Leetcode 969: Pancake Sorting
You are tasked with sorting an array of integers by performing a series of pancake flips. A pancake flip involves reversing the order of elements from the start of the array up to a specified index. The goal is to determine the sequence of flips needed to sort the array in ascending order.
Problem
Approach
Steps
Complexity
Input: An array of integers where each integer is unique and represents a permutation of numbers from 1 to n.
Example: Input: arr = [5, 1, 4, 2, 3]
Constraints:
• 1 <= arr.length <= 100
• 1 <= arr[i] <= arr.length
• All integers in arr are unique.
Output: An array of integers where each value corresponds to the position up to which a pancake flip is performed. The sequence of flips must sort the input array.
Example: Output: [5, 3, 4, 3, 2]
Constraints:
• The number of flips should not exceed 10 * arr.length.
Goal: Sort the array using a series of pancake flips.
Steps:
• Identify the largest unsorted element in the array.
• Perform a flip to bring the largest element to the beginning of the array.
• Perform another flip to move the largest element to its correct position in the sorted part of the array.
• Repeat until the entire array is sorted.
Goal: The constraints ensure that the problem is computationally feasible and the inputs are well-formed.
Steps:
• 1 <= arr.length <= 100
• 1 <= arr[i] <= arr.length
• All integers in arr are unique.
Assumptions:
• The array contains only integers.
• Each integer in the array is within the range [1, arr.length].
• The input array is a permutation of numbers from 1 to arr.length.
• Input: Input: arr = [3, 1, 4, 2]
• Explanation: We perform the following flips:
1. Flip at k = 3: arr = [4, 1, 3, 2]
2. Flip at k = 4: arr = [2, 3, 1, 4]
3. Flip at k = 3: arr = [3, 1, 2, 4]
4. Flip at k = 2: arr = [1, 3, 2, 4]
5. Flip at k = 3: arr = [1, 2, 3, 4]
Final Output: [3, 4, 3, 2, 3]
• Input: Input: arr = [1, 2, 3]
• Explanation: The array is already sorted, so no flips are needed. Output: []
Approach: The solution involves iteratively moving the largest unsorted element to its correct position using at most two flips per element.
Observations:
• A flip only changes the order of elements in the prefix of the array.
• To sort the array, we can repeatedly position the largest unsorted element at its correct location.
• Identify the position of the largest unsorted element.
• Perform a flip to move it to the beginning of the array.
• Perform another flip to place it at its correct position.
Steps:
• Start with the entire array.
• For each iteration, find the position of the largest unsorted element.
• Perform a flip to bring it to the beginning of the array.
• Perform another flip to move it to its correct position in the sorted part of the array.
• Repeat until the array is sorted.
Empty Inputs:
• An empty array would not require any flips.
Large Inputs:
• For an array with maximum length (100), ensure the algorithm completes within 1000 flips.
Special Values:
• An already sorted array should result in no flips.
Constraints:
• Verify that the input is a permutation of integers from 1 to n.
vector<int> pancakeSort(vector<int>& arr) {
// find next largest
// flip its index so that largest come first
// flip one more time, so that the first goes to end
vector<int> res;
int i;
for(int x = arr.size(); x > 0; x--) {
for(i = 0; arr[i] != x; i++) {};
reverse(arr.begin(), arr.begin() + i + 1);
res.push_back(i + 1);
reverse(arr.begin(), arr.begin() + x);
res.push_back(x);
}
return res;
}
1 : Function Definition
vector<int> pancakeSort(vector<int>& arr) {
The function definition begins the implementation of the pancake sorting algorithm.
2 : Variable Initialization
vector<int> res;
Initialize a vector `res` to store the sequence of flip operations.
3 : Variable Declaration
int i;
Declare an integer `i` for use in loops and iteration.
4 : Loop Construct
for(int x = arr.size(); x > 0; x--) {
Start a loop from the size of the array down to 1 to process all unsorted elements.
5 : Loop Iteration
for(i = 0; arr[i] != x; i++) {};
Find the index `i` of the current largest unsorted element `x`.
6 : Reverse Operation
reverse(arr.begin(), arr.begin() + i + 1);
Reverse the segment of the array from the start to index `i` to bring the largest element to the front.
7 : Vector Operation
res.push_back(i + 1);
Record the flip operation by adding `i + 1` (1-based index) to the result vector.
8 : Reverse Operation
reverse(arr.begin(), arr.begin() + x);
Reverse the segment of the array from the start to index `x`, moving the largest element to its final position.
9 : Vector Operation
res.push_back(x);
Record the second flip operation by adding `x` (the size of the remaining unsorted segment) to the result vector.
10 : Return Statement
return res;
Return the result vector containing the sequence of flip operations.
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The worst case involves performing two flips for each of the n elements, resulting in O(n^2) complexity.
Best Case: O(1)
Worst Case: O(1)
Description: The solution operates in-place, so the space complexity is O(1).
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