ans.toString();
}
class MultiChar {
int count;
char letter;
MultiChar(int ct, char ch) {
count = ct;
letter = ch;
}
}
// use 358. Rearrange String k Distance Apart
public String reorganizeString3(String S) {
return rearrangeString(S, 2);
}
public String rearrangeString(String str, int k) {
int length = str.length();
int[] count = new int[26];
int[] nextIndex = new int[26];
char[] chars = str.toCharArray();
for (char ch: chars){
count[ch-'a']++;
}
StringBuilder sb = new StringBuilder();
for (int i=0; i<length; i++) {
int idx = findValidMax(count, nextIndex, i);
if (idx == -1) return "";
sb.append((char) (idx + 'a'));
count[idx]--;
nextIndex[idx] = i+k;
}
return sb.toString();
}
private int findValidMax(int[] count, int[] nextIndex, int index){
int max = Integer.MIN_VALUE;
int idx = -1;
for (int i=0; i<26; i++) {
if (count[i] != 0 && count[i] > max && index >= nextIndex[i]) {
max = count[i];
idx = i;
}
}
return idx;
}
/**
* https://leetcode.com/problems/reorganize-string/discuss/113451/7-ms-Java-O(n)-Solution.-no-Sorting
*/
public String reorganizeString4(String S) {
int n = S.length();
int[] cnt = new int[128];
char mc = 'a';
for (char c : S.toCharArray()) {
cnt[c]++;
mc = (cnt[c] > cnt[mc]) ? c : mc;
}
if (cnt[mc] == 1) {
return S;
}
if (n - cnt[mc] <= cnt[mc] - 2) {
return "";
}
StringBuilder[] sb = new StringBuilder[cnt[mc]];
for (int i = 0; i < sb.length; i ++) {
sb[i] = new StringBuilder();
sb[i].append(mc);
}
int k = 0;
for (char c = 'a'; c <= 'z'; c++) {
while (c != mc && cnt[c] > 0) {
sb[k++].append(c);
cnt[c]--;
k %= sb.length;
}
}
for (int i = 1; i < sb.length; i++) {
sb[0].append(sb[i]);
}
return sb[0].toString();
}
}