Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1089.html
1089. Duplicate Zeros (Easy)
Given a fixed length array arr of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.
Note that elements beyond the length of the original array are not written.
Do the above modifications to the input array in place, do not return anything from your function.
Example 1:
Input: [1,0,2,3,0,4,5,0] Output: null Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
Example 2:
Input: [1,2,3] Output: null Explanation: After calling your function, the input array is modified to: [1,2,3]
Note:
1 <= arr.length <= 100000 <= arr[i] <= 9
Related Topics:
Array
Solution 1.
// OJ: https://leetcode.com/problems/duplicate-zeros/
// Time: O(N)
// Space: O(1)
class Solution {
public:
void duplicateZeros(vector<int>& A) {
int cnt = 0;
for (int n : A) cnt += n == 0;
if (cnt == 0) return;
for (int i = A.size() - 1; i >= 0; --i) {
cnt -= A[i] == 0;
int j = i + cnt;
if (j < A.size()) A[j] = A[i];
if (j + 1 < A.size() && A[i] == 0) A[j + 1] = 0;
}
}
};
-
class Solution { public void duplicateZeros(int[] arr) { int length = arr.length; int slow = 0, fast = 0; while (fast < length) { int num = arr[slow]; slow++; if (num == 0) fast += 2; else fast++; } if (fast == slow) return; if (arr[slow] == 0) fast++; while (fast >= length) { int num = arr[slow]; slow--; if (num == 0) { fast -= 2; if (fast == length - 2) arr[length - 1] = 0; } else fast--; } while (fast > slow) { int num = arr[slow]; slow--; arr[fast] = num; fast--; if (num == 0) { arr[fast] = num; fast--; } } } } ############ class Solution { public void duplicateZeros(int[] arr) { int n = arr.length; int i = -1, k = 0; while (k < n) { ++i; k += arr[i] > 0 ? 1 : 2; } int j = n - 1; if (k == n + 1) { arr[j--] = 0; --i; } while (j >= 0) { arr[j] = arr[i]; if (arr[i] == 0) { arr[--j] = arr[i]; } --i; --j; } } } -
// OJ: https://leetcode.com/problems/duplicate-zeros/ // Time: O(N) // Space: O(1) class Solution { public: void duplicateZeros(vector<int>& A) { int cnt = 0; for (int n : A) cnt += n == 0; if (cnt == 0) return; for (int i = A.size() - 1; i >= 0; --i) { cnt -= A[i] == 0; int j = i + cnt; if (j < A.size()) A[j] = A[i]; if (j + 1 < A.size() && A[i] == 0) A[j + 1] = 0; } } }; -
class Solution: def duplicateZeros(self, arr: List[int]) -> None: """ Do not return anything, modify arr in-place instead. """ n = len(arr) i, k = -1, 0 while k < n: i += 1 k += 1 if arr[i] else 2 j = n - 1 if k == n + 1: arr[j] = 0 i, j = i - 1, j - 1 while ~j: if arr[i] == 0: arr[j] = arr[j - 1] = arr[i] j -= 1 else: arr[j] = arr[i] i, j = i - 1, j - 1 ############ # 1089. Duplicate Zeros # https://leetcode.com/problems/duplicate-zeros/ class Solution: def duplicateZeros(self, arr: List[int]) -> None: """ Do not return anything, modify arr in-place instead. """ zeroes = arr.count(0) n = len(arr) for i in range(n - 1, -1, -1): if i + zeroes < n: arr[i + zeroes] = arr[i] if arr[i] == 0: zeroes -= 1 if i + zeroes < n: arr[i + zeroes] = 0 -
func duplicateZeros(arr []int) { n := len(arr) i, k := -1, 0 for k < n { i, k = i+1, k+1 if arr[i] == 0 { k++ } } j := n - 1 if k == n+1 { arr[j] = 0 i, j = i-1, j-1 } for j >= 0 { arr[j] = arr[i] if arr[i] == 0 { j-- arr[j] = arr[i] } i, j = i-1, j-1 } } -
impl Solution { pub fn duplicate_zeros(arr: &mut Vec<i32>) { let n = arr.len(); let mut i = 0; let mut j = 0; while j < n { if arr[i] == 0 { j += 1; } j += 1; i += 1; } while i > 0 { if arr[i - 1] == 0 { if j <= n { arr[j - 1] = arr[i - 1]; } j -= 1; } arr[j - 1] = arr[i - 1]; i -= 1; j -= 1; } } }