Concatenate words in such a way as to obtain the longest possible sub-string of the same letter
1
$begingroup$
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 way 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.
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
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;
}
}
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm strings programming-challenge
edited 6 mins ago
200_success
129k15153415
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
$endgroup$
add a comment |
1
$begingroup$
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 way 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.
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
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;
}
}
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm strings programming-challenge
edited 6 mins ago
200_success
129k15153415
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
$endgroup$
add a comment |
1
1
1
$begingroup$
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 way 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.
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
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;
}
}
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm strings programming-challenge
edited 6 mins ago
200_success
129k15153415
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
$endgroup$
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 way 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.
- Given N=3 and words=['aabb','aaaa','bbab'], your function should return 6. One of the best concatenations is
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;
}
}
Is there a better approach to solve the above problem? Is there something I can improve on?
java algorithm strings programming-challenge
java algorithm strings programming-challenge
edited 6 mins ago
200_success
129k15153415
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
edited 6 mins ago
200_success
129k15153415
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
edited 6 mins ago
200_success
129k15153415
edited 6 mins ago
200_success
129k15153415
edited 6 mins ago
200_success
129k15153415
129k15153415
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
asked 16 mins ago
Maclean PintoMaclean Pinto
1275
1275
add a comment |
add a comment |
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