From 2b71f994be21e9dae028a32eb60451fb85f786ee Mon Sep 17 00:00:00 2001
From: Ajay-Dhangar <ajaydhangar49@gmail.com>
Date: Fri, 19 Jul 2024 09:04:27 +0530
Subject: [PATCH] update file

---
 .../0400-0499/0493-reverse-pairs.md           | 496 ++++++++++++------
 1 file changed, 324 insertions(+), 172 deletions(-)

diff --git a/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md b/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md
index ab99d358c..861effb08 100644
--- a/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md
+++ b/dsa-solutions/lc-solutions/0400-0499/0493-reverse-pairs.md
@@ -1,172 +1,324 @@
----
-id: reverse-pairs
-title: Reverse Pairs
-sidebar_label: 0493 - Reverse Pairs
-tags:
-  - Leetcode
-  - Merge Sort
----
-
-## Problem Statement
-
-Given an integer array `nums`, return the number of reverse pairs in the array.
-
-A reverse pair is defined as a pair `(i, j)` where:
-
-- $0 <= i < j < nums.length and$
-- $nums[i] > 2 * nums[j].$
-
-## Examples
-
-### Example 1
-
-- Input: `nums = [1,3,2,3,1]`
-- Output: `2`
-- Explanation: The reverse pairs are:
-  - `(1, 4)` -> `nums[1] = 3`, `nums[4] = 1`, `3 > 2 * 1`
-  - `(3, 4)` -> `nums[3] = 3`, `nums[4] = 1`, `3 > 2 * 1`
-
-### Example 2
-
-- Input: `nums = [2,4,3,5,1]`
-- Output: `3`
-- Explanation: The reverse pairs are:
-  - `(1, 4)` -> `nums[1] = 4`, `nums[4] = 1`, `4 > 2 * 1`
-  - `(2, 4)` -> `nums[2] = 3`, `nums[4] = 1`, `3 > 2 * 1`
-  - `(3, 4)` -> `nums[3] = 5`, `nums[4] = 1`, `5 > 2 * 1`
-
-## Constraints
-
-- $1 <= nums.length <= 5 * 104$
-- $-231 <= nums[i] <= 231 - 1$
-
-## Algorithm
-
-To solve this problem efficiently, we can use a modified merge sort algorithm, which not only sorts the array but also counts the number of reverse pairs during the merging process. The idea is to leverage the divide-and-conquer approach to count pairs in `O(n log n)` time complexity.
-
-### Steps:
-
-1. **Divide**: Split the array into two halves.
-2. **Count**: Count the reverse pairs in each half recursively.
-3. **Merge and Count**: While merging the two halves, count the reverse pairs where one element is from the left half and the other is from the right half.
-
-## Code
-
-### Python
-
-```python
-class Solution:
-    def reversePairs(self, nums: List[int]) -> int:
-        def merge_sort(nums, l, r):
-            if l >= r:
-                return 0
-
-            mid = (l + r) // 2
-            count = merge_sort(nums, l, mid) + merge_sort(nums, mid + 1, r)
-
-            j = mid + 1
-            for i in range(l, mid + 1):
-                while j <= r and nums[i] > 2 * nums[j]:
-                    j += 1
-                count += j - (mid + 1)
-
-            nums[l:r + 1] = sorted(nums[l:r + 1])
-            return count
-
-        return merge_sort(nums, 0, len(nums) - 1)
-```
-
-### C++
-
-```cpp
-#include <vector>
-using namespace std;
-
-class Solution {
-public:
-    int reversePairs(vector<int>& nums) {
-        return reversePairsSub(nums, 0, nums.size() - 1);
-    }
-
-private:
-    int reversePairsSub(vector<int>& nums, int l, int r) {
-        if (l >= r) return 0;
-
-        int m = l + ((r - l) >> 1);
-        int res = reversePairsSub(nums, l, m) + reversePairsSub(nums, m + 1, r);
-
-        int i = l, j = m + 1, k = 0, p = m + 1;
-        vector<int> merge(r - l + 1);
-
-        while (i <= m) {
-            while (p <= r && nums[i] > 2L * nums[p]) p++;
-            res += p - (m + 1);
-
-            while (j <= r && nums[i] >= nums[j]) merge[k++] = nums[j++];
-            merge[k++] = nums[i++];
-        }
-
-        while (j <= r) merge[k++] = nums[j++];
-
-        copy(merge.begin(), merge.end(), nums.begin() + l);
-
-        return res;
-    }
-};
-```
-
-### Java
-
-```java
-import java.util.Arrays;
-
-class Solution {
-    public int reversePairs(int[] nums) {
-        return mergeSort(nums, 0, nums.length - 1);
-    }
-
-    private int mergeSort(int[] nums, int left, int right) {
-        if (left >= right) return 0;
-
-        int mid = left + (right - left) / 2;
-        int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
-
-        int j = mid + 1;
-        for (int i = left; i <= mid; i++) {
-            while (j <= right && nums[i] > 2L * nums[j]) j++;
-            count += j - (mid + 1);
-        }
-
-        Arrays.sort(nums, left, right + 1);
-        return count;
-    }
-}
-```
-
-### JavaScript
-
-```javascript
-var reversePairs = function (nums) {
-  function mergeSort(nums, left, right) {
-    if (left >= right) return 0;
-
-    let mid = left + Math.floor((right - left) / 2);
-    let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
-
-    let j = mid + 1;
-    for (let i = left; i <= mid; i++) {
-      while (j <= right && nums[i] > 2 * nums[j]) j++;
-      count += j - (mid + 1);
-    }
-
-    nums.splice(
-      left,
-      right - left + 1,
-      ...nums.slice(left, right + 1).sort((a, b) => a - b)
-    );
-    return count;
-  }
-
-  return mergeSort(nums, 0, nums.length - 1);
-};
-```
+---
+id: reverse-pairs
+title: Reverse Pairs
+sidebar_label: 493-Reverse-Pairs
+tags:
+  - Array
+  - Binary Search
+  - Divide and conquer
+  - Binary Indexed Tree
+  - Segment Tree
+  - Merge Sort
+  - Ordered Set
+description: The problem is to reverse the pairs.
+sidebar_position: 2667
+---
+
+## Problem Statement 
+
+### Problem Description
+
+Given an integer array `nums`, return the number of reverse pairs in the array.
+
+A reverse pair is a pair `(i, j)` where:
+
+`0 <= i < j < nums.length` and
+`nums[i] > 2 * nums[j]`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: nums = [1,3,2,3,1]
+Output: 2
+Explanation: The reverse pairs are:
+(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1
+```
+
+**Example 2:**
+
+```
+Input: nums = [2,4,3,5,1]
+Output: 3
+Explanation: The reverse pairs are:
+(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1
+(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1
+```
+
+### Constraints
+
+- `1 <= nums.length <= 5 * 10^4`
+
+## Solution of Given Problem
+
+### Intuition and Approach
+
+The intuition behind this solution is to use a modified Merge Sort algorithm to count the number of reverse pairs in the array.
+
+
+### Approaches
+
+- Divide the array into smaller subarrays until each subarray has only one element.
+- Merge the subarrays back together, counting the number of reverse pairs between each pair of subarrays.
+- The merge step is done in a way that ensures the count of reverse pairs is accurate.
+
+
+#### Codes in Different Languages
+
+<Tabs>
+  <TabItem value="JavaScript" label="JavaScript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```javascript
+    function mergeSortedArrays(nums, left, mid, right, temp) {
+  let i = left, j = mid + 1, k = 0;
+  while (i <= mid && j <= right) {
+    if (nums[i] <= nums[j]) temp[k++] = nums[i++];
+    else temp[k++] = nums[j++];
+  }
+  while (i <= mid) temp[k++] = nums[i++];
+  while (j <= right) temp[k++] = nums[j++];
+}
+
+function merge(nums, left, mid, right) {
+  let count = 0;
+  let j = mid + 1;
+  for (let i = left; i <= mid; i++) {
+    while (j <= right && nums[i] > 2 * nums[j]) j++;
+    count += j - mid - 1;
+  }
+  let temp = new Array(right - left + 1);
+  mergeSortedArrays(nums, left, mid, right, temp);
+  for (let i = left; i <= right; i++) nums[i] = temp[i - left];
+  return count;
+}
+function mergeSort(nums, left, right) {
+  if (left >= right) return 0;
+  let mid = left + (right - left) / 2;
+  let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+  count += merge(nums, left, mid, right);
+  return count;
+}
+
+class Solution {
+  reversePairs(nums) {
+    return mergeSort(nums, 0, nums.length - 1);
+  }
+}
+
+
+    ```
+
+  </TabItem>
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@Ishitamukherjee2004"/> 
+   ```typescript
+    function mergeSortedArrays(nums: number[], left: number, mid: number, right: number, temp: number[]) {
+  let i = left, j = mid + 1, k = 0;
+  while (i <= mid && j <= right) {
+    if (nums[i] <= nums[j]) temp[k++] = nums[i++];
+    else temp[k++] = nums[j++];
+  }
+  while (i <= mid) temp[k++] = nums[i++];
+  while (j <= right) temp[k++] = nums[j++];
+}
+
+function merge(nums: number[], left: number, mid: number, right: number) {
+  let count = 0;
+  let j = mid + 1;
+  for (let i = left; i <= mid; i++) {
+    while (j <= right && nums[i] > 2 * nums[j]) j++;
+    count += j - mid - 1;
+  }
+  let temp = new Array(right - left + 1);
+  mergeSortedArrays(nums, left, mid, right, temp);
+  for (let i = left; i <= right; i++) nums[i] = temp[i - left];
+  return count;
+}
+function mergeSort(nums: number[], left: number, right: number) {
+  if (left >= right) return 0;
+  let mid = left + (right - left) / 2;
+  let count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+  count += merge(nums, left, mid, right);
+  return count;
+}
+
+class Solution {
+  reversePairs(nums: number[]) {
+    return mergeSort(nums, 0, nums.length - 1);
+  }
+}
+
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python"> 
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```python
+    def merge_sorted_arrays(nums, left, mid, right, temp):
+    i, j, k = left, mid + 1, 0
+    while i <= mid and j <= right:
+        if nums[i] <= nums[j]:
+            temp[k] = nums[i]
+            i += 1
+        else:
+            temp[k] = nums[j]
+            j += 1
+        k += 1
+    while i <= mid:
+        temp[k] = nums[i]
+        i += 1
+        k += 1
+    while j <= right:
+        temp[k] = nums[j]
+        j += 1
+        k += 1
+
+def merge(nums, left, mid, right):
+    count = 0
+    j = mid + 1
+    for i in range(left, mid + 1):
+        while j <= right and nums[i] > 2 * nums[j]:
+            j += 1
+        count += j - mid - 1
+    temp = [0] * (right - left + 1)
+    merge_sorted_arrays(nums, left, mid, right, temp)
+    for i in range(left, right + 1):
+        nums[i] = temp[i - left]
+    return count
+
+def merge_sort(nums, left, right):
+    if left >= right:
+        return 0
+    mid = left + (right - left) // 2
+    count = merge_sort(nums, left, mid) + merge_sort(nums, mid + 1, right)
+    count += merge(nums, left, mid, right)
+    return count
+
+class Solution:
+    def reversePairs(self, nums):
+        return merge_sort(nums, 0, len(nums) - 1)
+
+    ```
+
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```java
+    public class Solution {
+    public int reversePairs(int[] nums) {
+        return mergeSort(nums, 0, nums.length - 1);
+    }
+
+    public int mergeSort(int[] nums, int left, int right) {
+        if (left >= right) {
+            return 0;
+        }
+        int mid = left + (right - left) / 2;
+        int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+        count += merge(nums, left, mid, right);
+        return count;
+    }
+     public int merge(int[] nums, int left, int mid, int right) {
+        int count = 0;
+        int j = mid + 1;
+        for (int i = left; i <= mid; i++) {
+            while (j <= right && nums[i] > 2 * nums[j]) {
+                j++;
+            }
+            count += j - mid - 1;
+        }
+        int[] temp = new int[right - left + 1];
+        mergeSortedArrays(nums, left, mid, right, temp);
+        for (int i = left; i <= right; i++) {
+            nums[i] = temp[i - left];
+        }
+        return count;
+    }
+
+    public void
+    mergeSortedArrays(int[] nums, int left, int mid, int right, int[] temp) {
+        int i = left, j = mid + 1, k = 0;
+        while (i <= mid && j <= right) {
+            if (nums[i] <= nums[j]) {
+                temp[k++] = nums[i++];
+            } else {
+                temp[k++] = nums[j++];
+            }
+        }
+        while (i <= mid) {
+            temp[k++] = nums[i++];
+        }
+        while (j <= right) {
+            temp[k++] = nums[j++];
+        }
+    }
+}
+
+    ```
+  </TabItem>
+  <TabItem value="cpp" label="C++">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```cpp
+    void mergeSortedArrays(vector<int>& nums, int left, int mid, int right, vector<int>& temp) {
+    int i = left, j = mid + 1, k = 0;
+    while (i <= mid && j <= right) {
+        if (nums[i] <= nums[j]) temp[k++] = nums[i++];
+        else temp[k++] = nums[j++];
+    }
+    while (i <= mid) temp[k++] = nums[i++];
+    while (j <= right) temp[k++] = nums[j++];
+}
+int merge(vector<int>& nums, int left, int mid, int right) {
+    int count = 0;
+    int j = mid + 1;
+    for (int i = left; i <= mid; i++) {
+        while (j <= right && nums[i] > 2 * nums[j]) j++;
+        count += j - mid - 1;
+    }
+    vector<int> temp(right - left + 1);
+    mergeSortedArrays(nums, left, mid, right, temp);
+    for (int i = left; i <= right; i++) nums[i] = temp[i - left];
+    return count;
+}
+int mergeSort(vector<int>& nums, int left, int right) {
+    if (left >= right) return 0;
+    int mid = left + (right - left) / 2;
+    int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
+    count += merge(nums, left, mid, right);
+    return count;
+}
+
+
+class Solution {
+public:
+    int reversePairs(vector<int>& nums) {
+       return mergeSort(nums, 0, nums.size() - 1);
+
+
+    }
+};
+    ```
+  </TabItem>  
+</Tabs>
+
+### Complexity Analysis
+
+- **Time Complexity**: $$O(n*log(n))$$, where n is the length of the input array. This is because the solution uses a modified Merge Sort algorithm, which has a time complexity of O(n log n).
+
+
+- **Space Complexity**: $$O(n)$$, where n is the length of the input array. This is because the solution uses a temporary array to store the merged sorted arrays.
+
+
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['Ishitamukherjee2004', 'ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>