Below is the implementation of the pancake sorting:
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152/** * Optimized Greedy Approach * Time Complexity: O(n²) - For each position, we need to find the element and perform flips * Space Complexity: O(n) - We need space to store the result array * * @param {number[]} arr - The array to be sorted * @return {number[]} - The sequence of k values for pancake flips */function pancakeSort(arr) { const n = arr.length; const result = []; // Helper function to reverse the first k elements of the array function flip(k) { let left = 0; let right = k - 1; while (left < right) { // Swap elements at left and right indices const temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } } // Sort the array one element at a time, starting from the largest for (let i = n - 1; i > 0; i--) { // Find the index of the element with value i+1 let j = 0; while (j <= i && 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) { result.push(j + 1); flip(j + 1); } // Move the element to its correct position result.push(i + 1); flip(i + 1); } return result;}
Let's break down the implementation:
Implement the pancake sorting solution in different programming languages.
Below is the implementation of the pancake sorting in different programming languages. Select a language tab to view the corresponding code.
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152/** * Optimized Greedy Approach * Time Complexity: O(n²) - For each position, we need to find the element and perform flips * Space Complexity: O(n) - We need space to store the result array * * @param {number[]} arr - The array to be sorted * @return {number[]} - The sequence of k values for pancake flips */function pancakeSort(arr) { const n = arr.length; const result = []; // Helper function to reverse the first k elements of the array function flip(k) { let left = 0; let right = k - 1; while (left < right) { // Swap elements at left and right indices const temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } } // Sort the array one element at a time, starting from the largest for (let i = n - 1; i > 0; i--) { // Find the index of the element with value i+1 let j = 0; while (j <= i && 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) { result.push(j + 1); flip(j + 1); } // Move the element to its correct position result.push(i + 1); flip(i + 1); } return result;}
First, understand that we need to sort an array using only pancake flips, where a pancake flip reverses the first k elements of the array.
Create a helper function to perform a pancake flip, which reverses the first k elements of the array.
Sort the array one element at a time, starting from the largest element and working down to the smallest.
For each position, find the index of the element that should be in that position.
Use two pancake flips to move each element to its correct position: first flip to bring the element to the beginning, then flip to move it to its correct position.
Keep track of the k values used in each pancake flip and return them as the result.
Handle edge cases such as when an element is already in its correct position.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the pancake sorting problem using a different approach than shown above.
If the array is already sorted, no flips are needed, and an empty array is returned.
If the array has only one element, it's already sorted, and an empty array is returned.
If an element is already in its correct position, we skip it and don't perform any flips for it.
If the element we want to move is already at the beginning of the array, we only need one flip to move it to its correct position.
Below is the implementation of the pancake sorting:
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152/** * Optimized Greedy Approach * Time Complexity: O(n²) - For each position, we need to find the element and perform flips * Space Complexity: O(n) - We need space to store the result array * * @param {number[]} arr - The array to be sorted * @return {number[]} - The sequence of k values for pancake flips */function pancakeSort(arr) { const n = arr.length; const result = []; // Helper function to reverse the first k elements of the array function flip(k) { let left = 0; let right = k - 1; while (left < right) { // Swap elements at left and right indices const temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } } // Sort the array one element at a time, starting from the largest for (let i = n - 1; i > 0; i--) { // Find the index of the element with value i+1 let j = 0; while (j <= i && 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) { result.push(j + 1); flip(j + 1); } // Move the element to its correct position result.push(i + 1); flip(i + 1); } return result;}
Let's break down the implementation:
Implement the pancake sorting solution in different programming languages.
Below is the implementation of the pancake sorting in different programming languages. Select a language tab to view the corresponding code.
12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152/** * Optimized Greedy Approach * Time Complexity: O(n²) - For each position, we need to find the element and perform flips * Space Complexity: O(n) - We need space to store the result array * * @param {number[]} arr - The array to be sorted * @return {number[]} - The sequence of k values for pancake flips */function pancakeSort(arr) { const n = arr.length; const result = []; // Helper function to reverse the first k elements of the array function flip(k) { let left = 0; let right = k - 1; while (left < right) { // Swap elements at left and right indices const temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left++; right--; } } // Sort the array one element at a time, starting from the largest for (let i = n - 1; i > 0; i--) { // Find the index of the element with value i+1 let j = 0; while (j <= i && 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) { result.push(j + 1); flip(j + 1); } // Move the element to its correct position result.push(i + 1); flip(i + 1); } return result;}
First, understand that we need to sort an array using only pancake flips, where a pancake flip reverses the first k elements of the array.
Create a helper function to perform a pancake flip, which reverses the first k elements of the array.
Sort the array one element at a time, starting from the largest element and working down to the smallest.
For each position, find the index of the element that should be in that position.
Use two pancake flips to move each element to its correct position: first flip to bring the element to the beginning, then flip to move it to its correct position.
Keep track of the k values used in each pancake flip and return them as the result.
Handle edge cases such as when an element is already in its correct position.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the pancake sorting problem using a different approach than shown above.
If the array is already sorted, no flips are needed, and an empty array is returned.
If the array has only one element, it's already sorted, and an empty array is returned.
If an element is already in its correct position, we skip it and don't perform any flips for it.
If the element we want to move is already at the beginning of the array, we only need one flip to move it to its correct position.