Given a string S of lowercase English alphabets, the task is to check if there exists an arrangement of string S such that it doesn’t contain any monotonous substring.
A monotonous substring has the following properties:
- Length os such substring is 2.
- Both the characters are consecutive, For example – “ab”, “cd”, “dc”, “zy” etc.
Examples:
Input: S = “abcd”
Output: Yes
Explanation:
String S can be rearranged into “cadb” or “bdac”Input: string = “aab”
Output: No
Explanation:
Every arrangement of the string contains a monotonous substring.
Approach: The idea is group the characters into two different buckets, where one bucket contains the characters which are at the even places and another bucket contains the characters which are at the odd places. Finally, check for the concatenation point of both the group is not a monotonous substring.
Below is the implementation of the above approach:
C++
// C++ implementation such that there // are no monotonous // string in given string #include <bits/stdc++.h> using namespace std; // Function to check a string doesn't // contains a monotonous substring bool check(string s) { bool ok = true; // Loop to itrate over the string // and check that it doesn't contains // the monotonous substring for (int i = 0; i + 1 < s.size(); ++i) ok &= (abs(s[i] - s[i + 1]) != 1); return ok; } // Function to check that there exist // a arrangement of string such that // it doesn't contains monotonous substring string monotonousString(string s) { string odd = "", even = ""; // Loop to group the characters // of the string into two buckets for (int i = 0; i < s.size(); ++i) { if (s[i] % 2 == 0) odd += s[i]; else even += s[i]; } // Sorting the two buckets sort(odd.begin(), odd.end()); sort(even.begin(), even.end()); // Condition to check if the // concatenation point doesn't // contains the monotonous string if (check(odd + even)) return "Yes"; else if (check(even + odd)) return "Yes"; return "No"; } // Driver Code int main() { string str = "abcd"; string ans; ans = monotonousString(str); cout << ans << endl; return 0; } |
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Java
// Java implementation such that there // are no monotonous // string in given string import java.io.*; import java.util.*; class GFG{ // Function to check a string doesn't // contains a monotonous substring static boolean check(String s) { boolean ok = true; // Loop to itrate over the string // and check that it doesn't contains // the monotonous substring for (int i = 0; i + 1 < s.length(); ++i) ok &= (Math.abs(s.charAt(i) - s.charAt(i + 1)) != 1); return ok; } // Function to check that there exist // a arrangement of string such that // it doesn't contains monotonous substring static String monotonousString(String s) { String odd = "", even = ""; // Loop to group the characters // of the string into two buckets for (int i = 0; i < s.length(); ++i) { if (s.charAt(i) % 2 == 0) odd += s.charAt(i); else even += s.charAt(i); } // Sorting the two buckets char oddArray[] = odd.toCharArray(); Arrays.sort(oddArray); odd = new String(oddArray); char evenArray[] = even.toCharArray(); Arrays.sort(evenArray); even = new String(evenArray); // Condition to check if the // concatenation point doesn't // contains the monotonous string if (check(odd + even)) return "Yes"; else if (check(even + odd)) return "Yes"; return "No"; } // Driver Code public static void main(String []args) { String str = "abcd"; String ans; ans = monotonousString(str); System.out.println( ans); } } // This code is contributed by ChitraNayal |
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Python3
# Python3 implementation such that there # are no monotonous string in given string # Function to check a string doesn't # contains a monotonous substring def check(s): ok = True # Loop to itrate over the string # and check that it doesn't contains # the monotonous substring for i in range(0, len(s) - 1, 1): ok = (ok & (abs(ord(s[i]) - ord(s[i + 1])) != 1)) return ok # Function to check that there exist # a arrangement of string such that # it doesn't contains monotonous substring def monotonousString(s): odd = "" even = "" # Loop to group the characters # of the string into two buckets for i in range(len(s)): if (ord(s[i]) % 2 == 0): odd += s[i] else: even += s[i] # Sorting the two buckets odd = list(odd) odd.sort(reverse = False) odd = str(odd) even = list(even) even.sort(reverse = False) even = str(even) # Condition to check if the # concatenation point doesn't # contains the monotonous string if (check(odd + even)): return "Yes" elif (check(even + odd)): return "Yes" return "No" # Driver Code if __name__ == '__main__': str1 = "abcd" ans = monotonousString(str1) print(ans) # This code is contributed by Samarth |
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Output:
Yes
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