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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. 1 <= A.length <= 100
  2. 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
        }
    }
    
    

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