(Analysis by Benjamin Qi)

Given an input string $s$ and an output string $t$, construct a directed graph on vertex set $\{\texttt{a}\dots\texttt{z}\}\cup \{\texttt{A}\dots\texttt{Z}\}$. For every two characters $x$ and $y$, add a directed edge from $x$ to $y$ if there exists at least one index $i$ such that $s_i=x$ and $t_i=y$.

First, let's figure out when it is impossible to transform $s$ into $t$. If there is a vertex in our directed graph with more than one out-edge, the answer is $-1$. The answer is also $-1$ whenever $s\neq t$ and $t$ contains all $52$ distinct characters; any replacement will cause $s$ to contain fewer than $52$ distinct characters.

In all other cases, it is possible to transform $s$ to $t$. A single keystroke allows us to take any edge $x\to y$ and replace it with an edge $z\to y$ as long as $z$ does not have out-degree greater than one afterward. Any keystroke can remove at most one edge from our directed graph, so the number of required keystrokes is at least the number of edges in the directed graph excluding self-loops. Also, suppose that we treat the edges as undirected and divide the graph into connected components. Then every connected component of size greater than one that is a cycle increases the answer by one (for example, consider the cycle $\texttt{A}\to \texttt{B}\to \texttt{A}$ in the last test case of the sample). Note that cycles within larger connected components don't increase the answer by one. For example, if the edges in the directed graph are $\texttt{A}\to \texttt{B}, \texttt{B}\to \texttt{A}, \texttt{C}\to \texttt{B}$, then the answer is three, because we can use the first keystroke to replace $\texttt{A}$ with $\texttt{C}$, which breaks the cycle.

Thus, the answer is equal to the following quantity: the number of edges in the graph excluding self-loops plus the number of connected components with size greater than one that are cycles. To prove that this is correct, it suffices to check that in all cases where the answer is not $-1$, any keystroke will decrease this quantity by at most one, and that there exists a keystroke that decreases this quantity by one.

Danny Mittal's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class FindAndReplaceSilverFixed {

public static void main(String[] args) throws IOException {
StringBuilder out = new StringBuilder();
for (int t = Integer.parseInt(in.readLine()); t > 0; t--) {
int[] becomes = new int[52];
Arrays.fill(becomes, -1);
boolean possible = true;
Set<Character> set = new HashSet<>();
for (int j = 0; j < before.length(); j++) {
int b = letterToNode(before.charAt(j));
int a = letterToNode(after.charAt(j));
if (becomes[b] != -1 && becomes[b] != a) {
possible = false;
}
becomes[b] = a;
}

if (set.size() == 52) {
possible = false;
}
if (before.equals(after)) {
possible = true;
}

if (possible) {
int[] inDegree = new int[52];
for (int a = 0; a < 52; a++) {
if (becomes[a] != -1 && becomes[a] != a) {
inDegree[becomes[a]]++;
}
}

for (int a = 0; a < 52; a++) {
if (becomes[a] != -1 && becomes[a] != a) {
}
}

int[] seen = new int[52];
for (int r = 0; r < 52; r++) {
if (seen[r] == 0) {
int a = r;
while (a != -1 && seen[a] == 0) {
seen[a] = r + 1;
a = becomes[a];
}
if (a != -1 && a != becomes[a] && seen[a] == r + 1) {
int s = a;
boolean freePass = false;
do {
seen[a] = 2;
if (inDegree[a] > 1) {
freePass = true;
}
a = becomes[a];
} while (a != s);
if (!freePass) {
}
}
}
}
} else {
}

}
System.out.print(out);
}

static int letterToNode(char letter) {
if ('a' <= letter && letter <= 'z') {
return letter - 'a';
} else {
return 26 + (letter - 'A');
}
}
}


Andi Qu's code:

ALPHABET = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'

t = int(input())
for test_case in range(t):
before = input()
after = input()

becomes = {}
possible = True
for i in range(len(before)):
if before[i] in becomes and becomes[before[i]] != after[i]:
possible = False
becomes[before[i]] = after[i]
if len(set(after)) == 52 and before != after:
possible = False

if possible:
in_degree = {}
for r in ALPHABET:
if r in becomes and becomes[r] != r:
in_degree[becomes[r]] = in_degree.get(becomes[r], 0) + 1
seen = {}
for r in ALPHABET:
if r not in seen:
a = r
while a not in seen:
seen[a] = r
a = becomes.get(a, a)
if a in becomes and a != becomes[a] and seen[a] == r:
s = a
cycle = True
while True:
seen[a] = 'moo'
if in_degree.get(a, 0) > 1:
cycle = False
a = becomes[a]
if a == s:
break
if cycle: