From 977837655ec24fdf1ead2079b5dbc3b40529dfc7 Mon Sep 17 00:00:00 2001
From: ImmidiSivani <147423543+ImmidiSivani@users.noreply.github.com>
Date: Sun, 28 Jul 2024 15:39:45 +0530
Subject: [PATCH] solution added to 969
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+---
+id: pancake-sorting
+title: Pancake Sorting
+sidebar_label: 969-Pancake Sorting
+tags:
+ - Sorting
+ - Array
+ - LeetCode
+ - Java
+ - Python
+ - C++
+description: "This is a solution to the Pancake Sorting problem on LeetCode."
+sidebar_position: 13
+---
+
+## Problem Description
+
+Given an array of integers `arr`, sort the array by performing a series of pancake flips.
+
+In one pancake flip, we do the following steps:
+1. Choose an integer `k` where `1 <= k <= arr.length`.
+2. Reverse the sub-array `arr[0...k-1]` (0-indexed).
+
+For example, if `arr = [3,2,1,4]` and we performed a pancake flip choosing `k = 3`, we reverse the sub-array `[3,2,1]`, so `arr = [1,2,3,4]` after the pancake flip at `k = 3`.
+
+Return an array of the `k`-values corresponding to a sequence of pancake flips that sort `arr`. Any valid answer that sorts the array within `10 * arr.length` flips will be judged as correct.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: arr = [3,2,4,1]
+Output: [4,2,4,3]
+Explanation:
+We perform 4 pancake flips, with k values 4, 2, 4, and 3.
+Starting state: arr = [3, 2, 4, 1]
+After 1st flip (k = 4): arr = [1, 4, 2, 3]
+After 2nd flip (k = 2): arr = [4, 1, 2, 3]
+After 3rd flip (k = 4): arr = [3, 2, 1, 4]
+After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted.
+```
+
+**Example 2:**
+
+```
+Input: arr = [1,2,3]
+Output: []
+Explanation: The input is already sorted, so there is no need to flip anything.
+Note that other answers, such as [3, 3], would also be accepted.
+```
+
+### Constraints
+
+- `1 <= arr.length <= 100`
+- `1 <= arr[i] <= arr.length`
+- All integers in `arr` are unique (i.e., `arr` is a permutation of the integers from 1 to `arr.length`).
+
+---
+
+## Solution for Pancake Sorting Problem
+
+To solve this problem, we perform a series of pancake flips to sort the array in ascending order. A pancake flip reverses the sub-array from the start to a chosen index `k`. The goal is to bring the largest unsorted element to its correct position iteratively.
+
+### Approach
+
+1. **Identify Largest Element:** Find the largest unsorted element in the array.
+2. **Bring to Front:** If this largest element is not already at the front, flip it to bring it to the front.
+3. **Move to Correct Position:** Flip the entire sub-array up to the correct position of the largest element to place it at its sorted position.
+4. **Repeat:** Repeat the above steps for the next largest unsorted elements, excluding the already sorted part of the array.
+
+### Code in Different Languages
+
+
+
+
+
+```cpp
+class Solution {
+public:
+ vector pancakeSort(vector& arr) {
+ vector res;
+ int n = arr.size();
+ for (int i = n; i > 1; --i) {
+ int maxIdx = max_element(arr.begin(), arr.begin() + i) - arr.begin();
+ if (maxIdx == i - 1) continue;
+ if (maxIdx != 0) {
+ res.push_back(maxIdx + 1);
+ reverse(arr.begin(), arr.begin() + maxIdx + 1);
+ }
+ res.push_back(i);
+ reverse(arr.begin(), arr.begin() + i);
+ }
+ return res;
+ }
+};
+```
+
+
+
+
+
+```java
+class Solution {
+ public List pancakeSort(int[] arr) {
+ List res = new ArrayList<>();
+ int n = arr.length;
+ for (int i = n; i > 1; --i) {
+ int maxIdx = findMaxIndex(arr, i);
+ if (maxIdx == i - 1) continue;
+ if (maxIdx != 0) {
+ res.add(maxIdx + 1);
+ flip(arr, maxIdx + 1);
+ }
+ res.add(i);
+ flip(arr, i);
+ }
+ return res;
+ }
+
+ private int findMaxIndex(int[] arr, int k) {
+ int maxIdx = 0;
+ for (int i = 1; i < k; i++) {
+ if (arr[i] > arr[maxIdx]) {
+ maxIdx = i;
+ }
+ }
+ return maxIdx;
+ }
+
+ private void flip(int[] arr, int k) {
+ for (int i = 0, j = k - 1; i < j; i++, j--) {
+ int temp = arr[i];
+ arr[i] = arr[j];
+ arr[j] = temp;
+ }
+ }
+}
+```
+
+
+
+
+
+```python
+class Solution:
+ def pancakeSort(self, arr: List[int]) -> List[int]:
+ res = []
+ n = len(arr)
+ for i in range(n, 1, -1):
+ max_idx = arr.index(max(arr[:i]))
+ if max_idx == i - 1:
+ continue
+ if max_idx != 0:
+ res.append(max_idx + 1)
+ arr[:max_idx + 1] = arr[:max_idx + 1][::-1]
+ res.append(i)
+ arr[:i] = arr[:i][::-1]
+ return res
+```
+
+
+
+
+#### Complexity Analysis
+
+- **Time Complexity**: $O(n^2)$, where `n` is the length of the array. Finding the maximum and flipping can each take up to `O(n)`, and we do this for each of the `n` elements.
+- **Space Complexity**: $O(1)$, excluding the output list.
+
+---
+
+