Formatted question description: https://leetcode.ca/all/969.html

969. Pancake Sorting (Medium)

Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A. We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A.

Return the k-values corresponding to a sequence of pancake flips that sort A. Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.

Example 1:

Input: [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: A = [3, 2, 4, 1]
After 1st flip (k=4): A = [1, 4, 2, 3]
After 2nd flip (k=2): A = [4, 1, 2, 3]
After 3rd flip (k=4): A = [3, 2, 1, 4]
After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted.

Example 2:

Input: [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.

Note:

1 <= A.length <= 100
A[i] is a permutation of [1, 2, ..., A.length]

Companies:
Square

Related Topics:
Array, Sort

Solution 1.

class Solution {
    public List<Integer> pancakeSort(int[] A) {
        List<Integer> list = new ArrayList<Integer>();
        int length = A.length;
        for (int i = length; i >= 1; i--) {
            if (A[i - 1] != i) {
                int flipIndex = 0;
                for (int j = length - 1; j >= 0; j--) {
                    if (A[j] == i) {
                        flipIndex = j;
                        break;
                    }
                }
                if (flipIndex > 0) {
                    pancakeFlip(A, flipIndex);
                    list.add(flipIndex + 1);
                }
                pancakeFlip(A, i - 1);
                list.add(i);
            }
        }
        return list;
    }

    public void pancakeFlip(int[] array, int index) {
        int left = 0, right = index;
        while (left < right) {
            int temp = array[left];
            array[left] = array[right];
            array[right] = temp;
            left++;
            right--;
        }
    }
}

############

class Solution {
    public List<Integer> pancakeSort(int[] arr) {
        int n = arr.length;
        List<Integer> ans = new ArrayList<>();
        for (int i = n - 1; i > 0; --i) {
            int j = i;
            for (; j > 0 && arr[j] != i + 1; --j)
                ;
            if (j < i) {
                if (j > 0) {
                    ans.add(j + 1);
                    reverse(arr, j);
                }
                ans.add(i + 1);
                reverse(arr, i);
            }
        }
        return ans;
    }

    private void reverse(int[] arr, int j) {
        for (int i = 0; i < j; ++i, --j) {
            int t = arr[i];
            arr[i] = arr[j];
            arr[j] = t;
        }
    }
}

// OJ: https://leetcode.com/problems/pancake-sorting
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    vector<int> pancakeSort(vector<int>& A) {
        vector<int> ans;
        for (int i = A.size(); i > 0; --i) {
            int j = i - 1;
            for (; j >= 0 && A[j] != i; --j);
            reverse(A.begin(), A.begin() + j + 1);
            ans.push_back(j + 1);
            reverse(A.begin(), A.begin() + i);
            ans.push_back(i);
        }
        return ans;
    }
};

class Solution:
    def pancakeSort(self, arr: List[int]) -> List[int]:
        def reverse(arr, j):
            i = 0
            while i < j:
                arr[i], arr[j] = arr[j], arr[i]
                i, j = i + 1, j - 1

        n = len(arr)
        ans = []
        for i in range(n - 1, 0, -1):
            j = i
            while j > 0 and arr[j] != i + 1:
                j -= 1
            if j < i:
                if j > 0:
                    ans.append(j + 1)
                    reverse(arr, j)
                ans.append(i + 1)
                reverse(arr, i)
        return ans

############

class Solution(object):
    def pancakeSort(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        N = len(A)
        res = []
        for x in range(N, 0, -1):
            i = A.index(x)
            res.extend([i + 1, x])
            A = A[:i:-1] + A[:i]
        return res

func pancakeSort(arr []int) []int {
	var ans []int
	n := len(arr)
	reverse := func(j int) {
		for i := 0; i < j; i, j = i+1, j-1 {
			arr[i], arr[j] = arr[j], arr[i]
		}
	}
	for i := n - 1; i > 0; i-- {
		j := i
		for ; j > 0 && arr[j] != i+1; j-- {
		}
		if j < i {
			if j > 0 {
				ans = append(ans, j+1)
				reverse(j)
			}
			ans = append(ans, i+1)
			reverse(i)
		}
	}
	return ans
}

function pancakeSort(arr: number[]): number[] {
    let ans = [];
    for (let n = arr.length; n > 1; n--) {
        let index = 0;
        for (let i = 1; i < n; i++) {
            if (arr[i] >= arr[index]) {
                index = i;
            }
        }
        if (index == n - 1) continue;
        reverse(arr, index);
        reverse(arr, n - 1);
        ans.push(index + 1);
        ans.push(n);
    }
    return ans;
}

function reverse(nums: Array<number>, end: number): void {
    for (let i = 0, j = end; i < j; i++, j--) {
        [nums[i], nums[j]] = [nums[j], nums[i]];
    }
}

impl Solution {
    pub fn pancake_sort(mut arr: Vec<i32>) -> Vec<i32> {
        let mut res = vec![];
        for n in (1..arr.len()).rev() {
            let mut max_idx = 0;
            for idx in 0..=n {
                if arr[max_idx] < arr[idx] {
                    max_idx = idx;
                }
            }
            if max_idx != n {
                if max_idx != 0 {
                    arr[..=max_idx].reverse();
                    res.push(max_idx as i32 + 1);
                }
                arr[..=n].reverse();
                res.push(n as i32 + 1);
            }
        }
        res
    }
}

969 - Pancake Sorting

969. Pancake Sorting (Medium)

Solution 1.

All Problems

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969. Pancake Sorting (Medium)

Solution 1.

All Problems

All Solutions