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VRV security is scam or not?
I think yes coz they are asking money to purchase some software. My friend has paid the amount but not get any response from them everyone is curious about this company and no proper response get from them. One of my friend get offer letter without giving interview how is it possible and after thatRead more
I think yes coz they are asking money to purchase some software. My friend has paid the amount but not get any response from them everyone is curious about this company and no proper response get from them. One of my friend get offer letter without giving interview how is it possible and after that they told to purchase one software for 2.5k
See lessHow 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 less