C++ Program to Next_permutation for All Arrangements with Explanation
C++
Hard
STL
34 views
2 min read
220 words
This problem helps you practice core C++ fundamentals in a practical way. It builds intuition around permutation, next, all. Let’s break it down step by step so you can implement it confidently.
Problem Statement
Generate next lexicographic permutation.
Real Life: Finding all possible arrangements.
Input Format
An integer n (size of array)
Then n integers (current permutation).
Output Format
Next lexicographic permutation of the array.
If no next permutation exists, print the smallest permutation.
Constraints
• n ≥ 1
• Elements are integers
• Array represents a valid permutation
Concept Explanation
A lexicographic permutation means dictionary order.
The “next” permutation is the next greater arrangement using the same elements.
This is useful when generating all possible arrangements one by one.
Step-by-Step Logic
1.Start from the right end of the array.
2.Find the first index i such that arr[i] < arr[i+1].
• If no such index exists, the array is the last permutation.
3.Again from the right, find index j such that arr[j] > arr[i].
4.Swap arr[i] and arr[j].
5.Reverse the elements from index i+1 to the end of the array.
6.The resulting array is the next lexicographic permutation.
Code Solution
This explanation is written for learning purposes and to help beginners understand the concept clearly.
void stl_q15_next_permutation() {
vector<int> numbers = {1, 2, 3};
cout << "All permutations:" << endl;
do {
for(int num : numbers) {
cout << num << " ";
}
cout << endl;
} while(next_permutation(numbers.begin(), numbers.end()));
}
int main() {
cout << "=== C++ PRACTICE QUESTIONS ===" << endl;
cout << "All topics covered with detailed explanations!" << endl;
return 0;
}
Common Mistakes
- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).
Solution Guide
Problem
Generate next lexicographic permutation.
Real Life: Finding all possible arrangements.
Input / Output
Input
An integer n (size of array)
Then n integers (current permutation).
Output
Next lexicographic permutation of the array.
If no next permutation exists, print the smallest permutation.
Constraints
• n ≥ 1
• Elements are integers
• Array represents a valid permutation
Explanation
Concept Explanation
A lexicographic permutation means dictionary order.
The “next” permutation is the next greater arrangement using the same elements.
This is useful when generating all possible arrangements one by one.
Step-by-Step Explanation
1.Start from the right end of the array.
2.Find the first index i such that arr[i] < arr[i+1].
• If no such index exists, the array is the last permutation.
3.Again from the right, find index j such that arr[j] > arr[i].
4.Swap arr[i] and arr[j].
5.Reverse the elements from index i+1 to the end of the array.
6.The resulting array is the next lexicographic permutation.
Details
Common Mistakes
- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).
Official Solution
void stl_q15_next_permutation() {
vector<int> numbers = {1, 2, 3};
cout << "All permutations:" << endl;
do {
for(int num : numbers) {
cout << num << " ";
}
cout << endl;
} while(next_permutation(numbers.begin(), numbers.end()));
}
int main() {
cout << "=== C++ PRACTICE QUESTIONS ===" << endl;
cout << "All topics covered with detailed explanations!" << endl;
return 0;
}
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