Count Occurrences(String) ----------------------------------------- Problem Description:
Find the number of occurrences of bob in string A consisting of lowercase English alphabets.
Problem Constraints:
1 <= |A| <= 1000
Input Format:
The first and only argument contains the string A, consisting of lowercase English alphabets.
Output Format:
Return an integer containing the answer.
Example Input
Input 1:
"abobc"
Input 2:
"bobob"
Example Output
Output 1:
1
Output 2:
2
Example Explanation
Explanation 1:
The only occurrence is at second position.
Explanation 2:
Bob occures at first and third position.
class Solution:
# @param A : string
# @return an integer
def solve(self, A):
count = 0
for i in range(len(A) - 2): # Loop until len(A) - 2 to avoid index out of range
if A[i:i+3] == "bob": # Check if substring from i to i+3 is "bob"
count += 1
return count
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Count unique elements
--------------------------------------------
Problem Description:
You are given an array A of N integers. Return the count of elements with frequncy 1 in the given array.
Problem Constraints
1 <= N <= 105
1 <= A[i] <= 109
Input Format:
First argument A is an array of integers.
Output Format:
Return an integer.
Example Input
Input 1:
A = [3, 4, 3, 6, 6]
Input 2:
A = [3, 3, 3, 9, 0, 1, 0]
Example Output
Output 1:
1
Output 2:
2
def solve(self, A):
fm={}
for i in A:
if i in fm:
fm[i]+=1
else:
fm[i]=1
count=0
for i in A:
if fm[i]==1:
count+=1
return count
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Count Occurrences(String)
-----------------------------------------
Problem Description:
Find the number of occurrences of bob in string A consisting of lowercase English alphabets.
Problem Constraints:
1 <= |A| <= 1000
Input Format:
The first and only argument contains the string A, consisting of lowercase English alphabets.
Output Format:
Return an integer containing the answer.
Example Input
Input 1:
"abobc"
Input 2:
"bobob"
Example Output
Output 1:
1
Output 2:
2
Example Explanation
Explanation 1:
The only occurrence is at second position.
Explanation 2:
Bob occures at first and third position.
class Solution:
# @param A : string
# @return an integer
def solve(self, A):
count = 0
for i in range(len(A) - 2): # Loop until len(A) - 2 to avoid index out of range
if A[i:i+3] == "bob": # Check if substring from i to i+3 is "bob"
count += 1
return count
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