Find the length of longest chain formed using given words in String
Tag : java , By : Justin Bowers
Date : March 29 2020, 07:55 AM
I hope this helps . okk friends here the logic and core part which I had made and my puzzle got solved import java.util.Map;
import java.util.Stack;
public class CandidateCode
{
public static int chainLength=0;
public static void main(String[] args) {
String s= "peas,sugar,rice,soup";
int chainLengthfinal=wordChain(s);
System.out.println("final length:"+chainLengthfinal);
}
public static int wordChain(String input1)
{
List<String> stringList = new ArrayList<String>();
stringList= Arrays.asList(input1.split(","));
boolean ischain = new CandidateCode().hasChain(stringList);
if (ischain) {
return chainLength;
}
return 0;
}
Map<Character, List<String>> startsWith = new HashMap<Character, List<String>>();
Map<Character, List<String>> endsWith = new HashMap<Character, List<String>>();
private Character getFirstChar(String str) {
return str.charAt(0);
}
private Character getLastChar(String str) {
return str.charAt(str.length() - 1);
}
boolean hasChain(List<String> stringList) {
for (String str : stringList) {
Character start = getFirstChar(str);
Character end = getLastChar(str);
List<String> startsWithList;
List<String> endsWithList;
if (startsWith.containsKey(start)) {
startsWithList = startsWith.get(start);
} else {
startsWithList = new ArrayList<String>();
startsWith.put(start, startsWithList);
}
if (endsWith.containsKey(end)) {
endsWithList = endsWith.get(end);
} else {
endsWithList = new ArrayList<String>();
endsWith.put(end, endsWithList);
}
startsWithList.add(str);
endsWithList.add(str);
}
Stack<String> stringStack = new Stack<String>();
for (String str : stringList) {
if (hasChain(stringList.size(), str, stringStack)) {
System.out.println(stringStack);
System.out.println("size "+stringStack.size());
chainLength= stringStack.size();
return true;
}
}
return false;
}
private boolean hasChain(int size, String startString, Stack<String> stringStack) {
if (size == stringStack.size()) return true;
Character last = getLastChar(startString);
if (startsWith.containsKey(last)) {
List<String> stringList = startsWith.get(last);
for (int i = 0; i < stringList.size(); i++) {
String candidate = stringList.remove(i--);
stringStack.push(candidate);
if (hasChain(size, candidate, stringStack)) {
return true;
}
stringStack.pop();
stringList.add(++i, candidate);
}
}
return false;
}
}
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How to find the longest consecutive chain of numbers in an array
Tag : python , By : Topher Cyll
Date : March 29 2020, 07:55 AM
With these it helps Group the items into subsequences using itertools.groupby based on constant differences from an increasing count (provided by an itertools.count object), and then take the longest subsequence using the built-in max on key parameter len: from itertools import groupby, count
lst = [0, 1, 3, 5, 7, 8, 9, 10, 12, 13]
c = count()
val = max((list(g) for _, g in groupby(lst, lambda x: x-next(c))), key=len)
print(val)
# [7, 8, 9, 10]
|
Longest word chain from a list of words
Date : March 29 2020, 07:55 AM
may help you . You can use recursion to explore every "branch" that emerges when every possible letter containing the proper initial character is added to a running list: words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse']
def get_results(_start, _current, _seen):
if all(c in _seen for c in words if c[0] == _start[-1]):
yield _current
else:
for i in words:
if i[0] == _start[-1]:
yield from get_results(i, _current+[i], _seen+[i])
new_d = [list(get_results(i, [i], []))[0] for i in words]
final_d = max([i for i in new_d if len(i) == len(set(i))], key=len)
['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']
words = ['giraffe', 'elephant', 'ant', 'ning', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse',]
new_d = [list(get_results(i, [i], []))[0] for i in words]
final_d = max([i for i in new_d if len(i) == len(set(i))], key=len)
['ant', 'tiger', 'racoon', 'ning', 'giraffe', 'elephant']
|
ClickHouse array - find a longest chain of repeating number in array
Date : March 29 2020, 07:55 AM
hope this fix your issue In Clickhouse I have a column with array of Int16 elements. I'm looking for a way to find a longest chain of repeating number 1. For example, in array [0,1,1,1,5,1,1,1,1,1,2] longest chain of repeating 1 is 5 elements. Is there any way do do it with existing functions ? , Try this query: SELECT
/* The source number. */
data.1 AS number,
/* The source array. */
data.2 AS array,
/* Number the values in each chain. */
arrayCumSumNonNegative((x, index) -> x = number ? 1 : -index, array, arrayEnumerate(array)) AS partiallySumArray,
arrayReduce('max', partiallySumArray) AS result
FROM
(
/* test data set */
SELECT arrayJoin([
/**/
(1, []),
(1, [0, 2, 2, 2, 5]),
(1, [0, 1, 1, 1, 5, 1, 1, 1, 1, 1,2]),
(1, [1, 1, 1, 2, 3, 4, 5, 1, 1]),
(1, [-5, 100, 1, 1, 0, 1, 1, 1]),
(1, [1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0]),
/**/
(5, []),
(5, [0, 2, 2, 2, 55]),
(5, [5, 5, 10, 300, 5, 77, 5])
]) AS data
)
FORMAT Vertical
/* Result:
Row 1:
──────
number: 1
array: []
partiallySumArray: []
result: 0
Row 2:
──────
number: 1
array: [0,2,2,2,5]
partiallySumArray: [0,0,0,0,0]
result: 0
Row 3:
──────
number: 1
array: [0,1,1,1,5,1,1,1,1,1,2]
partiallySumArray: [0,1,2,3,0,1,2,3,4,5,0]
result: 5
Row 4:
──────
number: 1
array: [1,1,1,2,3,4,5,1,1]
partiallySumArray: [1,2,3,0,0,0,0,1,2]
result: 3
Row 5:
──────
number: 1
array: [-5,100,1,1,0,1,1,1]
partiallySumArray: [0,0,1,2,0,1,2,3]
result: 3
Row 6:
──────
number: 1
array: [1,1,0,1,1,1,1,1,1,0,0]
partiallySumArray: [1,2,0,1,2,3,4,5,6,0,0]
result: 6
Row 7:
──────
number: 5
array: []
partiallySumArray: []
result: 0
Row 8:
──────
number: 5
array: [0,2,2,2,55]
partiallySumArray: [0,0,0,0,0]
result: 0
Row 9:
───────
number: 5
array: [5,5,10,300,5,77,5]
partiallySumArray: [1,2,0,0,1,0,1]
result: 2
*/
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Find the longest word/string in an array
Date : March 29 2020, 07:55 AM
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