ShaysXIA/CodeInsightBench huggingface dataset:
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 186 ms):
t=int(input())
for _ in range(t):
  a=input()
  b=input()
  c=input()
  ans=float('inf')
  dp=[[-1 for i in range(len(b)+1)] for j in range(len(a)+1)]
  dp[0][0]=0
  for i in range(1,len(b)+1):
    dp[0][i]=dp[0][i-1]+ (b[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    dp[i][0]=dp[i-1][0]+(a[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    for j in range(1,len(b)+1):
      p=dp[i-1][j]+(a[i-1]!=c[i+j-1])
      q=dp[i][j-1]+(b[j-1]!=c[i+j-1])
      dp[i][j]=min(p,q)
  print(dp[len(a)][len(b)])

Solution 2 (runtime: 187 ms):
t=int(input())
for _ in range(t):
    a=input()
    b=input()
    c=input()
    dp=[[10**9 for j in range(len(b)+1)]for i in range(len(a)+1)]
    dp[0][0]=0
    i=0
    j=0
    for i in range(len(a)+1):
        for j in range(len(b)+1):
            
            if i==0 and j==0 :
                continue
            cost1=0
            cost2=0
            if i != 0 and a[i-1]!=c[i+j-1]:
                cost1=1
            if j!=0 and b[j-1]!=c[i+j-1]:
                cost2=1
            if i==0:
                dp[i][j]=dp[i][j-1]+cost2
            if j==0:
                dp[i][j]=dp[i-1][j]+cost1
            else:
                dp[i][j]=min(dp[i-1][j]+cost1,dp[i][j-1]+cost2)

    print(dp[len(a)][len(b)])
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 186 ms):
t=int(input())
for _ in range(t):
  a=input()
  b=input()
  c=input()
  ans=float('inf')
  dp=[[-1 for i in range(len(b)+1)] for j in range(len(a)+1)]
  dp[0][0]=0
  for i in range(1,len(b)+1):
    dp[0][i]=dp[0][i-1]+ (b[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    dp[i][0]=dp[i-1][0]+(a[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    for j in range(1,len(b)+1):
      p=dp[i-1][j]+(a[i-1]!=c[i+j-1])
      q=dp[i][j-1]+(b[j-1]!=c[i+j-1])
      dp[i][j]=min(p,q)
  print(dp[len(a)][len(b)])

Solution 2 (runtime: 671 ms):
for _ in range(int(input())):
    a=input()
    b=input()
    c=input()
    A=len(a)
    B=len(b)
    C=len(c)
    dp=[[10**10 for _ in range(A+1)]for _ in range(C+1)]
    dp[0][0]=0
    for i in range(C):
        for j in range(A+1):
            if j<A:
                if c[i]==a[j]:
                    dp[i+1][j+1]=min(dp[i][j],dp[i+1][j+1])
                else:
                    dp[i+1][j+1]=min(dp[i][j]+1,dp[i+1][j+1])
            if 0<=i-j<B:
                if c[i]==b[i-j]:
                    dp[i+1][j]=min(dp[i][j],dp[i+1][j])
                else:
                    dp[i+1][j]=min(dp[i][j]+1,dp[i+1][j])
    print(min(dp[-1]))
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 186 ms):
t=int(input())
for _ in range(t):
  a=input()
  b=input()
  c=input()
  ans=float('inf')
  dp=[[-1 for i in range(len(b)+1)] for j in range(len(a)+1)]
  dp[0][0]=0
  for i in range(1,len(b)+1):
    dp[0][i]=dp[0][i-1]+ (b[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    dp[i][0]=dp[i-1][0]+(a[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    for j in range(1,len(b)+1):
      p=dp[i-1][j]+(a[i-1]!=c[i+j-1])
      q=dp[i][j-1]+(b[j-1]!=c[i+j-1])
      dp[i][j]=min(p,q)
  print(dp[len(a)][len(b)])

Solution 2 (runtime: 280 ms):
# import sys
# sys.setrecursionlimit(2*(10**3))
# for _ in range(int(input())):
#     a = input()
#     b = input()
#     c = input()
#     dic = {}

#     def calculate(i,j,k,dic):
#         if (i,j,k) in dic:
#             return dic[(i,j,k)]
#         if i == len(a):
#             count = 0
#             while k <len(c):
#                 if c[k] != b[j]:
#                     count+=1
#                 k+=1
#                 j+=1
#             dic[(i,j,k)] = count
#             return count
#         if j == len(b):
#             count = 0
#             while k < len(c):
#                 if c[k] != a[i]:
#                     count+=1
#                 i+=1
#                 k+=1
#             dic[(i,j,k)] = count
#             return count 

#         if c[k]!=a[i] and c[k]!=b[j]:
#             count = 1 + min(calculate(i+1,j,k+1,dic),calculate(i,j+1,k+1,dic))
#         elif c[k]!=a[i] and c[k] == b[j]:
#             count = min(calculate(i,j+1,k+1,dic),calculate(i+1,j,k+1,dic)+1)
#         elif c[k]==a[i] and c[k] != b[j]:
#             count = min(calculate(i,j+1,k+1,dic)+1,calculate(i+1,j,k+1,dic))
#         else:
#             count = min(calculate(i+1,j,k+1,dic),calculate(i,j+1,k+1,dic))

#         dic[(i,j,k)] = count
#         return count    
            
#     print(calculate(0,0,0,dic))

for _ in range(int(input())):
    a = input()
    b = input()
    c = input()

    dp = [[float('inf')] * (len(b) + 1) for _ in range(len(a) + 1)]

    for i in range(len(a) + 1):
        for j in range(len(b) + 1):
            k = i + j
            if k > len(c):
                continue

            if i == 0 and j == 0:
                dp[i][j] = 0
            else:
                if i > 0 and c[k - 1] == a[i - 1]:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j])
                if j > 0 and c[k - 1] == b[j - 1]:
                    dp[i][j] = min(dp[i][j], dp[i][j - 1])
                if i > 0 and c[k - 1] != a[i - 1]:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1)
                if j > 0 and c[k - 1] != b[j - 1]:
                    dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1)

    print(dp[len(a)][len(b)])
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 186 ms):
t=int(input())
for _ in range(t):
  a=input()
  b=input()
  c=input()
  ans=float('inf')
  dp=[[-1 for i in range(len(b)+1)] for j in range(len(a)+1)]
  dp[0][0]=0
  for i in range(1,len(b)+1):
    dp[0][i]=dp[0][i-1]+ (b[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    dp[i][0]=dp[i-1][0]+(a[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    for j in range(1,len(b)+1):
      p=dp[i-1][j]+(a[i-1]!=c[i+j-1])
      q=dp[i][j-1]+(b[j-1]!=c[i+j-1])
      dp[i][j]=min(p,q)
  print(dp[len(a)][len(b)])

Solution 2 (runtime: 390 ms):
def min_changes(a, b, c):

    n, m = len(a), len(b)

    

    # DP array

    dp = [[float('inf')] * (m + 1) for _ in range(n + 1)]

    dp[0][0] = 0

    

    for i in range(n + 1):

        for j in range(m + 1):

            # Skip invalid states

            if dp[i][j] == float('inf'):

                continue

            

            k = i + j  # Position in c

            

            # We can take the next character from a

            if i < n:

                cost = 0 if a[i] == c[k] else 1

                dp[i+1][j] = min(dp[i+1][j], dp[i][j] + cost)

            

            # We can take the next character from b

            if j < m:

                cost = 0 if b[j] == c[k] else 1

                dp[i][j+1] = min(dp[i][j+1], dp[i][j] + cost)

    

    return dp[n][m]



def main():

    t = int(input())

    

    for _ in range(t):

        a = input().strip()

        b = input().strip()

        c = input().strip()

        

        print(min_changes(a, b, c))



if __name__ == "__main__":

    main()
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 186 ms):
t=int(input())
for _ in range(t):
  a=input()
  b=input()
  c=input()
  ans=float('inf')
  dp=[[-1 for i in range(len(b)+1)] for j in range(len(a)+1)]
  dp[0][0]=0
  for i in range(1,len(b)+1):
    dp[0][i]=dp[0][i-1]+ (b[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    dp[i][0]=dp[i-1][0]+(a[i-1]!=c[i-1])
  for i in range(1,len(a)+1):
    for j in range(1,len(b)+1):
      p=dp[i-1][j]+(a[i-1]!=c[i+j-1])
      q=dp[i][j-1]+(b[j-1]!=c[i+j-1])
      dp[i][j]=min(p,q)
  print(dp[len(a)][len(b)])

Solution 2 (runtime: 186 ms):
t = int(input())

for _ in range(t):

    a = input()

    b = input()

    c = input()

    n = len(a)

    m = len(b)



    ## 背包？？ 区间？？X

    ## 搜索 2**n X

    ##状态转移 dp？？题解 QWQ

    dp = [[0for i in range(m+1)]for i in range(n+1)]

    for i in range(n):

        dp[i+1][0] = dp[i][0]+(c[i]!=a[i])

    for j in range(m):

        dp[0][j+1] = dp[0][j]+(c[j]!=b[j])

    for i in range(n):

        for j in range(m):

            dp[i+1][j+1] = min(dp[i][j+1]+(a[i]!=c[i+j+1]),dp[i+1][j]+(b[j]!=c[i+j+1]))

    print(dp[n][m])
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 2389 ms):
az = int(input())
for _ in range(az):
    s1 = input()
    s2 = input()
    s3 = input()

    if len(s1) + len(s2) != len(s3):
        print(-1)
        continue

    # DP table to store minimum cost
    dp = [[float('inf')] * (len(s2) + 1) for _ in range(len(s1) + 1)]
    dp[0][0] = 0

    # Fill the DP table
    for i in range(len(s1) + 1):
        for j in range(len(s2) + 1):
            if i > 0:  # Taking a character from s1
                dp[i][j] = min(dp[i][j], dp[i - 1][j] + (s1[i - 1] != s3[i + j - 1]))
            if j > 0:  # Taking a character from s2
                dp[i][j] = min(dp[i][j], dp[i][j - 1] + (s2[j - 1] != s3[i + j - 1]))

    # Output the result for this test case
    print(dp[len(s1)][len(s2)])
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 218 ms):
import os
import sys
import math
from collections import defaultdict




IS_LOCAL = False
try:
    current_directory = os.path.dirname(os.path.abspath(__file__))
    if 'cp_cursed' in current_directory:
        IS_LOCAL = True
except NameError:
    pass
if IS_LOCAL:
    sys.stdout=open('outputf.in', 'w')
    sys.stdin=open('inputf.in', 'r')

lol=lambda func=int: list(map(func, sys.stdin.readline().split()))
lol1=lambda: sys.stdin.readline().split()


T=lol()[0]
for _ in range(T):
    s1=lol1()[0]
    s2=lol1()[0]
    s3=lol1()[0]
    dp=[[i+j for j in range(len(s2)+1)] for i in range(len(s1)+1)]
    for i in range(-1, len(s1)):
        for j in range(-1, len(s2)):            
            if i!=-1:
                if s1[i]==s3[i+j+1]:
                    dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j+1])
                dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j+1]+1)
            if j!=-1:
                if s2[j]==s3[i+j+1]:
                    dp[i+1][j+1]=min(dp[i+1][j+1],dp[i+1][j])
                dp[i+1][j+1]=min(dp[i+1][j+1],dp[i+1][j]+1)
    print(dp[-1][-1])
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 1467 ms):

def solve(a,b,c):
	dp = [[0] * (len(b) + 1) for _ in (range(len(a)+1))]

	d = 0
	for i in reversed(range(0,len(a))):
		if a[i] != c[i+len(b)]:
			d += 1
		dp[i][len(b)] = d
	d = 0
	for j in reversed(range(0,len(b))):
		if b[j] != c[len(a)+j]:
			d += 1
		dp[len(a)][j] = d

	for k in reversed(range(0,len(a)+len(b)-1)):
		x0 = max(0,k-(len(b)-1))
		x1 = min(k,(len(a)-1))
		for i in range(x0,x1+1):
			j = k-i
			res = None
			if i == len(a):
				res = sum(map(lambda x: x[0]!=x[1], zip(iter(b[j:]),iter(c[i+j:]))))
			elif j == len(b):
				res = sum(map(lambda x: x[0]!=x[1], zip(iter(a[i:]),iter(c[i+j:]))))
			else:
				res = min(
					dp[i+1][j] + (0 if a[i] == c[i+j] else 1),
					dp[i][j+1] + (0 if b[j] == c[i+j] else 1)
				)

			dp[i][j] = res

	return dp[0][0]

def test():
	print(solve("a","b","cb"))
	print(solve("ab","cd","acbd"))
	print(solve("ab","ba","aabb"))
	print(solve("xxx","yyy","xyxyxy"))
	print(solve("a","bcd","decf"))
	print(solve("codes","horse","codeforces"))
	print(solve("egg","annie","egaegaeg"))
	exit()

for _ in range(int(input())):
	print(solve(input(),input(),input()))
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 1687 ms):
def min_changes(a, b, c):
    n, m = len(a), len(b)
    dp = [[float('inf')] * (m + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    for i in range(n + 1):
        for j in range(m + 1):
            if i > 0:
                dp[i][j] = min(dp[i][j], dp[i-1][j] + (a[i-1] != c[i + j - 1]))
            if j > 0:
                dp[i][j] = min(dp[i][j], dp[i][j-1] + (b[j-1] != c[i + j - 1]))
    
    return dp[n][m]

t = int(input())
for _ in range(t):
    a = input().strip()
    b = input().strip()
    c = input().strip()
    print(min_changes(a, b, c))
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 1702 ms):
def min_changes(t, cases):
    results = []
    
    for a, b, c in cases:
        len_a, len_b = len(a), len(b)
        
        # Создаем DP таблицу
        dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
        dp[0][0] = 0  # Базовый случай: ничего не нужно менять, если строка пустая
        
        for i in range(len_a + 1):
            for j in range(len_b + 1):
                if i < len_a:  # Если можно использовать следующий символ из a
                    change_cost = 0 if c[i + j] == a[i] else 1
                    dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + change_cost)
                    
                if j < len_b:  # Если можно использовать следующий символ из b
                    change_cost = 0 if c[i + j] == b[j] else 1
                    dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + change_cost)
        
        results.append(dp[len_a][len_b])
    
    return results

# Чтение входных данных
t = int(input())
cases = [tuple(input().strip() for _ in range(3)) for _ in range(t)]

# Получение результатов
results = min_changes(t, cases)

# Печать результатов
for result in results:
    print(result)
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 2171 ms):
def min_changes_to_interleave(test_cases):
    results = []
    
    for a, b, c in test_cases:
        n, m = len(a), len(b)
        dp = [[float('inf')] * (m + 1) for _ in range(n + 1)]
        
        dp[0][0] = 0
        
        for i in range(n + 1):
            for j in range(m + 1):
                if i + j > len(c):
                    continue
                
                if i > 0:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j] + (c[i + j - 1] != a[i - 1]))
                
                if j > 0:
                    dp[i][j] = min(dp[i][j], dp[i][j - 1] + (c[i + j - 1] != b[j - 1]))
        
        results.append(dp[n][m])
    
    return results

# Example usage
t = int(input())
test_cases = [tuple(input().strip() for _ in range(3)) for _ in range(t)]
results = min_changes_to_interleave(test_cases)

for result in results:
    print(result)
Problem: 2050E - Three Strings
Language: python

Solution 1 (runtime: 281 ms):
def min_changes(a, b, c):
    len_a = len(a)
    len_b = len(b)
    len_c = len(c)
    
    # Initialize the DP table
    dp = [[float('inf')] * (len_b + 1) for _ in range(len_a + 1)]
    dp[0][0] = 0  # Base case, no change needed for empty strings
    
    # Fill the DP table
    for i in range(len_a + 1):
        for j in range(len_b + 1):
            if i + j > len_c:
                continue
            if i < len_a and a[i] == c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            if j < len_b and b[j] == c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
            if i < len_a and a[i] != c[i + j]:
                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + 1)
            if j < len_b and b[j] != c[i + j]:
                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1)
    
    return dp[len_a][len_b]

# Input
t = int(input())  # Number of test cases
for _ in range(t):
    a = input().strip()  # First string
    b = input().strip()  # Second string
    c = input().strip()  # Third string
    
    # Compute the result
    result = min_changes(a, b, c)
    
    # Output the result
    print(result)


Solution 2 (runtime: 374 ms):
for _ in range(int(input())):
    a = input()
    b = input()
    c = input()

    m = len(a)
    n = len(b)
    
    a = '$' + a
    b = '$' + b
    c = '$' + c

    dp = [[None for j in range(n + 1)] for i in range(m + 1)]
    dp[0][0] = 0

    for i in range(1, m + 1):
        if a[i] == c[i]:
            dp[i][0] = dp[i - 1][0]
        else:
            dp[i][0] = 1 + dp[i - 1][0]

    for j in range(1, n + 1):
        if b[j] == c[j]:
            dp[0][j] = dp[0][j - 1]
        else:
            dp[0][j] = 1 + dp[0][j - 1]

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if a[i] == c[i + j]:
                if b[j] == c[i + j]:
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1])
                else:
                    dp[i][j] = min(dp[i - 1][j], 1 + dp[i][j - 1])
            else:
                if b[j] == c[i + j]:
                    dp[i][j] = min(1 + dp[i - 1][j], dp[i][j - 1])
                else:
                    dp[i][j] = min(1 + dp[i - 1][j], 1 + dp[i][j - 1])

    print(dp[m][n])