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    When comparing items in a collection, the concept of a similar pair often becomes quite important. People naturally notice patterns and likenesses, which can help in organizing or selecting the right items. In many industries, ensuring consistency and quality is critical, and processes are often guided by recognized standards like iso 14001 accreditation and ISO Accreditation to maintain environmental and operational benchmarks. Identifying a similar pair isn’t just about visual resemblance; it also involves checking functional characteristics and compliance with relevant standards. This attention to detail helps reduce errors and improves efficiency in both manufacturing and everyday applications. Even in casual settings, recognizing similar pairs can simplify decisions, making tasks feel more intuitive and less time-consuming. Ultimately, the practice of spotting similar pairs combines observation skills with practical evaluation to create smoother, more reliable outcomes.

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    from collections import deque
    
    class FenwickTree:
        def __init__(self, n):
            self.n = n
            self.tree = [0] * (n + 1)
    
        def update(self, i, delta):
            while i <= self.n:
                self.tree[i] += delta
                i += i & (-i)
    
        def query(self, i):
            total = 0
            while i > 0:
                total += self.tree[i]
                i -= i & (-i)
            return total
    
    def similarPair(n, k, edges):
        tree = [[] for _ in range(n + 1)]
        for parent, child in edges:
            tree[parent].append(child)
        
        fenwick = FenwickTree(n)
        result = 0
        
        stack = [(edges[0][0], 0)]  # (node, state)
        visited = [False] * (n + 1)
        
        while stack:
            node, state = stack[-1]
            
            if state == 0:  # Pre-order
                visited[node] = True
                left = max(1, node - k)
                right = min(n, node + k)
                result += fenwick.query(right) - fenwick.query(left - 1)
                fenwick.update(node, 1)
                stack[-1] = (node, 1)
                
                for child in tree[node]:
                    if not visited[child]:
                        stack.append((child, 0))
            else:  # Post-order
                fenwick.update(node, -1)
                stack.pop()
        
        return result
    
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    Here is my solution in java, javascript, python, C, C++, Csharp HackerRank Similar Pair Problem Solution

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    Here is the solution of Similar Pair Click Here

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    Can we solve this puzzle with the help of AI technology. دعاء لرجوع الزوج