The Reverse Pair Problem is a common coding challenge where you’re required to count the number of reverse pairs in an array. A reverse pair is a pair of indices (i, j) such that i < j and nums[i] > 2 * nums[j].
Here’s a detailed step-by-step guide on how to solve the Reverse Pair Problem in Java:
Step 1: Understand the Problem
Before you begin coding, it’s important to understand the problem statement and constraints. Make sure you know what a reverse pair is and the conditions that need to be satisfied.
Step 2: Plan the Approach
Think about a strategy to solve the problem. One common approach is to use a modified version of the Merge Sort algorithm. The idea is to count the reverse pairs while merging two sorted subarrays.
Step 3: Implement Merge Sort
Start by implementing the Merge Sort algorithm, which involves dividing the array into smaller subarrays, sorting them, and then merging them back together. Here’s a high-level overview of the merge sort algorithm:
public class ReversePairProblem {
public int reversePairs(int[] nums) {
return mergeSortAndCount(nums, 0, nums.length - 1);
}
private int mergeSortAndCount(int[] nums, int left, int right) {
if (left >= right) {
return 0;
}
int mid = left + (right - left) / 2;
int count = mergeSortAndCount(nums, left, mid) + mergeSortAndCount(nums, mid + 1, right);
int[] merged = new int[right - left + 1];
int i = left, j = mid + 1, k = 0;
while (i <= mid && j <= right) {
if ((long) nums[i] > 2 * (long) nums[j]) {
count += mid - i + 1;
j++;
} else {
i++;
}
}
i = left;
j = mid + 1;
while (i <= mid && j <= right) {
if (nums[i] <= nums[j]) {
merged[k++] = nums[i++];
} else {
merged[k++] = nums[j++];
}
}
while (i <= mid) {
merged[k++] = nums[i++];
}
while (j <= right) {
merged[k++] = nums[j++];
}
System.arraycopy(merged, 0, nums, left, merged.length);
return count;
}
public static void main(String[] args) {
int[] nums = { 3, 1, 2, 4 };
ReversePairProblem solution = new ReversePairProblem();
int count = solution.reversePairs(nums);
System.out.println("Number of reverse pairs: " + count);
}
}Step 4: Test Your Solution
Create various test cases to verify the correctness of your solution. Make sure to test edge cases and random inputs.
Step 5: Explain Your Code
Write a detailed explanation of your code in your blog post. Break down the key components of your solution, such as the merge sort algorithm, the counting of reverse pairs, and the merging process. Use comments in your code to explain critical sections.
Step 6: Optimize if Necessary
Consider optimizations if your initial solution is not efficient enough. For example, you might explore using a Binary Indexed Tree (BIT) to optimize the counting process.
Step 7: Conclude
Summarize the problem, your approach, and the effectiveness of your solution. Mention any trade-offs or alternative solutions that you considered.
By following these steps, you can create a comprehensive and informative blog post explaining the Reverse Pair Problem solution in Java.
Here’s how you can implement the Reverse Pair Problem solution in both Java and Python.
Java Implementation:
public class ReversePairProblem {
public int reversePairs(int[] nums) {
return mergeSortAndCount(nums, 0, nums.length - 1);
}
private int mergeSortAndCount(int[] nums, int left, int right) {
if (left >= right) {
return 0;
}
int mid = left + (right - left) / 2;
int count = mergeSortAndCount(nums, left, mid) + mergeSortAndCount(nums, mid + 1, right);
int[] merged = new int[right - left + 1];
int i = left, j = mid + 1, k = 0;
while (i <= mid && j <= right) {
if ((long) nums[i] > 2 * (long) nums[j]) {
count += mid - i + 1;
j++;
} else {
i++;
}
}
i = left;
j = mid + 1;
while (i <= mid && j <= right) {
if (nums[i] <= nums[j]) {
merged[k++] = nums[i++];
} else {
merged[k++] = nums[j++];
}
}
while (i <= mid) {
merged[k++] = nums[i++];
}
while (j <= right) {
merged[k++] = nums[j++];
}
System.arraycopy(merged, 0, nums, left, merged.length);
return count;
}
public static void main(String[] args) {
int[] nums = { 3, 1, 2, 4 };
ReversePairProblem solution = new ReversePairProblem();
int count = solution.reversePairs(nums);
System.out.println("Number of reverse pairs: " + count);
}
}Python Implementation:
class ReversePairProblem:
def reversePairs(self, nums):
self.temp = [0] * len(nums)
return self.mergeSortAndCount(nums, 0, len(nums) - 1)
def mergeSortAndCount(self, nums, left, right):
if left >= right:
return 0
mid = left + (right - left) // 2
count = self.mergeSortAndCount(nums, left, mid) + self.mergeSortAndCount(nums, mid + 1, right)
i, j = left, mid + 1
while i <= mid and j <= right:
if nums[i] > 2 * nums[j]:
count += mid - i + 1
j += 1
else:
i += 1
i, j, k = left, mid + 1, 0
merged = [0] * (right - left + 1)
while i <= mid and j <= right:
if nums[i] <= nums[j]:
merged[k] = nums[i]
i += 1
else:
merged[k] = nums[j]
j += 1
k += 1
while i <= mid:
merged[k] = nums[i]
i += 1
k += 1
while j <= right:
merged[k] = nums[j]
j += 1
k += 1
nums[left:right+1] = merged
return count
nums = [3, 1, 2, 4]
solution = ReversePairProblem()
count = solution.reversePairs(nums)
print("Number of reverse pairs:", count)Both of these implementations follow the same basic approach: they use a modified merge sort algorithm to count the reverse pairs in the given array. The Java code uses explicit array copying, while the Python code slices the original array to update it during merging.
Thank You!
Read More–
Equilibrium Index of an Array Problem | LeetCode Solution in Java-https://kamleshsingad.com/equilibrium-index-of-an-array-problem-leetcode-solution-in-java/
Reverse An Array | Program to Reverse an Array https://kamleshsingad.com/reverse-an-array-program-to-reverse-an-array/
Count Inversions in an Array | Coding Interview Question | Count Inversions Merge Sort – https://kamleshsingad.com/count-inversions-in-an-array-count-inversions-merge-sort/
