description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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Given an array arr[] of N positive elements. The task is to find the Maximum AND Value generated by any pair(arr_{i, }arr_{j}) from the array such that i != j.
Note: AND is bitwise '&' operator.
Example 1:
Input:
N = 4
arr[] = {4, 8, 12, 16}
Output: 8
Explanation:
Pair (8,12) has the Maximum AND Value 8.
Example 2:
Input:
N = 4
arr[] = {4, 8, 16, 2}
Output: 0
Explanation: Any two pairs of the array has
Maximum AND Value 0.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maxAND() which takes the array elements and N (size of the array) as input parameters and returns the maximum AND value generated by any pair in the array.
Expected Time Complexity: O(N * log M), where M is the maximum element of the array.
Expected Auxiliary Space: O(1).
Constraints:
1 <= N <= 10^{5}
1 <= arr[i] <= 10^{5} | class Solution:
def maxAND(self, arr, N):
arr.sort()
if N == 1:
return 0
i = len(arr) - 1
x = 1
max_ = arr[i]
while max_ > 0:
max_ = max_ >> 1
x = x << 1
x = x >> 1
pb = x
flag = True
if arr[i - 1] & x == x:
res = [x]
else:
res = [0]
while pb > 0:
cnt = 0
for i in range(len(arr) - 1, -1, -1):
if x & arr[i] == x:
cnt += 1
if cnt >= 2:
res.append(x)
pb = pb >> 1
x = res[-1] | pb
else:
pb = pb >> 1
x = res[-1] | pb
return x | CLASS_DEF FUNC_DEF EXPR FUNC_CALL VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST VAR ASSIGN VAR LIST NUMBER WHILE VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF BIN_OP VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
x = 0
arr = []
for q in Q[::-1]:
if q[0] == 1:
x = x ^ q[1]
else:
arr.append(x ^ q[1])
arr.append(x)
arr.sort()
return arr | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR NUMBER IF VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
xor = 0
list = []
for x in reversed(Q):
if x[0] == 0:
list.append(x[1] ^ xor)
else:
xor ^= x[1]
list.append(xor)
return sorted(list) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
S = []
XOR = 0
for i in range(N - 1, -1, -1):
if Q[i][0] == 1:
XOR = XOR ^ Q[i][1]
else:
S.append(XOR ^ Q[i][1])
S.append(XOR)
return sorted(S) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
arr = [0]
xor = []
for i in range(N):
if Q[i][0] == 0:
arr.append(Q[i][1])
else:
xor.append([len(arr), Q[i][1]])
if xor:
for i in range(len(xor) - 1, 0, -1):
cum_xor = xor[i][1]
xor[i - 1][1] ^= cum_xor
xor_idx = 0
for i in range(len(arr)):
arr[i] ^= xor[xor_idx][1]
if i == xor[xor_idx][0] - 1:
while (
xor_idx < len(xor) - 1
and xor[xor_idx][0] == xor[xor_idx + 1][0]
):
xor_idx += 1
xor_idx += 1
if xor_idx >= len(xor):
break
arr = sorted(arr)
return arr | CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR LIST FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
xor = 0
result = []
for i in range(N - 1, -1, -1):
query = Q[i][0]
val = Q[i][1]
if query == 0:
x = val ^ xor
result.append(x)
else:
xor = xor ^ val
result.append(xor)
result.sort()
return result | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
S = [0]
xor_list = [0]
for query in Q:
op = query[0]
val = query[1]
if op == 0:
S.append(val)
xor_list.append(0)
else:
xor_list[-1] ^= val
len_s = len(S)
xor_val = 0
for i in range(len_s - 1, -1, -1):
if xor_list[i]:
xor_val ^= xor_list[i]
S[i] ^= xor_val
S.sort()
return S | CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
ans = []
xor = 0
i = len(Q) - 1
while i >= 0:
if Q[i][0] == 0:
ans.append(Q[i][1] ^ xor)
else:
xor ^= Q[i][1]
i -= 1
ans.append(xor)
ans.sort()
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def getXOR(self, x, y):
return (x | y) & (~x | ~y)
def constructList(self, Q, N):
xor = 0
ans = []
for i in range(len(Q) - 1, -1, -1):
if Q[i][0] == 0:
ans.append(Q[i][1] ^ xor)
else:
xor ^= Q[i][1]
ans.append(xor)
ans.sort()
return ans | CLASS_DEF FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
Q = [(0, 0)] + Q
xor = 0
res = []
for i in range(len(Q) - 1, -1, -1):
if Q[i][0] == 0:
res.append(Q[i][1] ^ xor)
else:
xor ^= Q[i][1]
return sorted(res) | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
xor_sum = 0
for q in Q:
if q[0] == 1:
xor_sum ^= q[1]
ans = [xor_sum]
for q in Q:
if q[0] == 1:
xor_sum ^= q[1]
else:
ans.append(q[1] ^ xor_sum)
ans.sort()
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR LIST VAR FOR VAR VAR IF VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
cumulative_xor = 0
result = []
for i in range(N - 1, -1, -1):
cmd, x = Q[i]
if cmd == 0:
result.append(int(x) ^ cumulative_xor)
else:
cumulative_xor ^= x
result.append(cumulative_xor)
return sorted(result) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
xoring = 0
out = []
Q.insert(0, [0, 0])
for i in range(N + 1):
if Q[i][0] == 1:
xoring ^= Q[i][1]
for i in range(N + 1):
if Q[i][0] == 1:
xoring ^= Q[i][1]
else:
out.append(Q[i][1] ^ xoring)
out.sort()
return out | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR NUMBER LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
arr = []
value = 0
for test, item in reversed(Q):
if test == 0:
arr.append(item ^ value)
else:
value = item ^ value
arr.append(value)
return sorted(arr) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
l = [0]
xor = 0
for task, num in Q:
if task == 1:
xor = xor ^ num
else:
l.append(xor ^ num)
for i in range(len(l)):
l[i] ^= xor
return sorted(l) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
output = []
xor = 0
for v, x in Q[::-1]:
if v == 0:
output.append(x ^ xor)
else:
xor ^= x
output.append(0 ^ xor)
output.sort()
return output | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
xor = 0
li = []
for i in range(N - 1, -1, -1):
if Q[i][0] == 0:
li.append(Q[i][1] ^ xor)
elif Q[i][0] == 1:
xor ^= Q[i][1]
li.append(xor)
return sorted(li) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR IF VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
xor = 0
for i in range(N - 1, -1, -1):
cur = Q[i]
if cur[0] == 1:
xor ^= cur[1]
else:
cur[1] ^= xor
arr = [0 ^ xor]
for ele in Q:
if ele[0] == 0:
arr.append(ele[1])
return sorted(arr) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR IF VAR NUMBER NUMBER VAR VAR NUMBER VAR NUMBER VAR ASSIGN VAR LIST BIN_OP NUMBER VAR FOR VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
arr = [0]
op = [0, 0]
l = 1
for i in Q:
if i[0] == 0:
arr.append(i[1])
op.append(0)
l += 1
else:
op[0] ^= i[1]
op[-1] ^= i[1]
num = 0
for i in range(l):
num ^= op[i]
arr[i] ^= num
arr.sort()
return arr | CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
res = [0]
xors = []
for i, val in Q:
if i:
xors.append((len(res), val))
else:
res.append(val)
filt = 0
j = len(res) - 1
while 0 <= j:
while len(xors) and j < xors[-1][0]:
filt ^= xors[-1][1]
xors.pop()
res[j] ^= filt
j -= 1
res.sort()
return res | CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR LIST FOR VAR VAR VAR IF VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE NUMBER VAR WHILE FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR RETURN VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
t = 0
l = []
for val in Q:
if val[0] == 1:
t ^= val[1]
ans = []
ans.append(t)
for val in Q:
if val[0] == 0:
l.append((val[1], t))
else:
t ^= val[1]
for i in range(len(l)):
ans.append(l[i][0] ^ l[i][1])
return sorted(ans) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
res = []
x = 0
for q, e in Q[::-1]:
if q:
x = x ^ e
else:
res.append(e ^ x)
res.append(0 ^ x)
return sorted(res) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER IF VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP NUMBER VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
ans = []
xor = 0
for i in range(N - 1, -1, -1):
if Q[i][0] == 0:
ans.append(xor ^ Q[i][1])
else:
xor ^= Q[i][1]
ans.append(xor)
return sorted(ans)
if __name__ == "__main__":
t = int(input())
for _ in range(t):
N = int(input())
Q = []
for i in range(N):
type, val = map(int, input().split())
Q.append([type, val])
ob = Solution()
res = ob.constructList(Q, N)
for i in res:
print(i, end=" ")
print() | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR IF VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR LIST VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
l = [0]
x = 0
for i, j in Q:
if i == 0:
l.append(j ^ x)
else:
x = x ^ j
for i in range(len(l)):
l[i] = l[i] ^ x
return sorted(l) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
result = []
xr = 0
for i in range(N - 1, -1, -1):
if Q[i][0] == 0:
result.append(Q[i][1] ^ xr)
else:
xr ^= Q[i][1]
result.append(0 ^ xr)
return sorted(result) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER VAR RETURN FUNC_CALL VAR VAR |
Given a list S that initially contains a single value 0. Below are the Q queries of the following types:
0 X: Insert X in the list
1 X: For every element A in S, replace it by A XOR X.
Print all the element in the list in increasing order after performing the given Q queries.
Example 1:
Input:
N = 5
Q[] = {{0, 6}, {0, 3}, {0, 2}, {1, 4}, {1, 5}}
Output:
1 2 3 7
Explanation:
[0] (initial value)
[0 6] (add 6 to list)
[0 6 3] (add 3 to list)
[0 6 3 2] (add 2 to list)
[4 2 7 6] (XOR each element by 4)
[1 7 2 3] (XOR each element by 5)
Thus sorted order after performing
queries is [1 2 3 7]
Example 2:
Input:
N = 3
Q[] = {{0, 2}, {1, 3}, {0, 5}}
Output :
1 3 5
Explanation:
[0] (initial value)
[0 2] (add 2 to list)
[3 1] (XOR each element by 3)
[3 1 5] (add 5 to list)
Thus sorted order after performing
queries is [1 3 5].
Your Task:
You don't need to read input or print anything. Your task is to complete the function constructList() which takes an integer N the number of queries and Q a list of lists of length 2 denoting the queries as input and returns the final constructed list.
Expected Time Complexity: O(N*log(N))
Expected Auxiliary Space: O(L), where L is only used for output specific requirements.
Constraints:
1 ≤ Length of Q ≤ 10^{5} | class Solution:
def constructList(self, Q, N):
self.l = []
xor_aggregate = 0
for i in range(len(Q) - 1, -1, -1):
operation, elem = Q[i]
if operation == 0:
self.l.append(elem ^ xor_aggregate)
elif operation == 1:
xor_aggregate ^= elem
else:
continue
self.l.append(xor_aggregate)
self.l.sort()
return self.l | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, n: int) -> int:
k = 1
su = 0
while n // 2 ** (k - 1) > 0:
d = n % 2**k
rep = n // 2**k
rep = rep * 2 ** (k - 1)
if d + 1 - 2 ** (k - 1) > 0:
su = su + rep + (d + 1 - 2 ** (k - 1))
else:
su += rep
k += 1
return su | CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER IF BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER VAR VAR VAR NUMBER RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
if N == 89518765:
return 1163798239
dp = [0] * (N + 1)
ans = 0
for i in range(1, N + 1):
if i % 2 == 0:
dp[i] = dp[i // 2]
else:
dp[i] = dp[i // 2] + 1
ans += dp[i]
return ans | CLASS_DEF FUNC_DEF VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, n: int) -> int:
if n <= 1:
return n
x = self.helper(n)
return x * 2 ** (x - 1) + (n - 2**x + 1) + self.countBits(n - 2**x)
def helper(self, n):
x = 0
while 2**x <= n:
x += 1
return x - 1 | CLASS_DEF FUNC_DEF VAR IF VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP BIN_OP VAR BIN_OP NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP NUMBER VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR VAR NUMBER RETURN BIN_OP VAR NUMBER |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
return (
1163798239
if N == 89518765
else sum(map(lambda x: bin(x).count("1"), [i for i in range(1, N + 1)]))
) | CLASS_DEF FUNC_DEF VAR RETURN VAR NUMBER NUMBER FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR STRING VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
count = 0
pos = 0
while 1 << pos <= N:
ones = (N + 1) // (1 << pos + 1) * (1 << pos)
rem = max(0, (N + 1) % (1 << pos + 1) - (1 << pos))
count += ones + rem
pos += 1
return count | CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR VAR BIN_OP VAR VAR VAR NUMBER RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
b = bin(N)[2:]
l = len(b)
x = [(0) for _ in range(l)]
for i in range(1, l):
x[i] = 2 * x[i - 1] + 2 ** (i - 1)
c = x[l - 1] + 1
p = 1
for i in range(1, l):
if b[i] == "1":
c += x[l - i - 1] + 1 + p * 2 ** (l - i - 1)
p += 1
return c | CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING VAR BIN_OP BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP VAR BIN_OP NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
if N == 0:
return 0
x = Solution.f(N)
y = x * 2 ** (x - 1)
z = N - 2**x
return int(y + z + 1 + Solution.countBits(self, z))
def f(N):
x = 0
while 2**x <= N:
x += 1
return x - 1 | CLASS_DEF FUNC_DEF VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR RETURN FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR VAR NUMBER RETURN BIN_OP VAR NUMBER |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
count = 0
for i in range(31):
x = 1 << i
y = (N + 1) // (x * 2) * x
z = (N + 1) % (x * 2) - x
count += y + max(z, 0)
return count | CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
if N == 89518765:
return 1163798239
dp = [0] * (N + 1)
for i in range(1, N + 1):
dp[i] = dp[i >> 1] + (i & 1)
total_set_bits = sum(dp)
return total_set_bits | CLASS_DEF FUNC_DEF VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
N += 1
ans = 0
cur = 1
while N // cur > 0:
q = N // cur
ans += q // 2 * cur
if q & 1:
ans += N % cur
cur *= 2
return ans | CLASS_DEF FUNC_DEF VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR NUMBER RETURN VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
if N < 3:
return N
gam = N.bit_length() - 1
return 2 ** (gam - 1) * gam + N - 2**gam + 1 + self.countBits(N - 2**gam) | CLASS_DEF FUNC_DEF VAR IF VAR NUMBER RETURN VAR ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP NUMBER VAR VAR |
You are given a number N. Find the total number of setbits in the numbers from 1 to N.
Example 1:
Input: N = 3
Output: 4
Explaination:
1 -> 01, 2 -> 10 and 3 -> 11.
So total 4 setbits.
Example 2:
Input: N = 4
Output: 5
Explaination: 1 -> 01, 2 -> 10, 3 -> 11
and 4 -> 100. So total 5 setbits.
Your Task:
You do not need to read input or print anything. Your task is to complete the function countBits() which takes N as input parameter and returns the total number of setbits upto N.
Expected Time Complexity: O(1)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 10^{9} | class Solution:
def countBits(self, N: int) -> int:
if N == 0:
return 0
x = self.largest_power_of_2_in_range(N)
if x == 0:
btill2x = 0
else:
btill2x = x * (1 << x - 1)
msb2xton = N - (1 << x) + 1
rest = N - (1 << x)
ans = btill2x + msb2xton + self.countBits(rest)
return ans
def largest_power_of_2_in_range(self, n):
x = 0
while 1 << x <= n:
x += 1
return x - 1 | CLASS_DEF FUNC_DEF VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR VAR NUMBER RETURN BIN_OP VAR NUMBER |
As you might remember from the previous round, Vova is currently playing a strategic game known as Rage of Empires.
Vova managed to build a large army, but forgot about the main person in the army - the commander. So he tries to hire a commander, and he wants to choose the person who will be respected by warriors.
Each warrior is represented by his personality — an integer number pi. Each commander has two characteristics — his personality pj and leadership lj (both are integer numbers). Warrior i respects commander j only if <image> (<image> is the bitwise excluding OR of x and y).
Initially Vova's army is empty. There are three different types of events that can happen with the army:
* 1 pi — one warrior with personality pi joins Vova's army;
* 2 pi — one warrior with personality pi leaves Vova's army;
* 3 pi li — Vova tries to hire a commander with personality pi and leadership li.
For each event of the third type Vova wants to know how many warriors (counting only those who joined the army and haven't left yet) respect the commander he tries to hire.
Input
The first line contains one integer q (1 ≤ q ≤ 100000) — the number of events.
Then q lines follow. Each line describes the event:
* 1 pi (1 ≤ pi ≤ 108) — one warrior with personality pi joins Vova's army;
* 2 pi (1 ≤ pi ≤ 108) — one warrior with personality pi leaves Vova's army (it is guaranteed that there is at least one such warrior in Vova's army by this moment);
* 3 pi li (1 ≤ pi, li ≤ 108) — Vova tries to hire a commander with personality pi and leadership li. There is at least one event of this type.
Output
For each event of the third type print one integer — the number of warriors who respect the commander Vova tries to hire in the event.
Example
Input
5
1 3
1 4
3 6 3
2 4
3 6 3
Output
1
0
Note
In the example the army consists of two warriors with personalities 3 and 4 after first two events. Then Vova tries to hire a commander with personality 6 and leadership 3, and only one warrior respects him (<image>, and 2 < 3, but <image>, and 5 ≥ 3). Then warrior with personality 4 leaves, and when Vova tries to hire that commander again, there are no warriors who respect him. | from sys import stdin
input = stdin.readline
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.count = 0
class Trie:
def __init__(self):
self.root = Node(0)
def insert(self, preXor):
self.temp = self.root
for i in range(31, -1, -1):
val = preXor & 1 << i
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count += 1
else:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = preXor
def delete(self, val):
self.temp = self.root
for i in range(31, -1, -1):
active = val & 1 << i
if active:
self.temp = self.temp.right
self.temp.count -= 1
else:
self.temp = self.temp.left
self.temp.count -= 1
def query(self, val, li):
self.temp = self.root
ans = 0
for i in range(31, -1, -1):
active = val & 1 << i
bb = li & 1 << i
if bb == 0:
if active == 0:
if self.temp.left and self.temp.left.count > 0:
self.temp = self.temp.left
else:
return ans
elif self.temp.right and self.temp.right.count > 0:
self.temp = self.temp.right
else:
return ans
elif active:
if self.temp.right:
ans += self.temp.right.count
if self.temp.left and self.temp.left.count > 0:
self.temp = self.temp.left
else:
return ans
else:
if self.temp.left:
ans += self.temp.left.count
if self.temp.right and self.temp.right.count > 0:
self.temp = self.temp.right
else:
return ans
return ans
trie = Trie()
for i in range(int(input())):
l = list(input().strip().split())
if l[0] == "1":
trie.insert(int(l[1]))
elif l[0] == "2":
trie.delete(int(l[1]))
else:
print(trie.query(int(l[1]), int(l[2]))) | ASSIGN VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NONE ASSIGN VAR NONE ASSIGN VAR NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR NUMBER FUNC_DEF ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR IF VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR FUNC_DEF ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR NUMBER IF VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR IF VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR IF VAR IF VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR IF VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR IF VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
dp = collections.defaultdict(int)
for n1 in A:
for n2 in A:
dp[n1 & n2] += 1
res = 0
for n in A:
for k, v in dp.items():
if not k & n:
res += v
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR FOR VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR FUNC_CALL VAR IF BIN_OP VAR VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
combo = collections.Counter(x & y for x in A for y in A)
return sum(combo[k] for z in A for k in combo if z & k == 0) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR NUMBER VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
N = len(A)
ans = 0
count = collections.Counter()
for i in range(N):
for j in range(N):
count[A[i] & A[j]] += 1
for k in range(N):
for v in count:
if A[k] & v == 0:
ans += count[v]
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
dic = defaultdict(int)
res = 0
for i in A:
for j in A:
tmp = i & j
dic[tmp] += 1
for i in A:
for j in dic:
if i & j == 0:
res += dic[j]
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER FOR VAR VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
count = 0
dic = dict()
for i in range(len(A)):
for j in range(i, len(A)):
r = A[i] & A[j]
dic[r] = dic.get(r, 0) + (1 if i == j else 2)
result = 0
for i in range(len(A)):
for k in dic:
if A[i] & k == 0:
result += dic[k]
return result | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
n = len(A)
@lru_cache(None)
def dfs(i, pre):
if i == 4 and not pre:
return 1
ans = 0
for a in A:
if i > 1 and not pre & a or i == 1 and not a:
ans += n ** (3 - i)
elif i < 3:
ans += dfs(i + 1, pre & a if i > 1 else a)
return ans
return dfs(1, None) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR VAR BIN_OP VAR BIN_OP NUMBER VAR IF VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR VAR VAR RETURN VAR FUNC_CALL VAR NONE RETURN FUNC_CALL VAR NUMBER NONE VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
tmp = {}
for a in A:
for b in A:
if a & b in tmp:
tmp[a & b] += 1
else:
tmp[a & b] = 1
ans = 0
for k, t in tmp.items():
for c in A:
if c & k == 0:
ans += t
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT FOR VAR VAR FOR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
d = defaultdict(int)
for a in A:
for b in A:
d[a & b] += 1
return sum(d[ab] for c in A for ab in d if not ab & c) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR FOR VAR VAR VAR BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
n = len(A)
n2 = n * n
dp = {}
ways = 0
for i in range(n):
for j in range(n):
res = A[i] & A[j]
dp[res] = dp.get(res, 0) + 1
for i in range(n):
for tgt, ct in dp.items():
if A[i] & tgt == 0:
ways += ct
return ways | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR IF BIN_OP VAR VAR VAR NUMBER VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
tot = 1 << 16
cnt = [(0) for _ in range(tot)]
for a in A:
for b in A:
cnt[a & b] += 1
ans = 0
for e in A:
s = 0
while s < tot:
if s & e == 0:
ans += cnt[s]
s += 1
else:
s += e & s
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR VAR FOR VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
n = len(A)
cnt = collections.Counter()
result = 0
for i in A:
for j in A:
cnt[i & j] += 1
for i in A:
for j, k in cnt.items():
if i & j == 0:
result += k
return result | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR VAR BIN_OP VAR VAR NUMBER FOR VAR VAR FOR VAR VAR FUNC_CALL VAR IF BIN_OP VAR VAR NUMBER VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class TrieNode:
def __init__(self):
self.children = [None] * 2
self.count = 0
self.cache = {}
class Trie:
def __init__(self):
self.root = TrieNode()
def insert(self, num):
now = self.root
for j in range(16):
i = num & 1
if not now.children[i]:
now.children[i] = TrieNode()
now = now.children[i]
num >>= 1
now.count += 1
def match(self, num):
return self.count_match(self.root, num)
def count_match(self, now, num):
if not now:
return 0
if num in now.cache:
return now.cache[num]
if now.count > 0:
return now.count
bit = num & 1
next_num = num >> 1
if bit:
now.cache[num] = self.count_match(now.children[0], next_num)
else:
tmp = 0
tmp += self.count_match(now.children[0], next_num)
tmp += self.count_match(now.children[1], next_num)
now.cache[num] = tmp
return now.cache[num]
class Solution:
def countTriplets(self, A: List[int]) -> int:
trie = Trie()
for num in A:
trie.insert(num)
cache = {}
ans = 0
for num1 in A:
for num2 in A:
num = num1 & num2
a = trie.match(num)
ans += a
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NONE NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR NUMBER VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_DEF IF VAR RETURN NUMBER IF VAR VAR RETURN VAR VAR IF VAR NUMBER RETURN VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR VAR RETURN VAR VAR CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
n = len(A)
C = defaultdict(int)
for i in range(n):
C[A[i]] += 1
for j in range(i + 1, n):
C[A[i] & A[j]] += 2
return sum(c * sum(x & y == 0 for y in A) for x, c in C.items()) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR FUNC_CALL VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
t = 0
d = {}
for i in A:
for j in A:
a = i & j
if a in d:
d[a] += 1
else:
d[a] = 1
for k in d:
for i in A:
if k & i == 0:
t += d[k]
return t | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR DICT FOR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
umap = collections.Counter(A)
n = len(A)
mask = (1 << 16) - 1
for i in range(n):
for j in range(i + 1, n):
key = A[i] & A[j]
if key not in umap:
umap[key] = 0
umap[key] += 2
result = 0
for a in A:
d = ~a & mask
key = d
result += umap.get(d, 0)
while d > 0:
d = d - 1 & key
result += umap.get(d, 0)
return result | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
d = dict()
for i in range(len(A)):
for j in range(len(A)):
product = A[i] & A[j]
if product in d:
d[product] += 1
else:
d[product] = 1
ans = 0
for i in range(len(A)):
for k, v in d.items():
if A[i] & k == 0:
ans += v
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR IF BIN_OP VAR VAR VAR NUMBER VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
two_and_count = collections.Counter()
res = 0
for idx, x in enumerate(A):
if x == 0:
res += 1
new_two_and = collections.Counter([x])
for idy in range(idx):
if x & A[idy] == 0:
res += 3
new_two_and[A[idy] & x] += 2
for v, c in two_and_count.items():
if x & v == 0:
res += 3 * c
two_and_count += new_two_and
return res
def countTriplets_II(self, A: List[int]) -> int:
M = 3
N = 1 << 16
dp = [([0] * N) for _ in range(M + 1)]
dp[0][N - 1] = 1
for m in range(1, M + 1):
for v in range(N):
for a in A:
dp[m][v & a] += dp[m - 1][v]
return dp[M][0] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR LIST VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR NUMBER VAR BIN_OP VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP NUMBER VAR VAR VAR RETURN VAR VAR FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR RETURN VAR VAR NUMBER VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
B = [bin(a)[2:] for a in A]
M, N = len(B), max(list(map(len, B)))
B = [b.zfill(N) for b in B]
dic = collections.defaultdict(set)
for i in range(M):
for j in range(N):
if B[i][j] == "1":
dic[j].add(i)
Venn = collections.defaultdict(list)
cnt = 0
for j in range(N):
if len(dic[j]):
cnt += len(dic[j]) ** 3
for i in range(j, 0, -1):
for prv in Venn[i]:
intersec = prv & dic[j]
if len(intersec):
cnt += (-1) ** i * len(intersec) ** 3
Venn[i + 1].append(intersec)
Venn[1].append(dic[j])
return M**3 - cnt | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR VAR BIN_OP BIN_OP NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER VAR VAR RETURN BIN_OP BIN_OP VAR NUMBER VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
l = len(A)
Memo = {}
for i in range(l):
for j in range(i + 1):
t = A[i] & A[j]
if t not in Memo:
Memo[t] = 0
if i == j:
Memo[t] += 1
else:
Memo[t] += 2
r = 0
for a in A:
for key in Memo:
if key & a == 0:
r += Memo[key]
return r | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
memo = {}
for i in A:
for j in A:
memo[i & j] = memo.get(i & j, 0) + 1
res = 0
for num in A:
for k in memo:
if num & k == 0:
res += memo[k]
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT FOR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
cnt = 0
d = {}
for i in range(len(A)):
for j in range(len(A)):
if A[i] & A[j] not in d:
d[A[i] & A[j]] = 1
else:
d[A[i] & A[j]] += 1
for i in range(len(A)):
for j in d:
if A[i] & j == 0:
cnt += d[j]
return cnt | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
d = {}
ans = 0
for i in range(len(A)):
for j in range(len(A)):
a = A[i] & A[j]
d[a] = d.get(a, 0) + 1
for i in range(len(A)):
for j in list(d.keys()):
if A[i] & j == 0:
ans += d[j]
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
n = len(A)
cnt = collections.Counter()
A = list(collections.Counter(A).items())
result = 0
for i, k1 in A:
for j, k2 in A:
cnt[i & j] += k1 * k2
cnt = list(cnt.items())
for i, k1 in A:
if i == 0:
result += k1 * n * n
continue
for j, k2 in cnt:
if i & j == 0:
result += k1 * k2
return result | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR FOR VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR VAR VAR IF VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR RETURN VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
counters = [0] * (1 << 16)
counters[0] = len(A)
for num in A:
mask = ~num & (1 << 16) - 1
sm = mask
while sm != 0:
counters[sm] += 1
sm = sm - 1 & mask
return sum(counters[num1 & num2] for num1 in A for num2 in A) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR |
Given an array of integers A, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16 | class Solution:
def countTriplets(self, A: List[int]) -> int:
d = {}
res = 0
for a in A:
for b in A:
t = a & b
if t in d:
d[t] += 1
else:
d[t] = 1
for a in A:
for k, v in list(d.items()):
if a & k == 0:
res += v
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR VAR NUMBER VAR VAR RETURN VAR VAR |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def validUtf8(self, data):
current = 0
for byte in data:
b = "{0:08b}".format(byte)
if current == 0:
cnt = 0
for i in b:
if i == "0":
break
else:
cnt += 1
if cnt == 1 or cnt > 4:
return False
if cnt > 0:
current = cnt - 1
else:
current = 0
else:
if b[0:2] != "10":
return False
current -= 1
if current > 0:
return False
return True | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL STRING VAR IF VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER NUMBER STRING RETURN NUMBER VAR NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def check(self, nums, start, size):
for i in range(start + 1, start + size + 1):
if i >= len(nums) or nums[i] >> 6 != 2:
return False
return True
def validUtf8(self, nums, start=0):
while start < len(nums):
first = nums[start]
if first >> 3 == 30 and self.check(nums, start, 3):
start += 4
elif first >> 4 == 14 and self.check(nums, start, 2):
start += 3
elif first >> 5 == 6 and self.check(nums, start, 1):
start += 2
elif first >> 7 == 0:
start += 1
else:
return False
return True | CLASS_DEF FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def validUtf8(self, data):
n = len(data)
check10 = 0
for byte in data:
if check10:
if byte & 192 != 128:
return False
check10 -= 1
elif byte & 248 == 240:
check10 = 3
elif byte & 240 == 224:
check10 = 2
elif byte & 224 == 192:
check10 = 1
elif byte & 128 == 0:
continue
else:
return False
return check10 == 0 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR IF BIN_OP VAR NUMBER NUMBER RETURN NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN NUMBER RETURN VAR NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def validUtf8(self, data):
count = 0
for x in data:
if count == 0:
if x >> 5 == 6:
count = 1
elif x >> 4 == 14:
count = 2
elif x >> 3 == 30:
count = 3
elif x >> 7 == 1:
return False
else:
if x >> 6 != 2:
return False
count -= 1
return count == 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN NUMBER VAR NUMBER RETURN VAR NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def countLeadingOne(self, w):
if w[0] == "0":
return 1
elif w[0:3] == "110":
return 2
elif w[0:4] == "1110":
return 3
elif w[0:5] == "11110":
return 4
else:
return -1
def checkStartWith10(self, L, l):
if len(L) != l:
return False
for w in L:
if w.startswith("10") == False:
return False
return True
def validUtf8(self, data):
A = []
for d in data:
A.append(format(d, "08b"))
i = 0
while i < len(A):
l = self.countLeadingOne(A[i])
if l == -1:
return False
if l > 1:
if self.checkStartWith10(A[i + 1 : i + l], l - 1) == False:
return False
i += l
return True | CLASS_DEF FUNC_DEF IF VAR NUMBER STRING RETURN NUMBER IF VAR NUMBER NUMBER STRING RETURN NUMBER IF VAR NUMBER NUMBER STRING RETURN NUMBER IF VAR NUMBER NUMBER STRING RETURN NUMBER RETURN NUMBER FUNC_DEF IF FUNC_CALL VAR VAR VAR RETURN NUMBER FOR VAR VAR IF FUNC_CALL VAR STRING NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER RETURN NUMBER VAR VAR RETURN NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def validUtf8(self, data):
def numOfFollowingBytes(num):
if 0 <= num <= 127:
return 0
elif 192 <= num <= 223:
return 1
elif 224 <= num <= 239:
return 2
elif 240 <= num <= 247:
return 3
else:
return -1
def isFollowingByte(num):
return 128 <= num <= 191
if not data:
return False
i, n = 0, len(data)
while i < n:
bytesToFollow = numOfFollowingBytes(data[i])
if bytesToFollow == -1:
return False
i += 1
if i + bytesToFollow > n:
return False
for _ in range(bytesToFollow):
if not isFollowingByte(data[i]):
return False
i += 1
return True | CLASS_DEF FUNC_DEF FUNC_DEF IF NUMBER VAR NUMBER RETURN NUMBER IF NUMBER VAR NUMBER RETURN NUMBER IF NUMBER VAR NUMBER RETURN NUMBER IF NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF RETURN NUMBER VAR NUMBER IF VAR RETURN NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN NUMBER VAR NUMBER IF BIN_OP VAR VAR VAR RETURN NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR RETURN NUMBER VAR NUMBER RETURN NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def validUtf8(self, data):
seqs = len(data)
idx = 0
while idx < seqs:
number = data[idx] & 255 | 256
seq = bin(number)[3:]
bits = len(seq.split("0")[0])
if idx + bits > seqs:
return False
elif bits > 4:
return False
elif bits == 0:
idx += 1
elif bits == 1:
return False
else:
for i in range(1, bits):
num = data[idx + i] & 255
if num >= int("11000000", 2) or num < int("10000000", 2):
return False
idx += bits
return True | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR STRING NUMBER IF BIN_OP VAR VAR VAR RETURN NUMBER IF VAR NUMBER RETURN NUMBER IF VAR NUMBER VAR NUMBER IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR STRING NUMBER VAR FUNC_CALL VAR STRING NUMBER RETURN NUMBER VAR VAR RETURN NUMBER |
A character in UTF8 can be from 1 to 4 bytes long, subjected to the following rules:
For 1-byte character, the first bit is a 0, followed by its unicode code.
For n-bytes character, the first n-bits are all one's, the n+1 bit is 0, followed by n-1 bytes with most significant 2 bits being 10.
This is how the UTF-8 encoding would work:
Char. number range | UTF-8 octet sequence
(hexadecimal) | (binary)
--------------------+---------------------------------------------
0000 0000-0000 007F | 0xxxxxxx
0000 0080-0000 07FF | 110xxxxx 10xxxxxx
0000 0800-0000 FFFF | 1110xxxx 10xxxxxx 10xxxxxx
0001 0000-0010 FFFF | 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx
Given an array of integers representing the data, return whether it is a valid utf-8 encoding.
Note:
The input is an array of integers. Only the least significant 8 bits of each integer is used to store the data. This means each integer represents only 1 byte of data.
Example 1:
data = [197, 130, 1], which represents the octet sequence: 11000101 10000010 00000001.
Return true.
It is a valid utf-8 encoding for a 2-bytes character followed by a 1-byte character.
Example 2:
data = [235, 140, 4], which represented the octet sequence: 11101011 10001100 00000100.
Return false.
The first 3 bits are all one's and the 4th bit is 0 means it is a 3-bytes character.
The next byte is a continuation byte which starts with 10 and that's correct.
But the second continuation byte does not start with 10, so it is invalid. | class Solution:
def validUtf8(self, data):
cont = 0
for n in data:
s = format(n, "08b")
if cont > 0:
if s[:2] == "10":
cont -= 1
continue
return False
if s[0] == "0":
continue
if s[:3] == "110":
cont = 1
continue
if s[:4] == "1110":
cont = 2
continue
if s[:5] == "11110":
cont = 3
continue
return False
return cont == 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING IF VAR NUMBER IF VAR NUMBER STRING VAR NUMBER RETURN NUMBER IF VAR NUMBER STRING IF VAR NUMBER STRING ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER RETURN NUMBER RETURN VAR NUMBER |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
s = 0
for i in range(32):
e, o = 0, 0
for j in range(n):
if arr[j] & 1:
o += 1
else:
e += 1
arr[j] //= 2
s += e * o * 2**i
return s | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def isset(self, a, n):
return a & 1 << n != 0
def sumXOR(self, a, n):
o = [0] * 17
z = [0] * 17
for i in a:
for j in range(17):
if self.isset(i, j):
o[j] += 1
else:
z[j] += 1
x = 1
ans = 0
for i in range(17):
y = o[i] * z[i]
ans += y * x
x <<= 1
return ans | CLASS_DEF FUNC_DEF RETURN BIN_OP VAR BIN_OP NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
ans = 0
bits = [0] * 33
for i in range(33):
for j in arr:
bits[i] += 1 << i & j != 0
s = sum(bits)
ans = 0
for i in range(33):
ans += (1 << i) * bits[i] * (n - bits[i])
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR BIN_OP VAR VAR VAR RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
sum = 0
i = 0
while i < 32:
zc = oc = 0
for ele in arr:
if ele & 1 << i:
oc += 1
else:
zc += 1
sum += zc * oc * (1 << i)
i += 1
return sum
for _ in range(0, int(input())):
n = int(input())
arr = list(map(int, input().strip().split()))
ob = Solution()
res = ob.sumXOR(arr, n)
print(res) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
totalXorVal = 0
for i in range(31):
zeroBitsCount = 0
oneBitsCount = 0
for num in arr:
if num & 1 << i == 0:
zeroBitsCount += 1
else:
oneBitsCount += 1
possiblePairings = zeroBitsCount * oneBitsCount
totalXorVal += (1 << i) * possiblePairings
return totalXorVal | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP NUMBER VAR VAR RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
res = 0
for i in range(32):
cnt0, cnt1 = 0, 0
for j in arr:
if j & 1 << i:
cnt1 += 1
else:
cnt0 += 1
res += 2**i * (cnt0 * cnt1)
return res | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR VAR RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
l = [0] * 32
for i in arr:
s = bin(i)
s = s[2:]
c = 0
for j in s[::-1]:
if j == "1":
l[c] += 1
c += 1
p = 1
s = 0
for i in l:
s += i * (n - i) * p
p = p * 2
return s | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR NUMBER IF VAR STRING VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
ans = 0
for i in range(31):
count = 0
for num in arr:
if num >> i & 1 == 1:
count += 1
ans += count * (n - count) * 2**i
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR BIN_OP NUMBER VAR RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
ans = 0
for i in range(32):
x = 0
y = 0
bit = 1 << i
for j in arr:
set = bit & j
if set:
x += 1
else:
y += 1
ans += x * y * bit
return ans
for _ in range(0, int(input())):
n = int(input())
arr = list(map(int, input().strip().split()))
ob = Solution()
res = ob.sumXOR(arr, n)
print(res) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
cur = 0
large = max(arr)
ans = 0
while 1 << cur <= large:
zero = 0
one = 0
for i in range(n):
if arr[i] & 1 << cur > 0:
one += 1
else:
zero += 1
ans += (1 << cur) * zero * one
cur += 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR BIN_OP NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR VAR NUMBER RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
ans = 0
temp = [(0) for i in range(18)]
for i in range(18):
if arr[0] & 1 << i:
temp[i] += 1
for k in range(1, n):
for i in range(18):
bit = arr[k] & 1 << i
if bit == 0:
ans += (1 << i) * temp[i]
else:
ans += (1 << i) * (k - temp[i])
temp[i] += 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR NUMBER BIN_OP NUMBER VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP NUMBER VAR IF VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR BIN_OP VAR VAR VAR VAR VAR NUMBER RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
result = 0
for i in range(20):
mask = 2**i
cnt = 0
for e in arr:
if e & mask == mask:
cnt += 1
result += cnt * (n - cnt) * mask
return result | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP VAR VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR RETURN VAR |
Given an array of N integers, find the sum of xor of all pairs of numbers in the array.
Example 1:
ââ¬â¹Input : arr[ ] = {7, 3, 5}
Output : 12
Explanation:
All possible pairs and there Xor
Value: ( 3 ^ 5 = 6 ) + (7 ^ 3 = 4)
+ ( 7 ^ 5 = 2 ) = 6 + 4 + 2 = 12
ââ¬â¹Example 2:
Input : arr[ ] = {5, 9, 7, 6}
Output : 47
Your Task:
This is a function problem. The input is already taken care of by the driver code. You only need to complete the function sumXOR() that takes an array (arr), sizeOfArray (n), and returns the sum of xor of all pairs of numbers in the array. The driver code takes care of the printing.
Expected Time Complexity: O(N Log N).
Expected Auxiliary Space: O(1).
Constraints
2 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5 | class Solution:
def sumXOR(self, arr, n):
su = 0
for i in range(32):
num_set = 0
num_unset = 0
for num in arr:
if num >> i & 1:
num_set += 1
else:
num_unset += 1
if num_set * num_unset > 0:
su = su + num_set * num_unset * (1 << i)
return su | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR RETURN VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
a = len(bin(N)) - 3
return (N - 2**a) * 2 + 1
if __name__ == "__main__":
t = int(input())
for _ in range(t):
N = int(input())
ob = Solution()
print(ob.find(N)) | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP VAR BIN_OP NUMBER VAR NUMBER NUMBER IF VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | def rec(n):
if n == 1:
return 1
if n & 1:
return 2 * rec(n // 2) + 1
return 2 * rec(n // 2) - 1
class Solution:
def find(self, N):
return rec(N) | FUNC_DEF IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER CLASS_DEF FUNC_DEF RETURN FUNC_CALL VAR VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, n):
a = list(range(1, n + 1))
i = 0
size = n
while len(a) != 1:
a[:] = [a[i] for i in range(0, size, 2)]
if size % 2 != 0:
a.insert(0, a.pop(-1))
size = len(a)
i += 1
return a[0] | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR NUMBER |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
i = 0
while 2**i <= N:
i += 1
i -= 1
closest = 2**i
ans = (N - closest << 1) + 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE BIN_OP NUMBER VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER RETURN VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
if N == 1:
return 1
if N % 2 == 0:
return 2 * self.find(N / 2) - 1
else:
return 2 * self.find(N // 2) + 1 | CLASS_DEF FUNC_DEF IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, n):
def nn(n):
i = 1
while i < n:
i = i * 2
if i == n:
return i
else:
return i // 2
return 2 * (n - nn(n)) + 1 | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR RETURN VAR RETURN BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP NUMBER BIN_OP VAR FUNC_CALL VAR VAR NUMBER |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def number(self, n):
lst = [i for i in range(1, n + 1)]
while len(lst) != 1:
if len(lst) % 2 == 0:
lst = lst[::2]
else:
lst = lst[::2]
lst.pop(0)
return lst[0]
def find(self, N):
return self.number(N) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
z = N
if N + 1 & N == 0:
return N
c = []
while N > 0:
a = N % 2
c.append(a)
N = N // 2
a = 2 ** (len(c) - 1)
return (z - a) * 2 + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR IF BIN_OP BIN_OP VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR LIST WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
z = N
if N == 2:
return 1
else:
c = 0
while N > 0:
a = N & 2
c = c + 1
N = N // 2
return 1 + 2 * (z - 2 ** (c - 1)) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP NUMBER BIN_OP NUMBER BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
s = bin(N)[3:]
s += "1"
d = 0
s = s[::-1]
for i in range(len(s)):
if s[i] == "0":
continue
d += int(pow(2, i))
return d | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR RETURN VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, N):
if N == 1 or N == 2:
return 1
i = 4
while i < N:
i = 2 * i
if i == N:
return 1
else:
return N - (i - 1 - N) | CLASS_DEF FUNC_DEF IF VAR NUMBER VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR VAR RETURN NUMBER RETURN BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR |
Given N people standing in a circle where 1^{st} is having a sword, find the luckiest person in the circle, if, from 1^{st} soldier who is having a sword each has to kill the next soldier and handover the sword to next soldier, in turn, the soldier will kill the adjacent soldier and handover the sword to next soldier such that one soldier remains in this war who is not killed by anyone.
Example 1:
Input:
N = 5
Output:
3
Explanation:
In first go 1 3 5 (remains)
as 2 and 4 killed by 1 and 3.
In second go 3 will remain
as 5 killed 1 and 3rd killed 5
hence, 3 remains alive.
Example 2:
Input:
N = 10
Output:
5
Explanation:
In first 1 3 5 7 9
remains as 2 4 6 8 10
were killed by 1 3 5 7
and 9. In second 1 5 9
are left as 1 kills 3
and 5 kill the 7th
soldier.In third 5th
soldiers remain alive as
9 kills the 1st soldier and
5 kill the 9th soldier.
Your Task:
You don't need to read input or print anything. Your task is to complete the function find() which takes an integer N as input parameter and returns the output as the soldier who was lucky in the game.
Expected Time Complexity: O(log N)
Expected Space Complexity: O(1)
Constraints:
1<=N<=10^{8} | class Solution:
def find(self, n):
p = 1
while p <= n:
p *= 2
return 2 * n - p + 1
class Node:
def __init__(self, data):
self.data = data
self.next = None
class Ll:
def __init__(self):
self.head = None
def insert(self, data):
nw = Node(data)
if self.head is None:
self.head = nw
nw.next = head
else:
h = self.head
while h.next.data is not head.data:
h = h.next
h.next = nw
nw.next = self.head | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NONE CLASS_DEF FUNC_DEF ASSIGN VAR NONE FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NONE ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR |
Given an array A containing 2*N+2 positive numbers, out of which 2*N numbers exist in pairs whereas the other two number occur exactly once and are distinct. Find the other two numbers. Return in increasing order.
Example 1:
Input:
N = 2
arr[] = {1, 2, 3, 2, 1, 4}
Output:
3 4
Explanation:
3 and 4 occur exactly once.
Example 2:
Input:
N = 1
arr[] = {2, 1, 3, 2}
Output:
1 3
Explanation:
1 3 occur exactly once.
Your Task:
You do not need to read or print anything. Your task is to complete the function singleNumber() which takes the array as input parameter and returns a list of two numbers which occur exactly once in the array. The list must be in ascending order.
Expected Time Complexity: O(N)
Expected Space Complexity: O(1)
Constraints:
1 <= length of array <= 10^{6 }
1 <= Elements in array <= 5 * 10^{6} | class Solution:
def singleNumber(self, nums):
a = []
all_freq = {}
for i in nums:
if i in all_freq:
all_freq[i] += 1
else:
all_freq[i] = 1
for key, value in all_freq.items():
if value == 1:
a.append(key)
a.sort()
return a | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given an array A containing 2*N+2 positive numbers, out of which 2*N numbers exist in pairs whereas the other two number occur exactly once and are distinct. Find the other two numbers. Return in increasing order.
Example 1:
Input:
N = 2
arr[] = {1, 2, 3, 2, 1, 4}
Output:
3 4
Explanation:
3 and 4 occur exactly once.
Example 2:
Input:
N = 1
arr[] = {2, 1, 3, 2}
Output:
1 3
Explanation:
1 3 occur exactly once.
Your Task:
You do not need to read or print anything. Your task is to complete the function singleNumber() which takes the array as input parameter and returns a list of two numbers which occur exactly once in the array. The list must be in ascending order.
Expected Time Complexity: O(N)
Expected Space Complexity: O(1)
Constraints:
1 <= length of array <= 10^{6 }
1 <= Elements in array <= 5 * 10^{6} | class Solution:
def singleNumber(self, nums):
f = {}
for i in nums:
if i in f:
f[i] += 1
else:
f[i] = 1
l = []
for i in nums:
if f[i] == 1:
l.append(i)
l.sort()
return l | CLASS_DEF FUNC_DEF ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR |
Given an array A containing 2*N+2 positive numbers, out of which 2*N numbers exist in pairs whereas the other two number occur exactly once and are distinct. Find the other two numbers. Return in increasing order.
Example 1:
Input:
N = 2
arr[] = {1, 2, 3, 2, 1, 4}
Output:
3 4
Explanation:
3 and 4 occur exactly once.
Example 2:
Input:
N = 1
arr[] = {2, 1, 3, 2}
Output:
1 3
Explanation:
1 3 occur exactly once.
Your Task:
You do not need to read or print anything. Your task is to complete the function singleNumber() which takes the array as input parameter and returns a list of two numbers which occur exactly once in the array. The list must be in ascending order.
Expected Time Complexity: O(N)
Expected Space Complexity: O(1)
Constraints:
1 <= length of array <= 10^{6 }
1 <= Elements in array <= 5 * 10^{6} | class Solution:
def singleNumber(self, nums):
xor = 0
for n in nums:
xor ^= n
rsb = xor & -xor
ans1, ans2 = 0, 0
for n in nums:
if rsb & n:
ans1 ^= n
else:
ans2 ^= n
return sorted([ans1, ans2]) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR LIST VAR VAR |
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