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# Minimize swaps of same-indexed characters to make sum of ASCII value of characters of both the strings odd

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 ``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

 ``

Output:

`-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.