Concatenate the words in such a say as to obtain a single word with the longest possible sub-string
0
$begingroup$
An array of N words is given. Each word consists of small letters
('a'- 'z'). Our goal is to concatenate the words in such a say as to
obtain a single word with the longest possible sub-string composed of
one particular letter. Find the length of such a sub-string.
Examples:
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
words[1]+words[0]+words[2]='aaaaaabbbbab'. The longest sub-string is
composed of the letter 'a' and its length is 6. - Given N=3 and words=['xxbxx','xbx','x'], your function should return 4. One of the best concatenations is
words[0]+words[2]+words[1]='xxbxxxxbx'. The longest sub-string is
composed of letter 'x' and its length is 4.
public class DailyCodingProblem4 {
public static void main(String args) {
String arr = { "aabb", "aaaa", "bbab" };
int res = solution(arr);
System.out.println(res);
String arr2 = { "xxbxx", "xbx", "x" };
res = solution(arr2);
System.out.println(res);
}
private static int solution(String arr) {
Map<Integer, Integer> prefix = new HashMap<>();
Map<Integer, Integer> suffix = new HashMap<>();
Map<Integer, Integer> both = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
String word = arr[i];
int j = 1;
while (j < word.length() && word.charAt(0) == word.charAt(j)) {
j++;
}
int key = word.charAt(0);
if (j == word.length()) {
if (both.containsKey(key)) {
Integer temp = both.get(key);
if (j > temp) {
both.put(key, j);
}
} else {
both.put(key, j);
}
} else {
if (suffix.containsKey(key)) {
Integer temp = suffix.get(key);
if (j > temp) {
suffix.put(key, j);
}
} else {
suffix.put(key, j);
}
j = word.length() - 1;
while (j > 0 && word.charAt(word.length() - 1) == word.charAt(j - 1)) {
j--;
}
key = word.charAt(word.length() - 1);
if (prefix.containsKey(key)) {
Integer temp = prefix.get(key);
if (word.length() - j > temp) {
prefix.put(key, word.length() - j);
}
} else {
prefix.put(key, word.length() - j);
}
}
}
int res = 0;
for (Integer key : prefix.keySet()) {
if (suffix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : suffix.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : both.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + both.get(key);
if (temp > res) {
res = temp;
}
}
if (suffix.containsKey(key)) {
int temp = both.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
return res;
}
}
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
$endgroup$
add a comment |
0
$begingroup$
An array of N words is given. Each word consists of small letters
('a'- 'z'). Our goal is to concatenate the words in such a say as to
obtain a single word with the longest possible sub-string composed of
one particular letter. Find the length of such a sub-string.
Examples:
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
words[1]+words[0]+words[2]='aaaaaabbbbab'. The longest sub-string is
composed of the letter 'a' and its length is 6. - Given N=3 and words=['xxbxx','xbx','x'], your function should return 4. One of the best concatenations is
words[0]+words[2]+words[1]='xxbxxxxbx'. The longest sub-string is
composed of letter 'x' and its length is 4.
public class DailyCodingProblem4 {
public static void main(String args) {
String arr = { "aabb", "aaaa", "bbab" };
int res = solution(arr);
System.out.println(res);
String arr2 = { "xxbxx", "xbx", "x" };
res = solution(arr2);
System.out.println(res);
}
private static int solution(String arr) {
Map<Integer, Integer> prefix = new HashMap<>();
Map<Integer, Integer> suffix = new HashMap<>();
Map<Integer, Integer> both = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
String word = arr[i];
int j = 1;
while (j < word.length() && word.charAt(0) == word.charAt(j)) {
j++;
}
int key = word.charAt(0);
if (j == word.length()) {
if (both.containsKey(key)) {
Integer temp = both.get(key);
if (j > temp) {
both.put(key, j);
}
} else {
both.put(key, j);
}
} else {
if (suffix.containsKey(key)) {
Integer temp = suffix.get(key);
if (j > temp) {
suffix.put(key, j);
}
} else {
suffix.put(key, j);
}
j = word.length() - 1;
while (j > 0 && word.charAt(word.length() - 1) == word.charAt(j - 1)) {
j--;
}
key = word.charAt(word.length() - 1);
if (prefix.containsKey(key)) {
Integer temp = prefix.get(key);
if (word.length() - j > temp) {
prefix.put(key, word.length() - j);
}
} else {
prefix.put(key, word.length() - j);
}
}
}
int res = 0;
for (Integer key : prefix.keySet()) {
if (suffix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : suffix.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : both.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + both.get(key);
if (temp > res) {
res = temp;
}
}
if (suffix.containsKey(key)) {
int temp = both.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
return res;
}
}
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
$endgroup$
add a comment |
0
0
0
$begingroup$
An array of N words is given. Each word consists of small letters
('a'- 'z'). Our goal is to concatenate the words in such a say as to
obtain a single word with the longest possible sub-string composed of
one particular letter. Find the length of such a sub-string.
Examples:
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
words[1]+words[0]+words[2]='aaaaaabbbbab'. The longest sub-string is
composed of the letter 'a' and its length is 6. - Given N=3 and words=['xxbxx','xbx','x'], your function should return 4. One of the best concatenations is
words[0]+words[2]+words[1]='xxbxxxxbx'. The longest sub-string is
composed of letter 'x' and its length is 4.
public class DailyCodingProblem4 {
public static void main(String args) {
String arr = { "aabb", "aaaa", "bbab" };
int res = solution(arr);
System.out.println(res);
String arr2 = { "xxbxx", "xbx", "x" };
res = solution(arr2);
System.out.println(res);
}
private static int solution(String arr) {
Map<Integer, Integer> prefix = new HashMap<>();
Map<Integer, Integer> suffix = new HashMap<>();
Map<Integer, Integer> both = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
String word = arr[i];
int j = 1;
while (j < word.length() && word.charAt(0) == word.charAt(j)) {
j++;
}
int key = word.charAt(0);
if (j == word.length()) {
if (both.containsKey(key)) {
Integer temp = both.get(key);
if (j > temp) {
both.put(key, j);
}
} else {
both.put(key, j);
}
} else {
if (suffix.containsKey(key)) {
Integer temp = suffix.get(key);
if (j > temp) {
suffix.put(key, j);
}
} else {
suffix.put(key, j);
}
j = word.length() - 1;
while (j > 0 && word.charAt(word.length() - 1) == word.charAt(j - 1)) {
j--;
}
key = word.charAt(word.length() - 1);
if (prefix.containsKey(key)) {
Integer temp = prefix.get(key);
if (word.length() - j > temp) {
prefix.put(key, word.length() - j);
}
} else {
prefix.put(key, word.length() - j);
}
}
}
int res = 0;
for (Integer key : prefix.keySet()) {
if (suffix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : suffix.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : both.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + both.get(key);
if (temp > res) {
res = temp;
}
}
if (suffix.containsKey(key)) {
int temp = both.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
return res;
}
}
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
$endgroup$
An array of N words is given. Each word consists of small letters
('a'- 'z'). Our goal is to concatenate the words in such a say as to
obtain a single word with the longest possible sub-string composed of
one particular letter. Find the length of such a sub-string.
Examples:
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
words[1]+words[0]+words[2]='aaaaaabbbbab'. The longest sub-string is
composed of the letter 'a' and its length is 6. - Given N=3 and words=['xxbxx','xbx','x'], your function should return 4. One of the best concatenations is
words[0]+words[2]+words[1]='xxbxxxxbx'. The longest sub-string is
composed of letter 'x' and its length is 4.
public class DailyCodingProblem4 {
public static void main(String args) {
String arr = { "aabb", "aaaa", "bbab" };
int res = solution(arr);
System.out.println(res);
String arr2 = { "xxbxx", "xbx", "x" };
res = solution(arr2);
System.out.println(res);
}
private static int solution(String arr) {
Map<Integer, Integer> prefix = new HashMap<>();
Map<Integer, Integer> suffix = new HashMap<>();
Map<Integer, Integer> both = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
String word = arr[i];
int j = 1;
while (j < word.length() && word.charAt(0) == word.charAt(j)) {
j++;
}
int key = word.charAt(0);
if (j == word.length()) {
if (both.containsKey(key)) {
Integer temp = both.get(key);
if (j > temp) {
both.put(key, j);
}
} else {
both.put(key, j);
}
} else {
if (suffix.containsKey(key)) {
Integer temp = suffix.get(key);
if (j > temp) {
suffix.put(key, j);
}
} else {
suffix.put(key, j);
}
j = word.length() - 1;
while (j > 0 && word.charAt(word.length() - 1) == word.charAt(j - 1)) {
j--;
}
key = word.charAt(word.length() - 1);
if (prefix.containsKey(key)) {
Integer temp = prefix.get(key);
if (word.length() - j > temp) {
prefix.put(key, word.length() - j);
}
} else {
prefix.put(key, word.length() - j);
}
}
}
int res = 0;
for (Integer key : prefix.keySet()) {
if (suffix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : suffix.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
for (Integer key : both.keySet()) {
if (prefix.containsKey(key)) {
int temp = prefix.get(key) + both.get(key);
if (temp > res) {
res = temp;
}
}
if (suffix.containsKey(key)) {
int temp = both.get(key) + suffix.get(key);
if (temp > res) {
res = temp;
}
}
}
return res;
}
}
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm
java algorithm
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
asked 10 mins ago
Maclean PintoMaclean Pinto
1225
1225
add a comment |
add a comment |
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