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969. Pancake Sorting

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:

  • Choose an integer k where 1 <= k <= arr.length.
  • 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.

 

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).

Solutions

  • 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;
        }
    }
}

  • class Solution {
public:
    vector<int> pancakeSort(vector<int>& arr) {
        int n = arr.size();
        vector<int> ans;
        for (int i = n - 1; i > 0; --i) {
            int j = i;
            for (; j > 0 && arr[j] != i + 1; --j)
                ;
            if (j == i) continue;
            if (j > 0) {
                ans.push_back(j + 1);
                reverse(arr.begin(), arr.begin() + j + 1);
            }
            ans.push_back(i + 1);
            reverse(arr.begin(), arr.begin() + i + 1);
        }
        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


  • 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
    }
}

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