A common misconception is that practicing algorithms means memorizing many algorithms. This leads to ineffective practice, frustration, rejection of algorithms, etc. To rekindle your love for algorithms, here are a few simple coding problems that do not require much algorithmic knowledge, but reflect an important aspect often overlooked in algorithms.
Each problem comes with a link for you to submit your code and view the results. You should try solving them on your own before reading the solutions; if you don't get handson experience, you won't learn anything. Or, you might think they are too easy but end up making mistakes. 🤭
The solutions are written in Python. If you’re not familiar with Python, that’s fine; the main goal is to understand the ideas so you can implement them yourself. Alternatively, you can ask ChatGPT to convert the code into a language you're familiar with.
Leetcode 1688: Count of Matches in Tournament
Problem
Link: https://leetcode.com/problems/countofmatchesintournament/description/
You are given an integer n
, the number of teams in a tournament that has strange rules:
 If the current number of teams is even, each team gets paired with another team. A total of
n / 2
matches are played, andn / 2
teams advance to the next round.  If the current number of teams is odd, one team randomly advances in the tournament, and the rest gets paired. A total of
(n  1) / 2
matches are played, and(n  1) / 2 + 1
teams advance to the next round.
Return the number of matches played in the tournament until a winner is decided.
Example 1:

Input:
n = 7

Output:
6

Explanation: Details of the tournament:
 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 3 + 2 + 1 = 6.
Example 2:

Input:
n = 14

Output:
13

Explanation: Details of the tournament:
 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 7 + 3 + 2 + 1 = 13.
Constraints:
1 <= n <= 200
Solution
If you already know basic programming, this problem is quite simple, right? Each match involves 2 teams, so the number of matches each round is n / 2
. If the number of teams is odd, add 1 team to the next round. After each round, the number of teams is reduced approximately by half. Continue until only 1 team remains.
class Solution:
def numberOfMatches(self, n: int) > int:
matches = 0
while n > 1:
if n % 2 == 0:
matches += n // 2
n = n // 2
else:
matches += (n  1) // 2
n = ((n  1) // 2) + 1
return matches
However, if this problem was asked before you learned programming, you might have had a different answer.
There are n
teams in total, and they compete until only 1
team remains.
So there are n  1
teams eliminated.
Each match eliminates 1
team.
So, there are n  1
matches. Duh! 🤷♀️
The result is simply:
class Solution:
def numberOfMatches(self, n: int) > int:
return n  1
Leetcode 1332: Remove Palindromic Subsequences
Problem
Link: https://leetcode.com/problems/removepalindromicsubsequences/description/
You are given a string s
consisting only of letters 'a'
and 'b'
. In a single step you can remove one palindromic subsequence from s.
Return the minimum number of steps to make the given string empty.
A string is a subsequence of a given string if it is generated by deleting some characters of a given string without changing its order. Note that a subsequence does not necessarily need to be contiguous. In other words, a subsequence of a string is a sequence of elements taken from the string while preserving their order. For example, consider the string "abcd"
:

"acd"
is a subsequence. 
"bd"
is a subsequence. 
"ba"
is not a subsequence.
A string is called palindrome if is one that reads the same backward as well as forward. For example, abbcbba
is palindromic, while abcabc
is not.
Example 1:

Input:
s = "ababa"

Output:
1

Explanation:
s
is already a palindrome, so its entirety can be removed in a single step.
Example 2:

Input:
s = "abb"

Output:
2

Explanation:
"abb"
>"bb"
>""
. Remove palindromic subsequence"a"
then"bb"
.
Example 3:

Input:
s = "baabb"

Output:
2

Explanation:
"baabb"
>"b"
>""
. Remove palindromic subsequence"baab"
then"b"
.
Constraints:
1 <= s.length <= 1000

s[i]
is either'a'
or'b'
.
Solution
If s
is palindromic, then it can be removed in one step.
If not, in each step, you can choose a subsequence consisting of all the same characters, which should be palindromic because it consists of identical characters.
For example, s = "aabbbabab"
:
 Remove the palindromic subsequence
"bbbbb"
.  Remove the palindromic subsequence
"aaaa"
.
At most, it will take 2 steps.
class Solution:
def removePalindromeSub(self, s: str) > int:
return 1 if s == s[::1] else 2
Leetcode 1460: Make Two Arrays Equal by Reversing Subarrays
Problem
Link: https://leetcode.com/problems/maketwoarraysequalbyreversingsubarrays/description/.
You are given two integer arrays of equal length target
and arr
. In one step, you can select any nonempty subarray of arr
and reverse it. You are allowed to make any number of steps.
Return true
if you can make arr
equal to target
or false
otherwise.
Example 1:

Input:
target = [1,2,3,4]
,arr = [2,4,1,3]

Output:
true

Explanation: You can follow the next steps to convert
arr
totarget
: Reverse subarray
[2,4]
,arr
becomes[4,2,1,3]
 Reverse subarray
[1,3]
,arr
becomes[4,2,3,1]
 Reverse subarray
[2,3]
,arr
becomes[4,3,2,1]
 Reverse subarray
[4,3,2,1]
,arr
becomes[1,2,3,4]
There are multiple ways to convert
arr
totarget
, this is not the only way to do so.  Reverse subarray
Example 2:

Input:
target = [7]
,arr = [7]

Output:
true

Explanation:
arr
is equal totarget
without any reverses.
Example 3:

Input:
target = [3,7,9]
,arr = [3,7,11]

Output:
false

Explanation:
arr
does not have value9
and it can never be converted totarget
.
Constraints:
target.length == arr.length
1 <= target.length <= 1000
1 <= target[i] <= 1000
1 <= arr[i] <= 1000
Solution
Notice that with the subarray reversal operation, we can move an element to any position we want. For example, to move arr[5]
to arr[2]
, we reverse the subarray from 2 to 5. The other elements in the subarray will also change positions, but the elements outside the subarray are unaffected.
Taking advantage of this, we will transform each element of arr
, from left to right, to match the corresponding element in target
.
For each position i
, we find the element equal to target[i]
in arr
from i
to n  1
. If no such element is found, then it is impossible to transform arr
into target
because we cannot add new elements. If found, let the position of that element be j
, we reverse the subarray from i
to j
. Now arr[i]
is equal to target[i]
. Continue this process until the end of arr
.
Example: target = [1,2,3,4]
, arr = [2,4,1,3]
Thus, as long as we can find j
for each position i
, arr
will gradually be transformed into target
, from left to right. Therefore, the order of the elements does not matter, as long as arr
and target
contain the same elements, with the same quantities, arr
can always be transformed into target
.
In fact, you can come up with other algorithms to rearrange the elements. The important thing here is that you can always transform arr
to arrange it in any desired order.
So to determine if arr
can be transformed into target
, just compare if arr
and target
have the same elements. Same here means that the quantity of each value must also be the same, for example, if target
has 4 elements with value 3
and arr
has 5 elements with value 3
, then obviously it cannot be transformed because we cannot add or remove elements.
class Solution:
def canBeEqual(self, target: List[int], arr: List[int]) > bool:
return Counter(arr) == Counter(target)
If you are not familiar with Counter/Dict/Map, you can sort target
and arr
and then compare:
class Solution:
def canBeEqual(self, target: List[int], arr: List[int]) > bool:
return sorted(arr) == sorted(target)
Leetcode 1700: Number of Students Unable to Eat Lunch
Problem
Link: https://leetcode.com/problems/numberofstudentsunabletoeatlunch/description/.
The school cafeteria offers circular and square sandwiches at lunch break, referred to by numbers 0
and 1
respectively. All students stand in a queue. Each student either prefers square or circular sandwiches.
The number of sandwiches in the cafeteria is equal to the number of students. The sandwiches are placed in a stack. At each step:
 If the student at the front of the queue prefers the sandwich on the top of the stack, they will take it and leave the queue.
 Otherwise, they will leave it and go to the queue's end.
This continues until none of the queue students want to take the top sandwich and are thus unable to eat.
You are given two integer arrays students
and sandwiches
where sandwiches[i]
is the type of the i
th sandwich in the stack (i = 0
is the top of the stack) and students[j]
is the preference of the j
th student in the initial queue (j = 0
is the front of the queue).
Return the number of students that are unable to eat.
Example 1:

Input:
students = [1,1,0,0]
,sandwiches = [0,1,0,1]

Output:
0

Explanation:
 Front student leaves the top sandwich and returns to the end of the line making
students = [1,0,0,1]
.  Front student leaves the top sandwich and returns to the end of the line making
students = [0,0,1,1]
.  Front student takes the top sandwich and leaves the line making
students = [0,1,1]
andsandwiches = [1,0,1]
.  Front student leaves the top sandwich and returns to the end of the line making
students = [1,1,0]
.  Front student takes the top sandwich and leaves the line making
students = [1,0]
andsandwiches = [0,1]
.  Front student leaves the top sandwich and returns to the end of the line making
students = [0,1]
.  Front student takes the top sandwich and leaves the line making
students = [1]
andsandwiches = [1]
.  Front student takes the top sandwich and leaves the line making
students = []
andsandwiches = []
.
Hence all students are able to eat.
 Front student leaves the top sandwich and returns to the end of the line making
Example 2:

Input:
students = [1,1,1,0,0,1]
,sandwiches = [1,0,0,0,1,1]

Output:
3
Constraints:
1 <= students.length, sandwiches.length <= 100
students.length == sandwiches.length

sandwiches[i]
is0
or1
. 
students[i]
is0
or1
.
Solution
A naive algorithm is to simulate the process step by step:
class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) > int:
while len(students) > 0 and sandwiches[0] in students:
front = students.pop(0)
if front == sandwiches[0]:
sandwiches.pop(0)
else:
students.append(front)
return len(students)
However, continuously removing elements from the front of an array is very inefficient. Each time an element is removed from the front, the positions of all remaining elements must be updated. Thus, you have to traverse the students
list many, many times. If you know a bit about data structures, you might realize that you can replace the array with a queue to make the removal operation less costly.
But you can do even better.
Notice that for each sandwich on top, you will keep pushing the student at the front of the queue to the end of the queue until a student who wants that sandwich is at the front of the queue.
You only care whether there are any students left in the queue who want that sandwich. If not, the process ends. If there are, one student who wants that sandwich will eat it. Specifically which student... doesn't matter.
Thus, we don't even need to keep track of the list of students in the queue. We only need to know how many students want sandwich 0
and how many want sandwich 1
.
class Solution:
def countStudents(self, students: List[int], sandwiches: List[int]) > int:
# count_students[i] is the number of students who want sandwich type i
count_students = [0, 0]
for i in students:
count_students[i] += 1
not_eat = len(students)
for i in sandwiches:
if count_students[i] > 0:
count_students[i] = 1
not_eat = 1
else:
break
return not_eat
So you only need to traverse the students
list once and the sandwiches
list once. Much more efficient than the previous solution.
Leetcode 2073: Time Needed to Buy Tickets
Problem
Link: https://leetcode.com/problems/timeneededtobuytickets/description/.
There are n
people in a line queuing to buy tickets, where the 0
th person is at the front of the line and the n  1
th person is at the back of the line.
You are given a 0indexed integer array tickets of length n
where the number of tickets that the i
th person would like to buy is tickets[i]
.
Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.
Return the time taken for the person at position k
(0indexed) to finish buying tickets.
Example 1:

Input:
tickets = [2,3,2], k = 2

Output:
6

Explanation:
 In the first pass, everyone in the line buys a ticket and the line becomes
[1, 2, 1]
.  In the second pass, everyone in the line buys a ticket and the line becomes
[0, 1, 0]
.
The person at position 2 has successfully bought 2 tickets and it took
3 + 3 = 6
seconds.  In the first pass, everyone in the line buys a ticket and the line becomes
Example 2:

Input:
tickets = [5,1,1,1], k = 0

Output:
8

Explanation:
 In the first pass, everyone in the line buys a ticket and the line becomes
[4, 0, 0, 0]
.  In the next 4 passes, only the person in position 0 is buying tickets.
The person at position 0 has successfully bought 5 tickets and it took
4 + 1 + 1 + 1 + 1 = 8
seconds.  In the first pass, everyone in the line buys a ticket and the line becomes
Constraints:
n == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n
Solution
A naive algorithm is to simulate the entire ticketbuying process step by step until person k
finishes buying their tickets. You can use a queue/deque data structure for the ticketbuying line to make adding and removing elements more efficient.
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) > int:
t = 0
q = deque()
for i in range(len(tickets)):
q.append(i)
while tickets[k] > 0:
front = q.popleft()
tickets[front] = 1
t += 1
if tickets[front] > 0:
q.append(front)
return t
However, with this algorithm, you have to simulate the entire ticketbuying process step by step until it is complete. Let's try to find a more efficient way.
Each ticket purchase takes 1 second, so the elapsed time is equivalent to the total number of tickets everyone has bought. Let's try to calculate how many tickets everyone has bought.
Suppose we have a queue of 5 people: ["Alice", "Bob", "Carol", "Dave", "Eve"]
. They need to buy the corresponding number of tickets: [3, 10, 9, 13, 5]
.
Think carefully, when Carol buys her 9th ticket, how many tickets has each other person bought? Think about it and then read on.
The answer is:
 Alice bought 3 and left.
 Bob bought 9.
 Dave bought 8.
 Eve bought 5 and left.
In the initial order, the person in front can buy tickets faster than the person behind by 1 ticket, but never more than 1 ticket, because after buying, they have to go to the end of the line and wait for others to buy another ticket.
When a person buys their x
th ticket, all the people before them in the initial order have also bought x
tickets, or have finished buying if they needed fewer than x
tickets.
The people behind them in the initial order have also bought x  1
tickets, or have finished buying if they needed fewer.
When person k
has just finished buying all their tickets, we can also deduce how many tickets each other person has bought, thus knowing the total number of tickets bought and the elapsed time.
Code:
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) > int:
t = 0
for i in range(len(tickets)):
if i <= k:
t += min(tickets[i], tickets[k])
else:
t += min(tickets[i], tickets[k]  1)
return t
With this approach, you only traverse the number of tickets each person buys exactly once. Even if there are millions of people, each needing to buy millions of tickets, the program's runtime will still be less than 1 second.
Leetcode 1894: Find the Student that Will Replace the Chalk
Problem
Link: https://leetcode.com/problems/findthestudentthatwillreplacethechalk/description/
There are n
students in a class numbered from 0
to n  1
. The teacher will give each student a problem starting with the student number 0
, then the student number 1
, and so on until the teacher reaches the student number n  1
. After that, the teacher will restart the process, starting with the student number 0
again.
You are given a 0indexed integer array chalk
and an integer k
. There are initially k
pieces of chalk. When the student number i
is given a problem to solve, they will use chalk[i]
pieces of chalk to solve that problem. However, if the current number of chalk pieces is strictly less than chalk[i]
, then the student number i
will be asked to replace the chalk.
Return the index of the student that will replace the chalk pieces.
Example 1:

Input:
chalk = [5,1,5], k = 22

Output:
0

Explanation: The students go in turns as follows:
 Student number 0 uses 5 chalk, so k = 17.
 Student number 1 uses 1 chalk, so k = 16.
 Student number 2 uses 5 chalk, so k = 11.
 Student number 0 uses 5 chalk, so k = 6.
 Student number 1 uses 1 chalk, so k = 5.
 Student number 2 uses 5 chalk, so k = 0.
Student number 0 does not have enough chalk, so they will have to replace it.
Example 2:

Input:
chalk = [3,4,1,2], k = 25

Output:
1

Explanation: The students go in turns as follows:
 Student number 0 uses 3 chalk so k = 22.
 Student number 1 uses 4 chalk so k = 18.
 Student number 2 uses 1 chalk so k = 17.
 Student number 3 uses 2 chalk so k = 15.
 Student number 0 uses 3 chalk so k = 12.
 Student number 1 uses 4 chalk so k = 8.
 Student number 2 uses 1 chalk so k = 7.
 Student number 3 uses 2 chalk so k = 5.
 Student number 0 uses 3 chalk so k = 2.
Student number 1 does not have enough chalk, so they will have to replace it.
Constraints:
chalk.length == n
1 <= n <= 10^5
1 <= chalk[i] <= 10^5
1 <= k <= 10^9
Solution
A naive algorithm: Simulate the process of each student solving problems.
class Solution:
def chalkReplacer(self, chalk: List[int], k: int) > int:
i = 0
while True:
if k < chalk[i]:
return i
k = chalk[i]
if i + 1 < len(chalk):
i = i + 1
else:
i = 0
No good, right? Time Limit Exceeded.
The problem with the naive algorithm is that when k
is large, the process can take a very long time over many rounds (1 round is when it goes back to the first student's turn). The program has to traverse the chalk
array many times. For example, when k = 1000_000_000
, chalk = [1, 1, 1, 1, 1, 1...]
(100_000 ones), the algorithm will traverse chalk
10_000 times, and chalk
has 100_000 elements.
But pay attention to k
after each round.
 After 1 round, there are
1000_000_000  100_000
pieces left.  After 2 rounds, there are
1000_000_000  100_000  100_000
pieces left.  After 3 rounds, there are
1000_000_000  100_000  100_000  100_000
pieces left.  And so on...
After each round, the number of chalk pieces decreases by a fixed amount, which is the total amount of chalk the students use in one round.
Do you have to iterate through each round and subtract the chalk pieces bit by bit? You can calculate the number of rounds directly, right?
cycles = int(k / sum(chalk))
And you can also calculate the remaining chalk pieces after cycles
rounds:
k = k  cycles * sum(chalk)
Or just use the modulo operation:
k = k % sum(chalk)
Now the remaining pieces are not enough for one round, so you only need to go through the list one more time to run out of chalk.
Code:
class Solution:
def chalkReplacer(self, chalk: List[int], k: int) > int:
k = k % sum(chalk)
for i in range(len(chalk)):
if k < chalk[i]:
return i
k = chalk[i]
This algorithm only needs to traverse the chalk
array twice, once to calculate the total chalk for each round and once more to subtract the chalk until it runs out.
Leetcode 1509: Minimum Difference Between Largest and Smallest Value in Three Moves
Problem
You are given an integer array nums
.
In one move, you can choose one element of nums
and change it to any value.
Return the minimum difference between the largest and smallest value of nums
after performing at most three moves.
(In other words, you can perform at most 3 moves to change the difference between the largest and smallest values in nums
, return the smallest difference that can be achieved).
Example 1:

Input:
nums = [5,3,2,4]

Output:
0

Explanation: We can make at most 3 moves.
 In the first move, change
2
to3
.nums
becomes[5,3,3,4]
.  In the second move, change
4
to3
.nums
becomes[5,3,3,3]
.  In the third move, change
5
to3
.nums
becomes[3,3,3,3]
.
After performing 3 moves, the difference between the minimum and maximum is
3  3 = 0
.  In the first move, change
Example 2:

Input:
nums = [1,5,0,10,14]

Output:
1

Explanation: We can make at most 3 moves.
 In the first move, change
5
to0
.nums
becomes[1,0,0,10,14]
.  In the second move, change
10
to0
.nums
becomes[1,0,0,0,14]
.  In the third move, change
14
to1
.nums
becomes[1,0,0,0,1]
.
After performing 3 moves, the difference between the minimum and maximum is
1  0 = 1
.It can be shown that there is no way to make the difference 0 in 3 moves.
 In the first move, change
Example 3:

Input:
nums = [3,100,20]

Output:
0

Explanation: We can make at most 3 moves.
 In the first move, change
100
to7
.nums
becomes[3,7,20]
.  In the second move, change
20
to7
.nums
becomes[3,7,7]
.  In the third move, change
3
to7
.nums
becomes[7,7,7]
.
After performing 3 moves, the difference between the minimum and maximum is
7  7 = 0
.  In the first move, change
Constraints:
1 <= nums.length <= 10^5
10^9 <= nums[i] <= 10^9
Solution
Of course, you should not change an element in a way that decreases the minimum or increases the maximum of the array because it will increase the difference.
Changing an element that is neither the minimum nor the maximum also does not help. For example, nums = [1, 5, 0, 10, 14]
, changing 10
to 11
still results in a difference of 13
.
You should change the elements that are the maximum or minimum, and change them to a value within the new minimum and maximum range. Specifically, the exact value does not matter because we only care about the difference between the minimum and maximum.
In general, you only need to change the smallest, second smallest, third smallest, largest, second largest, and third largest elements. Now we have just a few cases that need to be checked:
 Change the 3 smallest elements.
 Change the 2 smallest elements and 1 largest element.
 Change 1 smallest element and 2 largest elements.
 Change the 3 largest elements.
Additionally, for arrays with 4 or fewer elements, you can change all elements to be equal to achieve a difference of 0.
class Solution:
def minDifference(self, nums: List[int]) > int:
if len(nums) <= 4:
return 0
nums = sorted(nums)
return min(
nums[1]  nums[3], # Change the 3 smallest elements so nums[3] becomes the smallest
nums[2]  nums[2], # Change 1 largest element so nums[2] becomes the largest
nums[3]  nums[1],
nums[4]  nums[0]
)
The array is sorted to easily access the smallest, second smallest, largest, second largest, etc.
In fact, you only need to focus on a few smallest and largest elements, so you do not need to sort the entire array. The algorithm can be further optimized. You can use a heap/priority queue to quickly find the largest and smallest elements (if you are a beginner, you can skip this part, the above method is sufficient for this problem).
class Solution:
def minDifference(self, nums: List[int]) > int:
if len(nums) <= 4:
return 0
smallest = heapq.nsmallest(4, nums)
largest = heapq.nlargest(4, nums)
return min(
largest[0]  smallest[3],
largest[1]  smallest[2],
largest[2]  smallest[1],
largest[3]  smallest[0]
)
Conclusion
Have the questions above changed your mindset about algorithms?
As you’ve seen, the problems discussed don’t require advanced knowledge of algorithms or data structures. Even someone wellversed in algorithms might struggle with these problems. There are countless problems, and it’s impossible to know every algorithm for every situation. The important thing is to develop skills in observation, thinking, reasoning, etc., to come up with algorithms to solve new problems.
However, I'm not saying you don't need to learn more algorithms. Common algorithms can help solve parts of problems. Sorting algorithms, for instance, are frequently used and often come prewritten, but other algorithms might not be as readily available. Sometimes, you’ll need to combine algorithms to tackle a problem or tweak an algorithm to fit a specific situation. It’s also important to understand the principles and ideas behind algorithms rather than just memorizing them. Personally, I don’t memorize algorithms, but I understand them so I can rewrite them. This way, you can tweak, learn the thinking, techniques, and tricks to apply in other cases. And who knows, understanding algorithms might even lead you to improve or invent better algorithms.
I hope this article helps you better understand algorithms, improve your programming skills, and inspires you to practice and develop your algorithmic abilities.
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