1089_Duplicate_Zeros

qiyuangong · qiyuangong · commit d90e3fe1cf42 · 2020-02-29T19:14:31.000+08:00
diff --git a/README.md b/README.md
@@ -217,6 +217,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 977 | [Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/977_Squares_of_a_Sorted_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/977_Squares_of_a_Sorted_Array.java) | 1. Sort, O(nlogn) and O(n)<br>2. Two point, O(n) and O(n) |
 | 973 | [K Closest Points to Origin](https://leetcode.com/problems/k-closest-points-to-origin/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/973_K_Closest_Points_to_Origin.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/973_K_Closest_Points_to_Origin.java) | 1. Sort and get 0-K, O(nlogn) and O(1)<br>2. Min Heap, O(nlogk) and O(k) |
 | 1064 | [Fixed Point](https://leetcode.com/problems/fixed-point/) &hearts; | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1064_Fixed_Point.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1064_Fixed_Point.java) | 1. Go through index and value, until find solution encounter index < value, O(n) and O(1)<br>2. Binary search, O(logn) and O(1) |
+| 1089 | [Duplicate Zeros](https://leetcode.com/problems/duplicate-zeros/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1089_Duplicate_Zeros.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1089_Duplicate_Zeros.java) | 2 Pass, store last position and final move steps, O(n) and O(1) |
 | 1260 | [Shift 2D Grid](https://leetcode.com/problems/shift-2d-grid/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1260_Shift_2D_Grid.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1260_Shift_2D_Grid.java) | Final position of each element can be computed according to k, m and n, e.g., k == mn, then don't move, O(mn) and O(mn) |
 | 1337 | [The K Weakest Rows in a Matrix](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1337_The_K_Weakest_Rows_in_a_Matrix.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1337_The_K_Weakest_Rows_in_a_Matrix.java) | Check by row, from left to right, until encount first zero, O(mn) and O(1) |
 
diff --git a/java/1089_Duplicate_Zeros.java b/java/1089_Duplicate_Zeros.java
@@ -0,0 +1,28 @@
+class Solution {
+    public void duplicateZeros(int[] arr) {
+        int movePos = 0;
+        int lastPos = arr.length - 1;
+        // Only check [0, lastPos - movePos]
+        for (int i = 0; i <= lastPos - movePos; i++) {
+            if (arr[i] == 0) {
+                // Special case
+                if (i == lastPos - movePos) {
+                    arr[lastPos] = 0;
+                    lastPos--;
+                    break;
+                }
+                movePos++;
+            }
+        }
+        lastPos = lastPos - movePos;
+        for (int i = lastPos; i >= 0; i--) {
+            if (arr[i] == 0) {
+                arr[i + movePos] = 0;
+                movePos--;
+                arr[i + movePos] = 0;
+            } else {
+                arr[i + movePos] = arr[i];
+            }
+        }
+    }
+}
diff --git a/python/1089_Duplicate_Zeros.py b/python/1089_Duplicate_Zeros.py
@@ -0,0 +1,26 @@
+class Solution:
+    def duplicateZeros(self, arr: List[int]) -> None:
+        """
+        Do not return anything, modify arr in-place instead.
+        """
+        move_pos = 0
+        last_pos = len(arr) - 1
+        for i in range(last_pos + 1):
+            # Only check [0, lastPos - movePos]
+            if i > last_pos - move_pos:
+                break
+            if arr[i] == 0:
+                # Special case
+                if i == last_pos - move_pos:
+                    arr[last_pos] = 0
+                    last_pos -= 1
+                    break
+                move_pos += 1
+        last_pos -= move_pos
+        for i in range(last, -1, -1):
+            if arr[i] == 0:
+                arr[i + move_pos] = 0
+                move_pos -= 1
+                arr[i + move_pos] = 0
+            else:
+                arr[i + move_pos] = arr[i]
