#include <bits/stdc++.
h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> arr(n);
vector<int> pos(n + 1); // 1-based index mapping
for (int i = 0; i < n; i++) {
cin >> arr[i]; pos[arr[i]] = i;
}
int rounds = 1; // At least one round
for (int i = 2; i <= n; i++) {
if (pos[i] < pos[i - 1]) {
rounds++;
}
}
cout << rounds << endl; return 0;
}
genralize this approach for me how can i use this further and what appraoch can be used
ChatGPT said:
Great — you're asking how to generalize this approach, understand when to use it, and
learn other ideas you can extend from it.
✅ Core Idea in This Code
The core concept is:
You're processing elements in natural order (1 to n) but they appear out of order in the
array, and you want to detect how often the order breaks (i.e., how many disjoint
"increasing runs" exist in the array when values are followed by their successors in increasing
index order).
You do this by:
1. Creating a pos[i] array — mapping values to their positions.
2. Scanning i = 2 to n, and counting how many times pos[i] < pos[i-1].
This means:
"The element i came before i-1 in the array, so I can't collect them in one pass — need a
new segment or round."
🔁 Generalization: What Problems Can Use This Pattern?
This is a powerful idea when dealing with ordering in permutations or sequences.
✅ Use Case 1: Counting Increasing Runs in a Permutation
You can use this approach to:
Split a permutation into minimal number of increasing subsequences (not
necessarily contiguous).
This is useful in greedy scheduling, stream processing, etc.
✅ Use Case 2: Segmenting Arrays Based on Ordered Constraints
Example:
You're given a sequence of tasks with dependencies (task i must come after i-1).
This approach helps you:
Count how many segments you'd need to schedule them without violating
dependencies.
✅ Use Case 3: Minimum Number of Passes to Sort
A variation of this idea is used in problems like:
How many passes over the array (left to right) are needed to sort it using adjacent swaps?
This approach detects breakpoints where the desired order is violated.
✅ Use Case 4: Compression of Monotonic Segments
In signal processing or compression:
Count how many runs of increasing values (in value or in position) exist in a stream.
Can be done using the same pos[i] < pos[i - 1] logic.
🚀 Generalized Algorithm Pattern
cpp
CopyEdit
// Input: arr[0...n-1] — a permutation of 1 to n
vector<int> pos(n + 1); // 1-based indexing
for (int i = 0; i < n; i++) {
pos[arr[i]] = i; // map value → index
}
int segments = 1;
for (int i = 2; i <= n; i++) {
if (pos[i] < pos[i - 1]) {
segments++; // order breaks
}
}
You can replace the loop condition pos[i] < pos[i - 1] with:
pos[i] > pos[i - 1] → count non-decreasing value segments
pos[i] == pos[i - 1] + 1 → count strictly adjacent segments
abs(pos[i] - pos[i - 1]) > 1 → count "skips"
🧠 Advanced Use: Tracking Any Custom Order
Let’s say you have a custom desired order target[], and an array arr[], and you want to
process arr in that order as efficiently as possible.
Build:
cpp
CopyEdit
unordered_map<int, int> order;
for (int i = 0; i < n; i++) {
order[target[i]] = i;
}
Now apply the same technique on arr by comparing:
cpp
CopyEdit
if (order[arr[i]] < order[arr[i-1]]) {
// ordering broken → need new group
}
This generalizes the pattern to any custom order, not just 1 → n.
Identification of greedy:
That's an excellent and important question — "When should I apply greedy, and when
should I avoid it?"
Greedy algorithms are powerful but not always correct, so let’s break it down:
✅ Checklist to Identify When Greedy Can Work
Ask yourself these questions:
1. ❓ Is the problem asking for an optimal solution?
E.g., minimize/maximize something (cost, time, distance, etc.)
👉 Greedy is not guaranteed to work unless a special property holds.
2. 🔁 Does the problem have an optimal substructure?
That means: solving a small part optimally helps solve the whole problem optimally.
Example: In Dijkstra’s algorithm (shortest path), choosing the smallest distance so far
always leads toward the correct solution.
3. 📉 Is the problem monotonic?
In greedy problems that work, once a choice is made, future choices are not hurt.
Think of Aggressive Cows: If distance d works, all smaller distances will also work
— a perfect setup for greedy + binary search.
4. 🔂 Does the problem allow irrevocable choices?
Greedy makes one decision at a time and never goes back.
If you need to "undo" or explore other possibilities → not greedy.
🚫 When Greedy Fails
If the answer to any of the following is yes, greedy probably won’t work:
✖️You need to explore multiple combinations of decisions to find the best.
✖️You need to backtrack or try all permutations.
✖️Choices are interdependent — one early decision affects future options.
✖️Problem explicitly mentions "reordering", "subsets", or "multiple ways" to
achieve something.
✅ Common Patterns Where Greedy Works
Problem Type Greedy Works? Notes
Interval Scheduling (non-overlapping) ✅ Sort by end time
Huffman Encoding ✅ Greedy + heap
Activity Selection ✅ Pick earliest finishing job
Fractional Knapsack ✅ Pick highest value/weight
Aggressive Cows ✅ Binary search + greedy placement
Minimum Number of Platforms ✅ Use sorted events
Dijkstra’s Shortest Path ✅ Greedy with priority queue
❌ Common Problems Where Greedy Fails
Problem Type Why Greedy Fails
Job Assignment (Min max load) Early bad decisions can ruin the schedule
Subset Sum / 0-1 Knapsack Must explore combinations, can’t take fractions
TSP (Traveling Salesman) Greedy may skip best routes
Maximum Subset / LIS Need dynamic programming to track all possibilities
Scheduling with deadlines Needs careful backtracking or sorting + logic
🧠 How to Decide in Practice?
1. Try Greedy First — see if a simple sort + selection works.
2. Test it on edge cases.
3. If it fails, ask:
o Can I reassign/reconsider choices later? → ❌ Greedy
o Are choices dependent on each other? → ❌ Greedy
o Do I need to track multiple states? → ❌ Greedy, try DP/backtracking
✅ Final Tip
Greedy is about staying shortsighted. If the problem rewards foresight, avoid greedy.
