allcoding1
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TATA steel

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Tata Steel

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def construct_graph(n):
    graph = [[] for _ in range(n+1)]
    for i in range(2, n+1):
        for j in range(2*i, n+1, i):
            graph[i].append(j)
            graph[j].append(i)
    return graph

def dfs(node, graph, visited):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor, graph, visited)

def count_groups(n):
    graph = construct_graph(n)
    visited = set()
    groups = 0
    for i in range(2, n+1):
        if i not in visited:
            groups += 1
            dfs(i, graph, visited)
    return groups

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Tata Steel

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Python
Simply Equal Qsn
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def construct_graph(n):
    graph = [[] for _ in range(n+1)]
    for i in range(2, n+1):
        for j in range(2*i, n+1, i):
            graph[i].append(j)
            graph[j].append(i)
    return graph

def dfs(node, graph, visited):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor, graph, visited)

def count_groups(n):
    graph = construct_graph(n)
    visited = set()
    groups = 0
    for i in range(2, n+1):
        if i not in visited:
            groups += 1
            dfs(i, graph, visited)
    return groups

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int countNecklaces(int maxPearls, int startMagnificence, int endMagnificence) {
    int count = 0;
    for (int pearls = 1; pearls <= maxPearls; pearls++) {
        for (int magnificence = startMagnificence; magnificence <= endMagnificence; magnificence++) {
            if (pearls == 1) {
                count++;
            } else {
                int prevCount = count;
                for (int prevMagnificence = startMagnificence; prevMagnificence <= magnificence; prevMagnificence++) {
                    count += prevCount;
                }
            }
        }
    }
    return count;
}

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Charles and Neckless

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