24. Last Moment Before All Ants Fall Out

✅ GFG solution to the Last Moment Before All Ants Fall Out problem: find when the last ant falls off the plank using optimal physics-based approach. 🚀

The problem can be found at the following link: 🔗 Question Link

🧩 Problem Description

You are given a wooden plank of length n units. Some ants are walking on the plank, some are walking left and some are walking right. Each ant moves with speed 1 unit per second.

When two ants moving in opposite directions meet at some point, they change their directions and continue moving again. Assume changing directions does not take any additional time. When an ant reaches one end of the plank at time t, it falls out of the plank immediately.

Given an integer n and two integer arrays left[] and right[], representing the positions of ants moving to the left and right respectively, return the time when the last ant(s) fall out of the plank.

📘 Examples

Example 1

Input: n = 4, left[] = [2], right[] = [0, 1, 3]
Output: 4
Explanation: As seen in the problem, the last ant falls off the plank at t = 4.

Example 2

Input: n = 4, left[] = [], right[] = [0, 1, 2, 3, 4]
Output: 4
Explanation: All ants are going to the right, the ant at index 0 needs 4 seconds to fall.

Example 3

Input: n = 3, left[] = [0], right[] = [3]
Output: 0
Explanation: The ants will fall off the plank as they are already on the end of the plank.

🔒 Constraints

$1 \le n \le 10^5$
$0 \le \text{left.length, right.length} \le n + 1$
$0 \le \text{left}[i], \text{right}[i] \le n$
$1 \le \text{left.length + right.length} \le n + 1$
All values are unique and each value appears in only one array

✅ My Approach

The key insight is that when two ants collide and change directions, we can think of them as passing through each other without changing direction. This is because the system behaves identically - the positions and movements remain the same, only the "identity" of ants changes.

Physics-Based Optimization

Key Insight:
- When ants collide, instead of tracking direction changes, we can imagine they pass through each other.
- The time for the last ant to fall is simply the maximum time any ant needs to reach either end.
Calculate Time for Each Ant:
- For ants moving left: Time = current position (distance to left edge = 0)
- For ants moving right: Time = n - current position (distance to right edge = n)
Find Maximum Time:
- The answer is the maximum of all individual times.
- This represents when the last ant will fall off the plank.
Why This Works:
- Collisions don't affect the overall system behavior.
- Each ant will eventually fall off at the same time as if there were no collisions.
- We only need to find the ant that takes the longest time to reach an edge.

📝 Time and Auxiliary Space Complexity

Expected Time Complexity: O(L + R), where L is the length of left array and R is the length of right array. We iterate through both arrays once to find the maximum values.
Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space for storing the maximum value and loop variables.

🧑‍💻 Code (C++)

class Solution {
public:
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int ans = 0;
        for (int x : left) ans = max(ans, x);
        for (int x : right) ans = max(ans, n - x);
        return ans;
    }
};

⚡ View Alternative Approaches with Code and Analysis

📊 2️⃣ Single Pass Maximum Calculation

💡 Algorithm Steps:

Calculate maximum from left array in single iteration
Calculate maximum from transformed right array simultaneously
Return the overall maximum without intermediate storage

class Solution {
public:
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int leftMax = left.empty() ? 0 : *max_element(left.begin(), left.end());
        int rightMax = 0;
        for (int pos : right) rightMax = max(rightMax, n - pos);
        return max(leftMax, rightMax);
    }
};

📝 Complexity Analysis:

Time: ⏱️ O(L + R) where L, R are array sizes
Auxiliary Space: 💾 O(1) - constant space

✅ Why This Approach?

Uses STL max_element for left array optimization
Minimal memory allocation
Clear separation of left and right processing

📊 3️⃣ Range-Based Iterator Approach

💡 Algorithm Steps:

Use range-based loops for cleaner syntax
Combine both calculations in minimal code
Leverage modern C++ features for optimization

class Solution {
public:
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int result = 0;
        for (auto& pos : left) result = max(result, pos);
        for (auto& pos : right) result = max(result, n - pos);
        return result;
    }
};

📝 Complexity Analysis:

Time: ⏱️ O(L + R)
Auxiliary Space: 💾 O(1)

✅ Why This Approach?

Modern C++11 syntax
Reference-based iteration prevents copying
Compiler-optimized range loops

📊 4️⃣ Branch-Free Computation

💡 Algorithm Steps:

Eliminate conditional branches for better CPU performance
Use arithmetic operations instead of max function
Better CPU pipeline utilization
Consistent execution time

class Solution {
public:
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int result = 0;
        for (int i = 0; i < left.size(); ++i) {
            result = result > left[i] ? result : left[i];
        }
        for (int i = 0; i < right.size(); ++i) {
            int val = n - right[i];
            result = result > val ? result : val;
        }
        return result;
    }
};

📝 Complexity Analysis:

Time: ⏱️ O(L + R) - branch-free execution
Auxiliary Space: 💾 O(1) - constant space

✅ Why This Approach?

Eliminates branch mispredictions
Better CPU cache utilization
Consistent performance profile

📊 5️⃣ Functional Programming Style

💡 Algorithm Steps:

Use STL algorithms for both arrays
Combine results using max function
Leverage compiler optimizations for STL

class Solution {
public:
    int getLastMoment(int n, vector<int>& left, vector<int>& right) {
        int leftMax = left.empty() ? 0 : *max_element(left.begin(), left.end());
        int rightMax = 0;
        if (!right.empty()) {
            vector<int> rightTimes(right.size());
            transform(right.begin(), right.end(), rightTimes.begin(), 
                     [n](int pos) { return n - pos; });
            rightMax = *max_element(rightTimes.begin(), rightTimes.end());
        }
        return max(leftMax, rightMax);
    }
};

📝 Complexity Analysis:

Time: ⏱️ O(L + R)
Auxiliary Space: 💾 O(R) - for transformed array

✅ Why This Approach?

Functional programming paradigm
STL algorithm optimization
Clear separation of concerns

🆚 🔍 Comparison of Approaches

🚀 Approach

⏱️ Time Complexity

💾 Space Complexity

✅ Pros

⚠️ Cons

🔍 Basic Two-Loop

🟡 O(L + R)

🟢 O(1)

🚀 Simple and direct

💾 Basic approach

🔺 Single Pass Maximum

🟡 O(L + R)

🟢 O(1)

🔧 STL optimization

💾 Slightly more complex

⏰ Range-Based Iterator

🟡 O(L + R)

🟢 O(1)

🚀 Modern C++ syntax

🔄 Compiler dependent optimization

📊 Branch-Free Computation

🟡 O(L + R)

🟢 O(1)

⚡ Better CPU utilization

🔧 More complex logic

🎯 Functional Programming

🟡 O(L + R)

🟡 O(R)

🔧 Clear functional style

💾 Extra space usage

🏆 Best Choice Recommendation

🎯 Scenario

🎖️ Recommended Approach

🔥 Performance Rating

⚡ Competitive Programming

🥇 Basic Two-Loop

★★★★★

📊 Production Code

🥈 Range-Based Iterator

★★★★☆

🎯 Performance Critical

🥉 Branch-Free Computation

★★★★☆

🚀 Interview Settings

🏅 Basic Two-Loop

★★★★★

🔧 Learning/Educational

🎖️ Single Pass Maximum

★★★☆☆

🧑‍💻 Code (Java)

class Solution {
    public int getLastMoment(int n, int left[], int right[]) {
        int ans = 0;
        for (int x : left) ans = Math.max(ans, x);
        for (int x : right) ans = Math.max(ans, n - x);
        return ans;
    }
}

🐍 Code (Python)

class Solution:
    def getLastMoment(self, n, left, right):
        ans = 0
        for x in left: ans = max(ans, x)
        for x in right: ans = max(ans, n - x)
        return ans

🧠 Contribution and Support

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: 📬 Any Questions?. Let's make this learning journey more collaborative!

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Last updated 6 months ago

hashtag🧩 Problem Description

hashtag📘 Examples

hashtagExample 1

hashtagExample 2

hashtagExample 3

hashtag🔒 Constraints

hashtag✅ My Approach

hashtagPhysics-Based Optimization

hashtag📝 Time and Auxiliary Space Complexity

hashtag🧑‍💻 Code (C++)

hashtag📊 2️⃣ Single Pass Maximum Calculation

hashtag💡 Algorithm Steps:

hashtag📝 Complexity Analysis:

hashtag✅ Why This Approach?

hashtag📊 3️⃣ Range-Based Iterator Approach

hashtag💡 Algorithm Steps:

hashtag📝 Complexity Analysis:

hashtag✅ Why This Approach?

hashtag📊 4️⃣ Branch-Free Computation

hashtag💡 Algorithm Steps:

hashtag📝 Complexity Analysis:

hashtag✅ Why This Approach?

hashtag📊 5️⃣ Functional Programming Style

hashtag💡 Algorithm Steps:

hashtag📝 Complexity Analysis:

hashtag✅ Why This Approach?

hashtag🆚 🔍 Comparison of Approaches

hashtag🏆 Best Choice Recommendation

hashtag🧑‍💻 Code (Java)

hashtag🐍 Code (Python)

hashtag🧠 Contribution and Support

hashtag📍Visitor Count

🧩 Problem Description

📘 Examples

Example 1

Example 2

Example 3

🔒 Constraints

✅ My Approach

Physics-Based Optimization

📝 Time and Auxiliary Space Complexity

🧑‍💻 Code (C++)

📊 2️⃣ Single Pass Maximum Calculation

💡 Algorithm Steps:

📝 Complexity Analysis:

✅ Why This Approach?

📊 3️⃣ Range-Based Iterator Approach

💡 Algorithm Steps:

📝 Complexity Analysis:

✅ Why This Approach?

📊 4️⃣ Branch-Free Computation

💡 Algorithm Steps:

📝 Complexity Analysis:

✅ Why This Approach?

📊 5️⃣ Functional Programming Style

💡 Algorithm Steps:

📝 Complexity Analysis:

✅ Why This Approach?

🆚 🔍 Comparison of Approaches

🏆 Best Choice Recommendation

🧑‍💻 Code (Java)

🐍 Code (Python)

🧠 Contribution and Support

📍Visitor Count