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chore: shell sort formatting
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@wingkwong
wingkwong committed Oct 10, 2023
1 parent 5fa0ae5 commit 6dfd7f6
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134 changes: 48 additions & 86 deletions solutions/0000-0099/0088-merge-sorted-array-easy.md
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Original file line number Diff line number Diff line change
Expand Up @@ -385,19 +385,15 @@ public:
## Approach 4: Shell sort
1. Calculate the gap value for merging the two arrays. The gap is determined as $\lceil \frac{{\text{size of arr1[]} + \text{size of arr2[]}}}{{2}} \rceil$.
1. Calculate the gap value for merging the two arrays. The gap is determined as $\lceil \frac{{\text{size of arr1} + \text{size of arr2}}}{{2}} \rceil$.
2. Initialize two pointers: the left pointer at index $0$ and the right pointer at index (left + gap).
3. Perform the following steps for each gap until the gap becomes $0$:
- Inside a loop that continues until the right pointer reaches the end (i.e., index $n + m$), handle three different cases:
- If the left pointer is inside $\text{arr1}[]$ and the right pointer is in $\text{arr2}[]$, compare $\text{arr1}[{\text{left}}]$ and $\text{arr2}[{\text{right}} - n]$. If $\text{arr1}[{\text{left}}] > \text{arr2}[{\text{right}} - n]$, swap them.
- If both pointers are in $\text{arr2}[]$, compare $\text{arr1}[{\text{left}} - n]$ and $\text{arr2}[{\text{right}} - n]$. If $\text{arr1}[{\text{left}} - n] > \text{arr2}[{\text{right}} - n]$, swap them.
- If both pointers are in $\text{arr1}[]$, compare $\text{arr1}[{\text{left}}]$ and $\text{arr2}[{\text{right}}]$. If $\text{arr1}[{\text{left}}] > \text{arr2}[{\text{right}}]$, swap them.
- If the left pointer is inside $\text{arr1}$ and the right pointer is in $\text{arr2}$, compare $\text{arr1}[{\text{left}}]$ and $\text{arr2}[{\text{right}} - n]$. If $\text{arr1}[{\text{left}}] > \text{arr2}[{\text{right}} - n]$, swap them.
- If both pointers are in $\text{arr2}$, compare $\text{arr1}[{\text{left}} - n]$ and $\text{arr2}[{\text{right}} - n]$. If $\text{arr1}[{\text{left}} - n] > \text{arr2}[{\text{right}} - n]$, swap them.
- If both pointers are in $\text{arr1}$, compare $\text{arr1}[{\text{left}}]$ and $\text{arr2}[{\text{right}}]$. If $\text{arr1}[{\text{left}}] > \text{arr2}[{\text{right}}]$, swap them.
4. After the right pointer reaches the end, decrease the value of the gap by setting it to $\lceil \frac{{\text{current gap}}}{{2}} \rceil$.
Expand All @@ -412,51 +408,39 @@ public:
```cpp
class Solution {
public:
void swapIfGreater(vector<int>& arr1, vector<int>& arr2, int ind1, int ind2) {
if (arr1[ind1] > arr2[ind2]) {
swap(arr1[ind1], arr2[ind2]);
}
public:
void swapIfGreater(vector<int>& arr1, vector<int>& arr2, int ind1, int ind2) {
if (arr1[ind1] > arr2[ind2]) {
swap(arr1[ind1], arr2[ind2]);
}
void merge(vector<int>& arr1, int n, vector<int>& arr2, int m) {
int len = n + m;
// Initial gap:
int gap = (len / 2) + (len % 2);
while (gap > 0) {
// Place 2 pointers:
int left = 0;
int right = left + gap;
while (right < len) {
// case 1: left in arr1[] and right in arr2[]:
if (left < n && right >= n) {
swapIfGreater(arr1, arr2, left, right - n);
}
// case 2: both pointers in arr2[]:
else if (left >= n) {
swapIfGreater(arr2, arr2, left - n, right - n);
}
// case 3: both pointers in arr1[]:
else {
swapIfGreater(arr1, arr1, left, right);
}
left++, right++;
}
// break if iteration gap=1 is completed:
if (gap == 1) break;
// Otherwise, calculate the new gap:
gap = (gap / 2) + (gap % 2);
}
int j = 0;
for (int i = n; i < n + m; i++) {
arr1[i] = arr2[j];
j++;
}
}
void merge(vector<int>& arr1, int n, vector<int>& arr2, int m) {
int len = n + m;
// initial gap:
int gap = (len / 2) + (len % 2);
while (gap > 0) {
// place 2 pointers:
int left = 0, right = left + gap;
while (right < len) {
// case 1: left in arr1 and right in arr2:
if (left < n && right >= n) swapIfGreater(arr1, arr2, left, right - n);
// case 2: both pointers in arr2:
else if (left >= n) swapIfGreater(arr2, arr2, left - n, right - n);
// case 3: both pointers in arr1:
else swapIfGreater(arr1, arr1, left, right);
left++, right++;
}
// break if iteration gap=1 is completed:
if (gap == 1) break;
// otherwise, calculate new gap:
gap = (gap / 2) + (gap % 2);
}
int j = 0;
for (int i = n; i < n + m; i++) {
arr1[i] = arr2[j];
j++;
}
}
};
```
</TabItem>
Expand All @@ -472,38 +456,25 @@ class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
# Calculate the total length
length = m + n

# Initial gap
gap = (length // 2) + (length % 2)

while gap > 0:
# Place two pointers
left = 0
right = left + gap

while right < length:
# case 1: left in nums1[] and right in nums2[]
if left < m and right >= m:
self.swapIfGreater(nums1, nums2, left, right - m)

# case 2: both pointers in nums2[]
elif left >= m:
self.swapIfGreater(nums2, nums2, left - m, right - m)

# case 3: both pointers in nums1[]
else:
self.swapIfGreater(nums1, nums1, left, right)

# case 1: left in nums1 and right in nums2
if left < m and right >= m: self.swapIfGreater(nums1, nums2, left, right - m)
# case 2: both pointers in nums2
elif left >= m: self.swapIfGreater(nums2, nums2, left - m, right - m)
# case 3: both pointers in nums1
else: self.swapIfGreater(nums1, nums1, left, right)
left += 1
right += 1

# Break if iteration with gap=1 is completed
if gap == 1:
break

if gap == 1: break
# Calculate the new gap
gap = (gap // 2) + (gap % 2)

j = 0
for i in range(m, length):
nums1[i] = nums2[j]
Expand All @@ -525,28 +496,19 @@ class Solution {

public void merge(int[] nums1, int m, int[] nums2, int n) {
int len = m + n;

// Initial gap
int gap = (len / 2) + (len % 2);

while (gap > 0) {
// Place two pointers
int left = 0;
int right = left + gap;

while (right < len) {
// case 1: left in nums1[] and right in nums2[]
if (left < m && right >= m) {
swapIfGreater(nums1, nums2, left, right - m);
}
// case 2: both pointers in nums2[]
else if (left >= m) {
swapIfGreater(nums2, nums2, left - m, right - m);
}
// case 3: both pointers in nums1[]
else {
swapIfGreater(nums1, nums1, left, right);
}
// case 1: left in nums1 and right in nums2
if (left < m && right >= m) swapIfGreater(nums1, nums2, left, right - m);
// case 2: both pointers in nums2
else if (left >= m) swapIfGreater(nums2, nums2, left - m, right - m);
// case 3: both pointers in nums1
else swapIfGreater(nums1, nums1, left, right);
left++;
right++;
}
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15 changes: 3 additions & 12 deletions tutorials/basic-topics/sorting/shell-sort.md
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Original file line number Diff line number Diff line change
Expand Up @@ -350,28 +350,19 @@ class Solution {

public void merge(int[] nums1, int m, int[] nums2, int n) {
int len = m + n;

// Initial gap
int gap = (len / 2) + (len % 2);

while (gap > 0) {
// Place two pointers
int left = 0;
int right = left + gap;

while (right < len) {
// case 1: left in nums1 and right in nums2
if (left < m && right >= m) {
swapIfGreater(nums1, nums2, left, right - m);
}
if (left < m && right >= m) swapIfGreater(nums1, nums2, left, right - m);
// case 2: both pointers in nums2
else if (left >= m) {
swapIfGreater(nums2, nums2, left - m, right - m);
}
else if (left >= m) swapIfGreater(nums2, nums2, left - m, right - m);
// case 3: both pointers in nums1
else {
swapIfGreater(nums1, nums1, left, right);
}
else swapIfGreater(nums1, nums1, left, right);
left++;
right++;
}
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