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How to get Netcore Backend developer coding answers
def count_leading_zeros(num): # Count leading zeros in a number str_num = str(num) count = 0 for digit in str_num: if digit == '0': count += 1 else: break return count def solve(n, tiles): # Create DP array to store minimum leading zeros dp = [[float('inf')] * n for _ in range(n)] # Create array toRead more
def count_leading_zeros(num):
# Count leading zeros in a number
str_num = str(num)
count = 0
for digit in str_num:
if digit == ‘0’:
count += 1
else:
break
return count
def solve(n, tiles):
# Create DP array to store minimum leading zeros
dp = [[float(‘inf’)] * n for _ in range(n)]
# Create array to store products
products = [[1] * n for _ in range(n)]
# Initialize first cell
dp[0][0] = count_leading_zeros(tiles[0][0])
products[0][0] = tiles[0][0]
# Fill first row
for j in range(1, n):
products[0][j] = products[0][j-1] * tiles[0][j]
dp[0][j] = count_leading_zeros(products[0][j])
# Fill first column
for i in range(1, n):
products[i][0] = products[i-1][0] * tiles[i][0]
dp[i][0] = count_leading_zeros(products[i][0])
# Fill rest of the DP table
for i in range(1, n):
for j in range(1, n):
# Calculate product from top
top_product = products[i-1][j] * tiles[i][j]
top_zeros = count_leading_zeros(top_product)
# Calculate product from left
left_product = products[i][j-1] * tiles[i][j]
left_zeros = count_leading_zeros(left_product)
# Choose minimum leading zeros path
if top_zeros <= left_zeros:
dp[i][j] = top_zeros
products[i][j] = top_product
else:
dp[i][j] = left_zeros
products[i][j] = left_product
return dp[n-1][n-1]
See lessNetcore hiring challenge coding answer
def count_leading_zeros(num): # Count leading zeros in a number str_num = str(num) count = 0 for digit in str_num: if digit == '0': count += 1 else: break return count def solve(n, tiles): # Create DP array to store minimum leading zeros dp = [[float('inf')] * n for _ in range(n)] # Create array toRead more
def count_leading_zeros(num):
# Count leading zeros in a number
str_num = str(num)
count = 0
for digit in str_num:
if digit == ‘0’:
count += 1
else:
break
return count
def solve(n, tiles):
# Create DP array to store minimum leading zeros
dp = [[float(‘inf’)] * n for _ in range(n)]
# Create array to store products
products = [[1] * n for _ in range(n)]
# Initialize first cell
dp[0][0] = count_leading_zeros(tiles[0][0])
products[0][0] = tiles[0][0]
# Fill first row
for j in range(1, n):
products[0][j] = products[0][j-1] * tiles[0][j]
dp[0][j] = count_leading_zeros(products[0][j])
# Fill first column
for i in range(1, n):
products[i][0] = products[i-1][0] * tiles[i][0]
dp[i][0] = count_leading_zeros(products[i][0])
# Fill rest of the DP table
for i in range(1, n):
for j in range(1, n):
# Calculate product from top
top_product = products[i-1][j] * tiles[i][j]
top_zeros = count_leading_zeros(top_product)
# Calculate product from left
left_product = products[i][j-1] * tiles[i][j]
left_zeros = count_leading_zeros(left_product)
# Choose minimum leading zeros path
if top_zeros <= left_zeros:
dp[i][j] = top_zeros
products[i][j] = top_product
else:
dp[i][j] = left_zeros
products[i][j] = left_product
return dp[n-1][n-1]
See lessNetcore Backend developer coding answer
def count_leading_zeros(num): # Count leading zeros in a number str_num = str(num) count = 0 for digit in str_num: if digit == '0': count += 1 else: break return count def solve(n, tiles): # Create DP array to store minimum leading zeros dp = [[float('inf')] * n for _ in range(n)] # Create array toRead more
def count_leading_zeros(num):
# Count leading zeros in a number
str_num = str(num)
count = 0
for digit in str_num:
if digit == ‘0’:
count += 1
else:
break
return count
def solve(n, tiles):
# Create DP array to store minimum leading zeros
dp = [[float(‘inf’)] * n for _ in range(n)]
# Create array to store products
products = [[1] * n for _ in range(n)]
# Initialize first cell
dp[0][0] = count_leading_zeros(tiles[0][0])
products[0][0] = tiles[0][0]
# Fill first row
for j in range(1, n):
products[0][j] = products[0][j-1] * tiles[0][j]
dp[0][j] = count_leading_zeros(products[0][j])
# Fill first column
for i in range(1, n):
products[i][0] = products[i-1][0] * tiles[i][0]
dp[i][0] = count_leading_zeros(products[i][0])
# Fill rest of the DP table
for i in range(1, n):
for j in range(1, n):
# Calculate product from top
top_product = products[i-1][j] * tiles[i][j]
top_zeros = count_leading_zeros(top_product)
# Calculate product from left
left_product = products[i][j-1] * tiles[i][j]
left_zeros = count_leading_zeros(left_product)
# Choose minimum leading zeros path
if top_zeros <= left_zeros:
dp[i][j] = top_zeros
products[i][j] = top_product
else:
dp[i][j] = left_zeros
products[i][j] = left_product
return dp[n-1][n-1]
See lessCoding answer of Netcore hiring challenge
def count_leading_zeros(num): # Count leading zeros in a number str_num = str(num) count = 0 for digit in str_num: if digit == '0': count += 1 else: break return count def solve(n, tiles): # Create DP array to store minimum leading zeros dp = [[float('inf')] * n for _ in range(n)] # Create array toRead more
def count_leading_zeros(num):
# Count leading zeros in a number
str_num = str(num)
count = 0
for digit in str_num:
if digit == ‘0’:
count += 1
else:
break
return count
def solve(n, tiles):
# Create DP array to store minimum leading zeros
dp = [[float(‘inf’)] * n for _ in range(n)]
# Create array to store products
products = [[1] * n for _ in range(n)]
# Initialize first cell
dp[0][0] = count_leading_zeros(tiles[0][0])
products[0][0] = tiles[0][0]
# Fill first row
for j in range(1, n):
products[0][j] = products[0][j-1] * tiles[0][j]
dp[0][j] = count_leading_zeros(products[0][j])
# Fill first column
for i in range(1, n):
products[i][0] = products[i-1][0] * tiles[i][0]
dp[i][0] = count_leading_zeros(products[i][0])
# Fill rest of the DP table
for i in range(1, n):
for j in range(1, n):
# Calculate product from top
top_product = products[i-1][j] * tiles[i][j]
top_zeros = count_leading_zeros(top_product)
# Calculate product from left
left_product = products[i][j-1] * tiles[i][j]
left_zeros = count_leading_zeros(left_product)
# Choose minimum leading zeros path
if top_zeros <= left_zeros:
dp[i][j] = top_zeros
products[i][j] = top_product
else:
dp[i][j] = left_zeros
products[i][j] = left_product
return dp[n-1][n-1]
See lessIs paid TCS NQT really worth it?
Yes , i paid in 2023 for tcs nqt and clear the exam so i think it is really worth it if you have a skills
Yes , i paid in 2023 for tcs nqt and clear the exam so i think it is really worth it if you have a skills
See lessIf any referal from Cognizant,Deloitte Please reply me
https://www.linkedin.com/posts/likitha-thalari-583838220_fullstackdeveloper-reactjs-nodejs-activity-7249316458915061760-Bv2x?utm_source=share&utm_medium=member_desktop Not from mentioned company but check this
https://www.linkedin.com/posts/likitha-thalari-583838220_fullstackdeveloper-reactjs-nodejs-activity-7249316458915061760-Bv2x?utm_source=share&utm_medium=member_desktop
Not from mentioned company but check this
See lessReferral
https://www.linkedin.com/posts/namitha-kolutthara-henry-56319a236_javadeveloper-javadeveloper-gamedeveloper-activity-7249302533632864256-EZhU?utm_source=share&utm_medium=member_desktop
https://www.linkedin.com/posts/namitha-kolutthara-henry-56319a236_javadeveloper-javadeveloper-gamedeveloper-activity-7249302533632864256-EZhU?utm_source=share&utm_medium=member_desktop
See less