给定一个字符串数组arr [] ,任务是在一个又一个的串联之后,从数组中找到由该字符串中的其他字符串组成的最大字符串。如果不存在这样的字符串,则打印-1 。
例子:
Input: arr[] = {“geeks”, “for”, “geeksfor”, “geeksforgeeks”}
Output: geeksforgeeks
“geeksforgeeks” is made up of (“geeks” + “for” + “geeks”).
Even though “geeksfor” is also made up of other strings
but it is not the largest string.
Input: arr[] = {“Hey”, “you”, “stop”, “right”, “there”}
Output : -1
方法:
- 根据字符串的长度以降序对所有字符串进行排序。
- 现在,从最长的字符串。检查字符串的所有可能前缀是否在给定数组中,以及字符串的其余部分,然后递归检查它是否可以由数组中的其他字符串组成。
- Map可用于检查数组中是否存在字符串。满足上述条件的第一个字符串就是答案。
- 如果不存在这样的字符串,则打印-1。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Comparator to sort the string by
// their lengths in decreasing order
bool compare(string s1, string s2)
{
return s1.size() > s2.size();
}
// Function that returns true if string s can be
// made up of by other two string from the array
// after concatenating one after another
bool canbuildword(string& s, bool isoriginalword,
map& mp)
{
// If current string has been processed before
if (mp.find(s) != mp.end() && mp[s] == 0)
return false;
// If current string is found in the map and
// it is not the string under consideration
if (mp.find(s) != mp.end() && mp[s] == 1
&& isoriginalword == 0) {
return true;
}
for (int i = 1; i < s.length(); i++) {
// Split the string into two
// contiguous sub-strings
string left = s.substr(0, i);
string right = s.substr(i);
// If left sub-string is found in the map and
// the right sub-string can be made from
// the strings from the given array
if (mp.find(left) != mp.end() && mp[left] == 1
&& canbuildword(right, 0, mp)) {
return true;
}
}
// If everything failed, we return false
mp[s] = 0;
return false;
}
// Function to return the longest string
// that can made be made up from the
// other string of the given array
string printlongestword(vector listofwords)
{
// Put all the strings in the map
map mp;
for (string s : listofwords) {
mp[s] = 1;
}
// Sort the string in decreasing
// order of their lengths
sort(listofwords.begin(), listofwords.end(), compare);
// Starting from the longest string
for (string s : listofwords) {
// If current string can be made
// up from other strings
if (canbuildword(s, 1, mp))
return s;
}
return "-1";
}
// Driver code
int main()
{
vector listofwords = { "geeks", "for", "geeksfor",
"geeksforgeeks" };
cout << printlongestword(listofwords);
return 0;
}
Python3
# Python implementation of the approach
# Function that returns true if string s can be
# made up of by other two string from the array
# after concatenating one after another
def canbuildword(s, isoriginalword, mp):
# If current string has been processed before
if s in mp and mp[s] == 0:
return False
# If current string is found in the map and
# it is not the string under consideration
if s in mp and mp[s] == 1 and isoriginalword == 0:
return True
for i in range(1, len(s)):
# Split the string into two
# contiguous sub-strings
left = s[:i]
right = s[i:]
# If left sub-string is found in the map and
# the right sub-string can be made from
# the strings from the given array
if left in mp and mp[left] == 1 and canbuildword(right, 0, mp):
return True
# If everything failed, we return false
mp[s] = 0
return False
# Function to return the longest string
# that can made be made up from the
# other string of the given array
def printlongestword(listofwords):
# Put all the strings in the map
mp = dict()
for i in listofwords:
mp[i] = 1
# Sort the string in decreasing
# order of their lengths
listofwords.sort(key=lambda x: len(x), reverse=True)
# Starting from the longest string
for i in listofwords:
# If current string can be made
# up from other strings
if canbuildword(i, 1, mp):
return i
return "-1"
# Driver code
if __name__ == "__main__":
listofwords = ["geeks", "for", "geeksfor",
"geeksforgeeks"]
print(printlongestword(listofwords))
# This code is contributed by
# sanjeev2552
输出:
geeksforgeeks
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