If you have been following my blog lately, you must have come across the name “David Goggins” from the book “Can’t Hurt Me.” The tough guy wrote a follow-up to that book, giving more recent updates from his life in his 2022 book called “Never Finished.”
Like the first book, this one continues to offer tips, tricks, and techniques on how to become mentally strong.
First, let me be honest about this guy — his story moved me. I mean, I think I was regular at the gym for over a month — lost 4 kg.
But I was also exploring other books, and there was this one called “Light From Many Lamps” — which was teaching almost the antithesis of what Goggins was trying to preach.
And as I sit down to think about the two books, trying to settle myself, I realize that maybe neither book is wrong — just right in different contexts. Like Newton’s Laws of Motion and Einstein’s Theory of Relativity.
“Light From Many Lamps” by Lillian Watson teaches about happiness and enjoyment, faith in God, courage and fear, among other things.
I feel you need Goggins’ book in the morning — when you need motivation to get out of your warm, comfortable bed — and Lillian’s at night — when you want the storm of thoughts about events and people to settle down, when you want to soothe your soul to sleep.
I feel you need Goggins’ book when you are leading people in a professional or academic space — and Lillian’s book when you are leading or guiding people in your relationships.
I feel you need Goggins’ book when you want to go to war with yourself — and Lillian’s book when you want to make peace with yourself.
Personally, I feel Goggins’ book has helped me a lot — no pain, no gain — but yes, at the same time, I need Lillian’s teachings — about happiness, about contentment, about faith, about the good in the world — when I am in bed at night, trying to fall asleep.
Note: I'd mark this problem for 'medium complexity', though it is rated as 'easy' on Hacker Earth -- probably because it tests the student for some well known tips, tricks and shortcuts.
Problem
You have an N x M matrix, where 1 to N are rows and 1 to M are columns.
Element at the intersection of
row and
column is equal to
, where
is equal to the number of prime divisors of x.
Determine the sum of all the elements of the matrix.
Input
The first line contains an integer T denoting the number of test cases.
For each test case, the first line contains two space-separated integers N, M.
Output
For each test case, print the sum of all the elements of the matrix in a new line.
Constraints
Sample Input
1
3 3
Sample Output
10
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation
The matrix and the corresponding values are:
=
Hence, the sum of all the elements is 10.
Code first (and explanation later)
import sys
input = sys.stdin.readline
def precompute_f(max_val):
F = [0] * (max_val + 1)
for i in range(2, max_val + 1):
if F[i] == 0: # prime
for j in range(i, max_val + 1, i):
F[j] += 1
return F
def solve():
T = int(input())
queries = []
max_sum = 0
for _ in range(T):
N, M = map(int, input().split())
queries.append((N, M))
max_sum = max(max_sum, N + M)
F = precompute_f(max_sum)
for N, M in queries:
total = 0
for s in range(2, N + M + 1):
low = max(1, s - M)
high = min(N, s - 1)
if low <= high:
count = high - low + 1
total += F[s] * count
print(total)
solve()
User Prompt
I am learning Data Structures and Algorithms.
I will provide a screenshot of a problem statement.
Your task is NOT to directly give me the final code solution unless I explicitly ask for it.
Instead, please do the following:
1. First, carefully restate the problem in your own words to confirm correct understanding.
- If anything is unclear in the screenshot, ask me clarification questions before proceeding.
2. Identify:
- What category of problem this is (e.g., prefix sum, greedy, DP, two pointers, binary search, graph, etc.)
- What core DSA concepts are being tested
- What mathematical observations (if any) are involved
- What constraints are important and how they affect the approach
3. Explain:
- What naive/brute-force solution would look like
- Why that may or may not work under the given constraints
- What optimization insight is required
4. Clearly explain the key trick or insight needed to solve it.
- Why does that trick work?
- What pattern should I remember for future problems?
5. Provide:
- The time complexity and space complexity of the optimal approach
- Why it satisfies the constraints
6. If there are multiple valid approaches, compare them.
7. Only after all conceptual explanation, provide clean and correct Python code.
8. Avoid guessing unclear details.
- If any part of the screenshot is ambiguous, ask me instead of assuming.
Focus on correctness over speed. Be explicit in reasoning.
Excellent — this is a very strong math + counting + precomputation problem. Let’s go step-by-step exactly in the structured way you requested.
1️⃣ Restating the Problem (Confirm Understanding)
We are given:
A matrix of size N × M
Each element at position (i, j) is:
A[i][j]=F(i+j)
Where:
F(x) = number of distinct prime divisors of x.
We must compute:
i=1∑Nj=1∑MF(i+j)
We do this for T test cases.
Example: N = 3, M = 3
Matrix becomes:
i+j
F(i+j)
2
1
3
1
4
1
5
1
6
2
Final sum = 10.
2️⃣ Identify Category & Concepts
📌 Category
This is:
Number Theory + Frequency Counting + Prefix / Diagonal Counting + Sieve
📌 Core DSA Concepts Being Tested
Prime factorization
Sieve of Eratosthenes
Frequency counting trick
Observing symmetry in matrix
Avoiding O(NM) brute force
📌 Mathematical Observation
Important:
The matrix depends only on:
i+j
So instead of thinking in 2D:
We count how many times each value of (i + j) appears.
Range of (i + j)
Minimum:
1+1=2
Maximum:
N+M
So we only care about numbers in range:
[2,N+M]
3️⃣ Naive / Brute Force
Brute force approach:
css
for i in 1..N:
for j in 1..M:
compute F(i+j)
add to answer
Now that the reasoning is clear, here is the correct implementation:
python
import sys
input = sys.stdin.readline
defprecompute_f(max_val):
F = [0] * (max_val + 1)
for i inrange(2, max_val + 1):
if F[i] == 0: # primefor j inrange(i, max_val + 1, i):
F[j] += 1return F
defsolve():
T = int(input())
queries = []
max_sum = 0for _ inrange(T):
N, M = map(int, input().split())
queries.append((N, M))
max_sum = max(max_sum, N + M)
F = precompute_f(max_sum)
for N, M in queries:
total = 0for s inrange(2, N + M + 1):
low = max(1, s - M)
high = min(N, s - 1)
if low <= high:
count = high - low + 1
total += F[s] * count
print(total)
solve()
🔥 Pattern to Remember for Interviews
If matrix element depends on:
i + j → count diagonals
i - j → count anti-diagonals
i × j → factor pair counting
Always convert 2D iteration into counting frequencies.
If you’d like next:
🔎 Optimization using prefix sums
⚡ Reducing constant factor
🧠More advanced diagonal counting problems
📊 Similar Codeforces level question
Tell me your goal (CP rating / interviews / placements).
can you please explain this more clearly, with more patience... 4️⃣ Optimization Insight
What is a diagonal
Perfect. Let’s slow down properly and build intuition from scratch.
You asked:
What is a diagonal?
And how does the optimization using diagonals work?
Let’s go very gently.
Step 1: Look at a Small Example
Take:
N = 3
M = 3
So indices look like this (we’ll write i+j in each cell):
