Given a **weighted tree** containing **N** nodes, and two nodes **u** and **v**, the task is to find the count of nodes having **prime weight** on the simple path between **u and v (both inclusive)**.

**Examples:**

Input:u = 3, v = 5

Output:2Explanation:

Prime weight on path 3 to 5 is [11, 5]. Hence, the answer is 2.

**Approach:** To solve the problem mentioned above, the idea is to use the basic concept when we find the LCA of two nodes.

- Precompute all the prime numbers till MAX using the Sieve method to check if a number is prime or not in O(1)
- Given two nodes u and v, we will make both nodes at the same level, by moving the greater level node move upwards. As we move up we will also check if the weight is prime or not.
- If
**v == u**then we will simply check the weight of the current node and return the count. - If
**v is not equal to u**then we will move**both u and v upward by 1**till they are not the same. - Now we will finally check the weight of the first ancestor of u or v and return the count.

Below is the implementation of the above approach:

## C++

`// C++ program Count prime weight` `// nodes between two nodes in the given tree` `#include <bits/stdc++.h>` `using` `namespace` `std;` `#define MAX 1000` `int` `weight[MAX];` `int` `level[MAX];` `int` `par[MAX];` `bool` `prime[MAX + 1];` `vector<` `int` `> graph[MAX];` `// Function to perform` `// Sieve Of Eratosthenes for prime number` `void` `SieveOfEratosthenes()` `{` ` ` `// Initialize all entries of prime it as true` ` ` `// A value in prime[i] will finally be false` ` ` `// if i is Not a prime, else true.` ` ` `memset` `(prime, ` `true` `, ` `sizeof` `(prime));` ` ` `for` `(` `int` `p = 2; p * p <= MAX; p++) {` ` ` `// Check if prime[p] is not changed,` ` ` `// then it is a prime` ` ` `if` `(prime[p] == ` `true` `) {` ` ` `// Update all multiples` ` ` `// of p greater than or` ` ` `// equal to the square of it` ` ` `// numbers which are multiple` ` ` `// of p and are less than p^2` ` ` `// are already been marked.` ` ` `for` `(` `int` `i = p * p; i <= MAX; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` `}` `// Function to perform dfs` `void` `dfs(` `int` `node, ` `int` `parent, ` `int` `h)` `{` ` ` `// Stores parent of each node` ` ` `par[node] = parent;` ` ` `// Stores level of each node from root` ` ` `level[node] = h;` ` ` `for` `(` `int` `child : graph[node]) {` ` ` `if` `(child == parent)` ` ` `continue` `;` ` ` `dfs(child, node, h + 1);` ` ` `}` `}` `// Function to perform prime` `// number between the path` `int` `findPrimeOnPath(` `int` `u, ` `int` `v)` `{` ` ` `int` `count = 0;` ` ` `// The node which is present farthest` ` ` `// from the root node is taken as v` ` ` `// If u is farther from root node` ` ` `// then swap the two` ` ` `if` `(level[u] > level[v])` ` ` `swap(u, v);` ` ` `int` `d = level[v] - level[u];` ` ` `// Find the ancestor of v` ` ` `// which is at same level as u` ` ` `while` `(d--) {` ` ` `// If Weight is prime` ` ` `// increment count` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `v = par[v];` ` ` `}` ` ` `// If u is the ancestor of v` ` ` `// then u is the LCA of u and v` ` ` `// Now check if weigh[v]` ` ` `// is prime or not` ` ` `if` `(v == u) {` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `return` `count;` ` ` `}` ` ` `// When v and u are on the same level but` ` ` `// are in different subtree. Now move both` ` ` `// u and v up by 1 till they are not same` ` ` `while` `(v != u) {` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `if` `(prime[weight[u]])` ` ` `count++;` ` ` `u = par[u];` ` ` `v = par[v];` ` ` `}` ` ` `// If weight of first ancestor` ` ` `// is prime` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `return` `count;` `}` `// Driver code` `int` `main()` `{` ` ` `// Precompute all the prime` ` ` `// numbers till MAX` ` ` `SieveOfEratosthenes();` ` ` `// Weights of the node` ` ` `weight[1] = 5;` ` ` `weight[2] = 10;` ` ` `weight[3] = 11;` ` ` `weight[4] = 8;` ` ` `weight[5] = 6;` ` ` `// Edges of the tree` ` ` `graph[1].push_back(2);` ` ` `graph[2].push_back(3);` ` ` `graph[2].push_back(4);` ` ` `graph[1].push_back(5);` ` ` `dfs(1, -1, 0);` ` ` `int` `u = 3, v = 5;` ` ` `cout << findPrimeOnPath(u, v)` ` ` `<< endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to count` `// prime weight nodes` `// between two nodes` `// in the given tree` `import` `java.util.*;` `class` `GFG{` ` ` `static` `final` `int` `MAX = ` `1000` `;` `static` `int` `[]weight =` ` ` `new` `int` `[MAX];` `static` `int` `[]level =` ` ` `new` `int` `[MAX];` `static` `int` `[]par =` ` ` `new` `int` `[MAX];` `static` `boolean` `[]prime =` ` ` `new` `boolean` `[MAX + ` `1` `];` `static` `Vector<Integer>[] graph =` ` ` `new` `Vector[MAX]; ` `// Function to perform` `// Sieve Of Eratosthenes` `// for prime number` `static` `void` `SieveOfEratosthenes()` `{` ` ` `// Initialize all entries of` ` ` `// prime it as true a value in` ` ` `// prime[i] will finally be false` ` ` `// if i is Not a prime, else true.` ` ` `for` `(` `int` `i = ` `0` `;` ` ` `i < prime.length; i++)` ` ` `prime[i] = ` `true` `;` ` ` `for` `(` `int` `p = ` `2` `;` ` ` `p * p <= MAX; p++)` ` ` `{` ` ` `// Check if prime[p]` ` ` `// is not changed,` ` ` `// then it is a prime` ` ` `if` `(prime[p] == ` `true` `)` ` ` `{` ` ` `// Update all multiples` ` ` `// of p greater than or` ` ` `// equal to the square of it` ` ` `// numbers which are multiple` ` ` `// of p and are less than p^2` ` ` `// are already been marked.` ` ` `for` `(` `int` `i = p * p;` ` ` `i <= MAX; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` `}` `// Function to perform dfs` `static` `void` `dfs(` `int` `node,` ` ` `int` `parent, ` `int` `h)` `{` ` ` `// Stores parent of each node` ` ` `par[node] = parent;` ` ` `// Stores level of each` ` ` `// node from root` ` ` `level[node] = h;` ` ` `for` `(` `int` `child : graph[node])` ` ` `{` ` ` `if` `(child == parent)` ` ` `continue` `;` ` ` `dfs(child, node, h + ` `1` `);` ` ` `}` `}` `// Function to perform prime` `// number between the path` `static` `int` `findPrimeOnPath(` `int` `u,` ` ` `int` `v)` `{` ` ` `int` `count = ` `0` `;` ` ` `// The node which is present` ` ` `// farthest from the root` ` ` `// node is taken as v` ` ` `// If u is farther from` ` ` `// root node then swap the two` ` ` `if` `(level[u] > level[v])` ` ` `{` ` ` `int` `temp = v;` ` ` `v = u;` ` ` `u = temp;` ` ` `}` ` ` `int` `d = level[v] - level[u];` ` ` `// Find the ancestor of v` ` ` `// which is at same level as u` ` ` `while` `(d-- > ` `0` `)` ` ` `{` ` ` `// If Weight is prime` ` ` `// increment count` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `v = par[v];` ` ` `}` ` ` `// If u is the ancestor of v` ` ` `// then u is the LCA of u and v` ` ` `// Now check if weigh[v]` ` ` `// is prime or not` ` ` `if` `(v == u)` ` ` `{` ` ` `if` `(prime[weight[v]])` ` ` `count++; ` ` ` `return` `count;` ` ` `}` ` ` `// When v and u are on the` ` ` `// same level but are in` ` ` `// different subtree. Now` ` ` `// move both u and v up by` ` ` `// 1 till they are not same` ` ` `while` `(v != u)` ` ` `{` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `if` `(prime[weight[u]])` ` ` `count++;` ` ` `u = par[u];` ` ` `v = par[v];` ` ` `}` ` ` ` ` `// If weight of first` ` ` `// ancestor is prime` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `for` `(` `int` `i = ` `0` `; i < graph.length; i++)` ` ` `graph[i] = ` `new` `Vector<Integer>();` ` ` ` ` `// Precompute all the prime` ` ` `// numbers till MAX` ` ` `SieveOfEratosthenes();` ` ` `// Weights of the node` ` ` `weight[` `1` `] = ` `5` `;` ` ` `weight[` `2` `] = ` `10` `;` ` ` `weight[` `3` `] = ` `11` `;` ` ` `weight[` `4` `] = ` `8` `;` ` ` `weight[` `5` `] = ` `6` `;` ` ` `// Edges of the tree` ` ` `graph[` `1` `].add(` `2` `);` ` ` `graph[` `2` `].add(` `3` `);` ` ` `graph[` `2` `].add(` `4` `);` ` ` `graph[` `1` `].add(` `5` `);` ` ` `dfs(` `1` `, -` `1` `, ` `0` `);` ` ` `int` `u = ` `3` `, v = ` `5` `;` ` ` `System.out.print(findPrimeOnPath(u, v));` `}` `}` `// This code is contributed by shikhasingrajput` |

## Python3

`# Python3 program count prime weight` `# nodes between two nodes in the given tree` `MAX` `=` `1000` `weight ` `=` `[` `0` `for` `i ` `in` `range` `(` `MAX` `)]` `level ` `=` `[` `0` `for` `i ` `in` `range` `(` `MAX` `)]` `par ` `=` `[` `0` `for` `i ` `in` `range` `(` `MAX` `)]` `prime ` `=` `[` `True` `for` `i ` `in` `range` `(` `MAX` `+` `1` `)]` `graph ` `=` `[[] ` `for` `i ` `in` `range` `(` `MAX` `)]` `# Function to perform` `# Sieve Of Eratosthenes` `# for prime number` `def` `SieveOfEratosthenes():` ` ` ` ` `# Initialize all entries of prime it` ` ` `# as true. A value in prime[i] will` ` ` `# finally be false if i is Not a prime,` ` ` `# else true. memset(prime, true,` ` ` `# sizeof(prime))` ` ` `for` `p ` `in` `range` `(` `2` `, ` `MAX` `+` `1` `):` ` ` `if` `p ` `*` `p > ` `MAX` `+` `1` `:` ` ` `break` ` ` `# Check if prime[p] is not changed,` ` ` `# then it is a prime` ` ` `if` `(prime[p] ` `=` `=` `True` `):` ` ` ` ` `# Update all multiples` ` ` `# of p greater than or` ` ` `# equal to the square of it` ` ` `# numbers which are multiple` ` ` `# of p and are less than p^2` ` ` `# are already been marked.` ` ` `for` `i ` `in` `range` `(p ` `*` `p, ` `MAX` `+` `1` `, p):` ` ` `prime[i] ` `=` `False` `# Function to perform dfs` `def` `dfs(node, parent, h):` ` ` ` ` `# Stores parent of each node` ` ` `par[node] ` `=` `parent` ` ` `# Stores level of each node from root` ` ` `level[node] ` `=` `h` ` ` `for` `child ` `in` `graph[node]:` ` ` `if` `(child ` `=` `=` `parent):` ` ` `continue` ` ` ` ` `dfs(child, node, h ` `+` `1` `)` `# Function to perform prime` `# number between the path` `def` `findPrimeOnPath(u, v):` ` ` ` ` `count ` `=` `0` ` ` `# The node which is present farthest` ` ` `# from the root node is taken as v` ` ` `# If u is farther from root node` ` ` `# then swap the two` ` ` `if` `(level[u] > level[v]):` ` ` `u, v ` `=` `v, u` ` ` `d ` `=` `level[v] ` `-` `level[u]` ` ` `# Find the ancestor of v` ` ` `# which is at same level as u` ` ` `while` `(d):` ` ` ` ` `# If Weight is prime` ` ` `# increment count` ` ` `if` `(prime[weight[v]]):` ` ` `count ` `+` `=` `1` ` ` `v ` `=` `par[v]` ` ` `d ` `-` `=` `1` ` ` `# If u is the ancestor of v` ` ` `# then u is the LCA of u and v` ` ` `# Now check if weigh[v]` ` ` `# is prime or not` ` ` `if` `(v ` `=` `=` `u):` ` ` `if` `(prime[weight[v]]):` ` ` `count ` `+` `=` `1` ` ` ` ` `return` `count` ` ` `# When v and u are on the same level but` ` ` `# are in different subtree. Now move both` ` ` `# u and v up by 1 till they are not same` ` ` `while` `(v !` `=` `u):` ` ` `if` `(prime[weight[v]]):` ` ` `count ` `+` `=` `1` ` ` `if` `(prime[weight[u]]):` ` ` `count ` `+` `=` `1` ` ` `u ` `=` `par[u]` ` ` `v ` `=` `par[v]` ` ` ` ` `# If weight of first ancestor` ` ` `# is prime` ` ` `if` `(prime[weight[v]]):` ` ` `count ` `+` `=` `1` ` ` `return` `count` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Precompute all the prime` ` ` `# numbers till MAX` ` ` `SieveOfEratosthenes()` ` ` `# Weights of the node` ` ` `weight[` `1` `] ` `=` `5` ` ` `weight[` `2` `] ` `=` `10` ` ` `weight[` `3` `] ` `=` `11` ` ` `weight[` `4` `] ` `=` `8` ` ` `weight[` `5` `] ` `=` `6` ` ` `# Edges of the tree` ` ` `graph[` `1` `].append(` `2` `)` ` ` `graph[` `2` `].append(` `3` `)` ` ` `graph[` `2` `].append(` `4` `)` ` ` `graph[` `1` `].append(` `5` `)` ` ` `dfs(` `1` `, ` `-` `1` `, ` `0` `)` ` ` `u ` `=` `3` ` ` `v ` `=` `5` ` ` ` ` `print` `(findPrimeOnPath(u, v))` ` ` `# This code is contributed by mohit kumar 29` |

## C#

`// C# program to count prime weight` `// nodes between two nodes in the` `// given tree` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `static` `readonly` `int` `MAX = 1000;` `static` `int` `[]weight = ` `new` `int` `[MAX];` `static` `int` `[]level = ` `new` `int` `[MAX];` `static` `int` `[]par = ` `new` `int` `[MAX];` `static` `bool` `[]prime = ` `new` `bool` `[MAX + 1];` `static` `List<` `int` `>[] graph = ` `new` `List<` `int` `>[MAX]; ` `// Function to perform` `// Sieve Of Eratosthenes` `// for prime number` `static` `void` `SieveOfEratosthenes()` `{` ` ` ` ` `// Initialize all entries of` ` ` `// prime it as true a value in` ` ` `// prime[i] will finally be false` ` ` `// if i is Not a prime, else true.` ` ` `for` `(` `int` `i = 0;` ` ` `i < prime.Length; i++)` ` ` `prime[i] = ` `true` `;` ` ` `for` `(` `int` `p = 2;` ` ` `p * p <= MAX; p++)` ` ` `{` ` ` ` ` `// Check if prime[p]` ` ` `// is not changed,` ` ` `// then it is a prime` ` ` `if` `(prime[p] == ` `true` `)` ` ` `{` ` ` ` ` `// Update all multiples` ` ` `// of p greater than or` ` ` `// equal to the square of it` ` ` `// numbers which are multiple` ` ` `// of p and are less than p^2` ` ` `// are already been marked.` ` ` `for` `(` `int` `i = p * p;` ` ` `i <= MAX; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` `}` `// Function to perform dfs` `static` `void` `dfs(` `int` `node, ` `int` `parent,` ` ` `int` `h)` `{` ` ` ` ` `// Stores parent of each node` ` ` `par[node] = parent;` ` ` `// Stores level of each` ` ` `// node from root` ` ` `level[node] = h;` ` ` `foreach` `(` `int` `child ` `in` `graph[node])` ` ` `{` ` ` `if` `(child == parent)` ` ` `continue` `;` ` ` ` ` `dfs(child, node, h + 1);` ` ` `}` `}` `// Function to perform prime` `// number between the path` `static` `int` `findPrimeOnPath(` `int` `u,` ` ` `int` `v)` `{` ` ` `int` `count = 0;` ` ` `// The node which is present` ` ` `// farthest from the root` ` ` `// node is taken as v` ` ` `// If u is farther from` ` ` `// root node then swap the two` ` ` `if` `(level[u] > level[v])` ` ` `{` ` ` `int` `temp = v;` ` ` `v = u;` ` ` `u = temp;` ` ` `}` ` ` `int` `d = level[v] - level[u];` ` ` `// Find the ancestor of v` ` ` `// which is at same level as u` ` ` `while` `(d-- > 0)` ` ` `{` ` ` ` ` `// If Weight is prime` ` ` `// increment count` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` ` ` `v = par[v];` ` ` `}` ` ` `// If u is the ancestor of v` ` ` `// then u is the LCA of u and v` ` ` `// Now check if weigh[v]` ` ` `// is prime or not` ` ` `if` `(v == u)` ` ` `{` ` ` `if` `(prime[weight[v]])` ` ` `count++; ` ` ` ` ` `return` `count;` ` ` `}` ` ` `// When v and u are on the` ` ` `// same level but are in` ` ` `// different subtree. Now` ` ` `// move both u and v up by` ` ` `// 1 till they are not same` ` ` `while` `(v != u)` ` ` `{` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `if` `(prime[weight[u]])` ` ` `count++;` ` ` `u = par[u];` ` ` `v = par[v];` ` ` `}` ` ` ` ` `// If weight of first` ` ` `// ancestor is prime` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `for` `(` `int` `i = 0; i < graph.Length; i++)` ` ` `graph[i] = ` `new` `List<` `int` `>();` ` ` ` ` `// Precompute all the prime` ` ` `// numbers till MAX` ` ` `SieveOfEratosthenes();` ` ` `// Weights of the node` ` ` `weight[1] = 5;` ` ` `weight[2] = 10;` ` ` `weight[3] = 11;` ` ` `weight[4] = 8;` ` ` `weight[5] = 6;` ` ` `// Edges of the tree` ` ` `graph[1].Add(2);` ` ` `graph[2].Add(3);` ` ` `graph[2].Add(4);` ` ` `graph[1].Add(5);` ` ` `dfs(1, -1, 0);` ` ` `int` `u = 3, v = 5;` ` ` ` ` `Console.Write(findPrimeOnPath(u, v));` `}` `}` `// This code is contributed by Amit Katiyar` |

## Javascript

`<script>` `// Javascript program Count prime weight` `// nodes between two nodes in the given tree` `var` `MAX = 1000;` `var` `weight = Array(MAX);` `var` `level = Array(MAX);` `var` `par = Array(MAX);` `var` `prime = Array(MAX+1).fill(` `true` `);` `var` `graph = Array.from(Array(MAX), ()=>Array());` `// Function to perform` `// Sieve Of Eratosthenes for prime number` `function` `SieveOfEratosthenes()` `{` ` ` ` ` `for` `(` `var` `p = 2; p * p <= MAX; p++) {` ` ` `// Check if prime[p] is not changed,` ` ` `// then it is a prime` ` ` `if` `(prime[p] == ` `true` `) {` ` ` `// Update all multiples` ` ` `// of p greater than or` ` ` `// equal to the square of it` ` ` `// numbers which are multiple` ` ` `// of p and are less than p^2` ` ` `// are already been marked.` ` ` `for` `(` `var` `i = p * p; i <= MAX; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` `}` `// Function to perform dfs` `function` `dfs(node, parent, h)` `{` ` ` `// Stores parent of each node` ` ` `par[node] = parent;` ` ` `// Stores level of each node from root` ` ` `level[node] = h;` ` ` `graph[node].forEach(child => {` ` ` ` ` `if` `(child != parent)` ` ` `dfs(child, node, h + 1);` ` ` `});` `}` `// Function to perform prime` `// number between the path` `function` `findPrimeOnPath(u, v)` `{` ` ` `var` `count = 0;` ` ` `// The node which is present farthest` ` ` `// from the root node is taken as v` ` ` `// If u is farther from root node` ` ` `// then swap the two` ` ` `if` `(level[u] > level[v])` ` ` `{` ` ` `[u,v] = [v,u]` ` ` `}` ` ` `var` `d = level[v] - level[u];` ` ` `// Find the ancestor of v` ` ` `// which is at same level as u` ` ` `while` `(d--) {` ` ` `// If Weight is prime` ` ` `// increment count` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `v = par[v];` ` ` `}` ` ` `// If u is the ancestor of v` ` ` `// then u is the LCA of u and v` ` ` `// Now check if weigh[v]` ` ` `// is prime or not` ` ` `if` `(v == u) {` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `return` `count;` ` ` `}` ` ` `// When v and u are on the same level but` ` ` `// are in different subtree. Now move both` ` ` `// u and v up by 1 till they are not same` ` ` `while` `(v != u) {` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `if` `(prime[weight[u]])` ` ` `count++;` ` ` `u = par[u];` ` ` `v = par[v];` ` ` `}` ` ` `// If weight of first ancestor` ` ` `// is prime` ` ` `if` `(prime[weight[v]])` ` ` `count++;` ` ` `return` `count;` `}` `// Driver code` `// Precompute all the prime` `// numbers till MAX` `SieveOfEratosthenes();` `// Weights of the node` `weight[1] = 5;` `weight[2] = 10;` `weight[3] = 11;` `weight[4] = 8;` `weight[5] = 6;` `// Edges of the tree` `graph[1].push(2);` `graph[2].push(3);` `graph[2].push(4);` `graph[1].push(5);` `dfs(1, -1, 0);` `var` `u = 3, v = 5;` `document.write( findPrimeOnPath(u, v));` `</script>` |

**Output:**

2

__Complexity Analysis:__

**Time Complexity:**O(N).

In dfs, every node of the tree is processed once, and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, for processing each node the SieveOfEratosthenes() function is used which has a complexity of O(sqrt(N)) too but since this function is executed only once, it does not affect the overall time complexity. Therefore, the time complexity is O(N).**Auxiliary Space:**O(N).

Extra space is used for the prime array, so the space complexity is O(N).

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