Solution Approach
Solution Overview
There are 2 main approaches to solve this problem:
- Brute Force Approach - Complex approach
- Optimized Greedy Approach - Complex approach
Approach 1: Brute Force Approach
Let's start by understanding the problem: we need to sort an array using only pancake flips, where a pancake flip reverses the first k elements of the array.
Thinking Process: The most straightforward approach would be to try all possible sequences of pancake flips and find one that sorts the array. However, this would be extremely inefficient, as there are many possible sequences.
Instead, we can use a greedy approach that sorts the array one element at a time, starting from the largest element. For each step, we find the position of the largest unsorted element, then use two flips to move it to its correct position at the end of the unsorted portion of the array.
Intuition: This approach is similar to selection sort, where we repeatedly find the largest element and move it to its correct position. The difference is that we're restricted to using pancake flips to move elements.
Algorithm:
- For each position i from n-1 down to 0 (where n is the length of the array):
- a. Find the index j of the largest element in the subarray arr[0...i].
- b. If j is not equal to i (the element is not already in its correct position):
- i. If j is not 0, perform a pancake flip with k=j+1 to move the largest element to the beginning of the array.
- ii. Perform a pancake flip with k=i+1 to move the largest element to position i.
- Return the list of k values used in the pancake flips.
Time Complexity:
Where n is the length of the array. For each of the n positions, we need to find the largest element in the subarray, which takes O(n) time.
Space Complexity:
We need O(n) space to store the list of k values used in the pancake flips.
Approach 2: Optimized Greedy Approach
We can optimize the brute force approach by using a more efficient way to find the largest element in each step.
Thinking Process: Since the problem states that all integers in the array are unique and form a permutation of the integers from 1 to n, we know that the largest element in the subarray arr[0...i] should be i+1. So, instead of finding the maximum element each time, we can directly search for the element with value i+1.
Intuition: This optimization reduces the time complexity of finding the largest element from O(n) to O(n) in the worst case, but it's more efficient on average because we can stop searching as soon as we find the element we're looking for.
Algorithm:
- For each position i from n-1 down to 0 (where n is the length of the array):
- a. Find the index j of the element with value i+1 in the subarray arr[0...i].
- b. If j is not equal to i (the element is not already in its correct position):
- i. If j is not 0, perform a pancake flip with k=j+1 to move the element to the beginning of the array.
- ii. Perform a pancake flip with k=i+1 to move the element to position i.
- Return the list of k values used in the pancake flips.
Time Complexity:
Where n is the length of the array. For each of the n positions, we need to find the element with value i+1 in the subarray, which takes O(n) time in the worst case.
Space Complexity:
We need O(n) space to store the list of k values used in the pancake flips.
Pseudocode
123456789101112131415161718192021222324252627282930313233function pancakeSort(arr): n = arr.length result = [] for i from n-1 down to 0: // Find the index of the largest element in arr[0...i] maxIndex = 0 for j from 1 to i: if arr[j] > arr[maxIndex]: maxIndex = j // If the largest element is already in the correct position, skip if maxIndex == i: continue // Move the largest element to the beginning of the array if maxIndex != 0: // Perform a pancake flip with k=maxIndex+1 reverse(arr, 0, maxIndex) result.append(maxIndex + 1) // Move the largest element to position i // Perform a pancake flip with k=i+1 reverse(arr, 0, i) result.append(i + 1) return result function reverse(arr, start, end): while start < end: arr[start], arr[end] = arr[end], arr[start] start++ end--String Problems
Learning Objective
Understand different approaches to solve the pancake sorting problem and analyze their efficiency.
Solution Approaches
Approach 1: Brute Force Approach
Let's start by understanding the problem: we need to sort an array using only pancake flips, where a pancake flip reverses the first k elements of the array.
Thinking Process: The most straightforward approach would be to try all possible sequences of pancake flips and find one that sorts the array. However, this would be extremely inefficient, as there are many possible sequences.
Instead, we can use a greedy approach that sorts the array one element at a time, starting from the largest element. For each step, we find the position of the largest unsorted element, then use two flips to move it to its correct position at the end of the unsorted portion of the array.
Intuition: This approach is similar to selection sort, where we repeatedly find the largest element and move it to its correct position. The difference is that we're restricted to using pancake flips to move elements.
Algorithm:
- For each position i from n-1 down to 0 (where n is the length of the array):
- a. Find the index j of the largest element in the subarray arr[0...i].
- b. If j is not equal to i (the element is not already in its correct position):
- i. If j is not 0, perform a pancake flip with k=j+1 to move the largest element to the beginning of the array.
- ii. Perform a pancake flip with k=i+1 to move the largest element to position i.
- Return the list of k values used in the pancake flips.
Approach 2: Optimized Greedy Approach
We can optimize the brute force approach by using a more efficient way to find the largest element in each step.
Thinking Process: Since the problem states that all integers in the array are unique and form a permutation of the integers from 1 to n, we know that the largest element in the subarray arr[0...i] should be i+1. So, instead of finding the maximum element each time, we can directly search for the element with value i+1.
Intuition: This optimization reduces the time complexity of finding the largest element from O(n) to O(n) in the worst case, but it's more efficient on average because we can stop searching as soon as we find the element we're looking for.
Algorithm:
- For each position i from n-1 down to 0 (where n is the length of the array):
- a. Find the index j of the element with value i+1 in the subarray arr[0...i].
- b. If j is not equal to i (the element is not already in its correct position):
- i. If j is not 0, perform a pancake flip with k=j+1 to move the element to the beginning of the array.
- ii. Perform a pancake flip with k=i+1 to move the element to position i.
- Return the list of k values used in the pancake flips.
Complexity Analysis
Brute Force Approach Approach
Time Complexity
Where n is the length of the array. For each of the n positions, we need to find the largest element in the subarray, which takes O(n) time.
Space Complexity
We need O(n) space to store the list of k values used in the pancake flips.
Optimized Greedy Approach Approach
Time Complexity
Where n is the length of the array. For each of the n positions, we need to find the element with value i+1 in the subarray, which takes O(n) time in the worst case.
Space Complexity
We need O(n) space to store the list of k values used in the pancake flips.
Pseudocode
Brute Force Approach Approach
123456789101112131415161718192021222324252627282930313233function pancakeSort(arr): n = arr.length result = [] for i from n-1 down to 0: // Find the index of the largest element in arr[0...i] maxIndex = 0 for j from 1 to i: if arr[j] > arr[maxIndex]: maxIndex = j // If the largest element is already in the correct position, skip if maxIndex == i: continue // Move the largest element to the beginning of the array if maxIndex != 0: // Perform a pancake flip with k=maxIndex+1 reverse(arr, 0, maxIndex) result.append(maxIndex + 1) // Move the largest element to position i // Perform a pancake flip with k=i+1 reverse(arr, 0, i) result.append(i + 1) return result function reverse(arr, start, end): while start < end: arr[start], arr[end] = arr[end], arr[start] start++ end--Optimized Greedy Approach Approach
1234567891011121314151617181920212223242526272829303132function pancakeSort(arr): n = arr.length result = [] for i from n-1 down to 0: // Find the index of the element with value i+1 j = 0 while j <= i and arr[j] != i+1: j++ // If the element is already in the correct position, skip if j == i: continue // Move the element to the beginning of the array if j != 0: // Perform a pancake flip with k=j+1 reverse(arr, 0, j) result.append(j + 1) // Move the element to position i // Perform a pancake flip with k=i+1 reverse(arr, 0, i) result.append(i + 1) return result function reverse(arr, start, end): while start < end: arr[start], arr[end] = arr[end], arr[start] start++ end--