Given two N-length strings S and T consisting of lowercase alphabets, the task is to minimize the number of swaps of same indexed elements required to make the sum of ASCII value of characters of both the strings odd. If it is not possible to make the sum of ASCII values odd, then print “-1”.
Examples:
Input:S = ”acd”, T = ”dbf”
Output: 1
Explanation:
Swapping S[1] and T[1] modifies S to “abd” and T to “dcf”.
Sum of ASCII value of characters of the string S = 97 + 98 + 100 = 297 (Odd).
Sum of ASCII value of characters of the string T = 100 + 99 + 102 = 301 (Odd).Input: S = “aey”, T = “cgj”
Output: -1
Approach: Follow the steps below to solve the problem:
- Calculate the sum of ASCII values of the characters of the string S and T and store it in variables sum1 and sum2 respectively.
- If sum1 and sum2 are already odd, then print 0, as no swaps are required.
- If sum1 and sum2 are of different parities, print -1, as the sum cannot be of same parity for both the strings.
- If sum1 and sum2 are both even, then traverse the given strings S and T. If there exists any character with odd ASCII value, sum of ASCII values of the characters of both the strings can be made odd by only 1 swap. Otherwise, print -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count the number of swaps // required to make the sum of ASCII values // of the characters of both strings odd void countSwaps(string S, string T) { // Initialize alphabets with value int value[26]; // Initialize values for each // alphabet for ( int i = 0; i < 26; i++) value[i] = i + 1; // Size of the string int N = S.size(); // Sum of string S int sum1 = 0; // Sum of string T int sum2 = 0; // Stores whether there is any // index i such that S[i] and // T[i] have different parities bool flag = false ; // Traverse the strings for ( int i = 0; i < N; i++) { // Update sum1 and sum2 sum1 += value[S[i] - 'a' ]; sum2 += value[T[i] - 'a' ]; // If S[i] and T[i] have // different parities if ((value[S[i] - 'a' ] % 2 == 0 && value[T[i] - 'a' ] % 2 == 1) || (value[S[i] - 'a' ] % 2 == 1 && value[T[i] - 'a' ] % 2 == 0)) flag = false ; } // If sum1 and sum2 are both odd if (sum1 % 2 == 1 && sum2 % 2 == 1) cout << "0\n" ; // If sum1 and sum2 are both even else if (sum1 % 2 == 0 && sum2 % 2 == 0) { // If exists print 1 if (flag) cout << "1" ; // Otherwise else cout << "-1" ; } // If sum1 and sum2 are // of different parities else { cout << "-1" ; } } // Driver Code int main() { string S = "acd" ; string T = "dbf" ; // Function Call countSwaps(S, T); return 0; } |
Java
// Java program for the above approach import java.io.*; class GFG { // Function to count the number of swaps // required to make the sum of ASCII values // of the characters of both strings odd static void countSwaps(String S, String T) { // Initialize alphabets with value int [] value = new int [ 26 ]; // Initialize values for each // alphabet for ( int i = 0 ; i < 26 ; i++) value[i] = i + 1 ; // Size of the string int N = S.length(); // Sum of string S int sum1 = 0 ; // Sum of string T int sum2 = 0 ; // Stores whether there is any // index i such that S[i] and // T[i] have different parities boolean flag = false ; // Traverse the strings for ( int i = 0 ; i < N; i++) { // Update sum1 and sum2 sum1 += value[S.charAt(i) - 'a' ]; sum2 += value[T.charAt(i) - 'a' ]; // If S[i] and T[i] have // different parities if ((value[S.charAt(i) - 'a' ] % 2 == 0 && value[T.charAt(i) - 'a' ] % 2 == 1 ) || (value[S.charAt(i) - 'a' ] % 2 == 1 && value[T.charAt(i) - 'a' ] % 2 == 0 )) flag = false ; } // If sum1 and sum2 are both odd if (sum1 % 2 == 1 && sum2 % 2 == 1 ) System.out.println( "0\n" ); // If sum1 and sum2 are both even else if (sum1 % 2 == 0 && sum2 % 2 == 0 ) { // If exists print 1 if (flag) System.out.println( "1" ); // Otherwise else System.out.println( "-1" ); } // If sum1 and sum2 are // of different parities else { System.out.println( "-1" ); } } // Driver Code public static void main(String[] args) { String S = "acd" ; String T = "dbf" ; // Function Call countSwaps(S, T); } } // This code is contributed by susmitakundugoaldanga. |
Python3
# Python3 program for the above approach # Function to count the number of swaps # required to make the sum of ASCII values # of the characters of both strings odd def countSwaps(S, T): # Initialize alphabets with value value = [ 0 ] * 26 # Initialize values for each # alphabet for i in range ( 26 ): value[i] = i + 1 # Size of the string N = len (S) # Sum of S sum1 = 0 # Sum of T sum2 = 0 # Stores whether there is any # index i such that S[i] and # T[i] have different parities flag = False # Traverse the strings for i in range (N): # Update sum1 and sum2 sum1 + = value[ ord (S[i]) - ord ( 'a' )] sum2 + = value[ ord (T[i]) - ord ( 'a' )] # If ord(S[i]) anord('a)rd(T[i]) haord('a) # different parities if (value[ ord (S[i]) - ord ( 'a' )] % 2 = = 0 and value[ ord (T[i]) - ord ( 'a' )] % 2 = = 1 or value[ ord (S[i]) - ord ( 'a' )] % 2 = = 1 and value[ ord (T[i]) - ord ( 'a' )] % 2 = = 0 ): flag = False # If sum1 and sum2 are both odd if (sum1 % 2 = = 1 and sum2 % 2 = = 1 ): print ( "0" ) # If sum1 and sum2 are both even elif (sum1 % 2 = = 0 and sum2 % 2 = = 0 ): # If exists pr1 if (flag): print ( "1" ) # Otherwise else : print ( "-1" ) # If sum1 and sum2 are # of different parities else : print ( "-1" ) # Driver Code if __name__ = = '__main__' : S = "acd" T = "dbf" # Function Call countSwaps(S, T) # This code is contributed by mohit kumar 29. |
C#
// C# program for the above approach using System; public class GFG { // Function to count the number of swaps // required to make the sum of ASCII values // of the characters of both strings odd static void countSwaps( string S, string T) { // Initialize alphabets with value int [] value = new int [26]; // Initialize values for each // alphabet for ( int i = 0; i < 26; i++) value[i] = i + 1; // Size of the string int N = S.Length; // Sum of string S int sum1 = 0; // Sum of string T int sum2 = 0; // Stores whether there is any // index i such that S[i] and // T[i] have different parities bool flag = false ; // Traverse the strings for ( int i = 0; i < N; i++) { // Update sum1 and sum2 sum1 += value[S[i] - 'a' ]; sum2 += value[T[i] - 'a' ]; // If S[i] and T[i] have // different parities if ((value[S[i] - 'a' ] % 2 == 0 && value[T[i] - 'a' ] % 2 == 1) || (value[S[i] - 'a' ] % 2 == 1 && value[T[i] - 'a' ] % 2 == 0)) flag = false ; } // If sum1 and sum2 are both odd if (sum1 % 2 == 1 && sum2 % 2 == 1) Console.Write( "0\n" ); // If sum1 and sum2 are both even else if (sum1 % 2 == 0 && sum2 % 2 == 0) { // If exists print 1 if (flag) Console.Write( "1" ); // Otherwise else Console.Write( "-1" ); } // If sum1 and sum2 are // of different parities else { Console.Write( "-1" ); } } // Driver Code public static void Main(String[] args) { string S = "acd" ; string T = "dbf" ; // Function Call countSwaps(S, T); } } // This code is contributed by code_hunt. |
Javascript
<script> // JavaScript program for the above approach // Function to count the number of swaps // required to make the sum of ASCII values // of the characters of both strings odd function countSwaps(S, T) { // Initialize alphabets with value var value = [...Array(26)]; // Initialize values for each // alphabet for ( var i = 0; i < 26; i++) value[i] = i + 1; // Size of the string var N = S.length; // Sum of string S var sum1 = 0; // Sum of string T var sum2 = 0; // Stores whether there is any // index i such that S[i] and // T[i] have different parities var flag = false ; // Traverse the strings for ( var i = 0; i < N; i++) { // Update sum1 and sum2 sum1 += value[S[i] - "a" ]; sum2 += value[T[i] - "a" ]; // If S[i] and T[i] have // different parities if ( (value[S[i] - "a" ] % 2 === 0 && value[T[i] - "a" ] % 2 === 1) || (value[S[i] - "a" ] % 2 === 1 && value[T[i] - "a" ] % 2 === 0) ) flag = false ; } // If sum1 and sum2 are both odd if (sum1 % 2 === 1 && sum2 % 2 === 1) document.write( "0 <br>" ); // If sum1 and sum2 are both even else if (sum1 % 2 === 0 && sum2 % 2 === 0) { // If exists print 1 if (flag) document.write( "1" ); // Otherwise else document.write( "-1" ); } // If sum1 and sum2 are // of different parities else { document.write( "-1" ); } } // Driver Code var S = "aey" ; var T = "cgj" ; // Function Call countSwaps(S, T); // This code is contributed by rdtank. </script> |
-1
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
Auxiliary Space: O(26), as we are using extra space for value array.
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