[Solved] Recalling Early Days GP with Trees solution in Hackerrank - Hacerrank solution C, C++, java,js, Python

Algorithm

Problem Name: Data Structures - Recalling Early Days GP with Trees

Problem Link: https://www.hackerrank.com/challenges/recalling-early-days-gp-with-trees/problem?isFullScreen=true

In this HackerRank in Data Structures - Recalling Early Days GP with Trees solutions,

You are given a tree with N nodes and each has a value associated with it. You are given Q queries, each of which is either an update or a retrieval operation.

The update query is of the format

i j X

This means you'd have to add a GP series to the nodes which lie in the path from node i to node j (both inclusive) with first term of the GP as X on node i and the common ratio as R (given in the input)

The retrieval query is of the format

i j

You need to return the sum of the node values (S) lying in the path from node i to node j modulo 100711433.

Input Format
The first line contains two integers (N and R respectively) separated by a space.
In the next N-1 lines, the i^th line describes the i^th edge: a line with two integers a b separated by a single space denotes an edge between a, b.
The next line contains 2 space separated integers (U and Q respectively) representing the number of Update and Query operations to follow.
U lines follow. Each of the next U lines contains 3 space separated integers (i,j, and X respectively).
Each of the next Q lines contains 2 space separated integers, i and j respectively.

Output Format
It contains exactly Q lines and each line containing the answer of the i^th query.

Constraints

2 <= N <= 100000
2 <= R <= 10⁹
1 <= U <= 100000
1 <= Q <= 100000
1 <= X <= 10
1 <= a, b, i, j <= N

Sample Input

Sample Output

31
18

Explanation

The node values after the first updation becomes :

3 6 0 0 0 12

The node values after second updation becomes :

3 6 20 10 5 12

Answer to Query #1: 12 + 6 + 3 + 10 = 31
Answer to Query #2: 5 + 10 +3 = 18

Code Examples

#1 Code Example with C Programming

Code - C Programming


#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MOD 100711433
typedef struct _lnode
{
    int x;
    int w;
    struct _lnode *next;
}lnode;
typedef struct _tree
{
    long long sum;
    long long offset1;
    long long offset2;
}tree;
int N, cn, level[100000], DP[18][100000], subtree_size[100000], special[100000], node_chain[100000], node_idx[100000], chain_head[100000], chain_len[100000] = {0};
long long p[100001], sp[100001];
lnode *table[100000] = {0};
tree *chain[100000];
void insert_edge(int x, int y, int w)
{
    lnode *t=malloc(sizeof(lnode));
    t -> x = y;
    t -> w = w;
    t -> next = table[x];
    table[x] = t;
    t = malloc(sizeof(lnode));
    t -> x = x;
    t -> w = w;
    t -> next = table[y];
    table[y] = t;
    return;
}
void dfs0(int u)
{
    lnode *x;
    subtree_size[u] = 1;
    special[u] = -1;
    for( x = table[u] ; x ; x = x -> next )
    {
        if( x -> x != DP[0][u] )
        {
            DP[0][x -> x] = u;
            level[x -> x] = level[u] + 1;
            dfs0(x -> x);
            subtree_size[u] += subtree_size[x -> x];
            if( special[u] == -1 || subtree_size[x -> x] > subtree_size[special[u]] )
            {
                special[u] = x -> x;
            }
        }
    }
    return;
}
void dfs1(int u, int c)
{
    lnode *x;
    node_chain[u] = c;
    node_idx[u] = chain_len[c]++;
    for( x = table[u] ; x ; x = x -> next )
    {
        if( x -> x != DP[0][u] )
        {
            if( x -> x == special[u] )
            {
                dfs1(x -> x, c);
            }
            else
            {
                chain_head[cn] = x -> x;
                dfs1(x -> x, cn++);
            }
        }
    }
    return;
}
void preprocess()
{
    int i, j;
    level[0] = 0;
    DP[0][0] = 0;
    dfs0(0);
    for( i = 1 ; i < 18 ; i++ )
    {
        for( j = 0 ; j  <  N ; j++ )
        {
            DP[i][j] = DP[i-1][DP[i-1][j]];
        }
    }
    cn = 1;
    chain_head[0] = 0;
    dfs1(0, 0);
    for( i = 0 ; i  <  cn ; i++ )
    {
        chain[i] = (tree*)malloc(4 * chain_len[i] * sizeof(tree));
        memset(chain[i], 0, 4*chain_len[i]*sizeof(tree));
    }
    return;
}
int lca(int a, int b>
{
    int i;
    if( level[a] > level[b] )
    {
        i = a;
        a = b;
        b = i;
    }
    int d = level[b] - level[a];
    for( i = 0 ; i < 18 ; i++ )
    {
        if( d & ( 1 << i ) )
        {
            b = DP[i][b];
        }
    }
    if( a == b >
    {
        return a;
    }
    for( i = 17 ; i >= 0 ; i-- )
    {
        if( DP[i][a] != DP[i][b] )
        {
            a = DP[i][a], b = DP[i][b];
        }
    }
    return DP[0][a];
}
long long sum(int v, int tl, int tr, int l, int r, tree *t)
{
    push(v, tl, tr, t);
    if( l > r )
    {
        return 0;
    }
    if( l == tl && r == tr )
    {
        return t[v].sum;
    }
    int tm = ( tl + tr ) / 2;
    return ( sum(v*2, tl, tm, l, min(r, tm), t) + sum(v*2+1, tm+1, tr, max(l, tm+1), r, t) ) % MOD;
}
void range_update(int v, int tl, int tr, int pos1, int pos2, long long o1, long long o2, tree *t)
{
    push(v, tl, tr, t);
    if( pos2  <  tl || pos1 > tr )
    {
        return;
    }
    int tm = ( tl + tr ) / 2;
    if( pos1  < = tl && pos2 >= tr )
    {
        t[v].offset1 = o1 * p[tl-pos1] % MOD;
        t[v].offset2 = o2 * p[pos2-tr] % MOD;
    }
    else
    {
        range_update(v*2, tl, tm, pos1, pos2, o1, o2, t);
        range_update(v*2+1, tm+1, tr, pos1, pos2, o1, o2, t);
        push(v*2, tl, tm, t);
        push(v*2+1, tm+1, tr, t);
        t[v].sum = ( t[v*2].sum + t[v*2+1].sum ) % MOD;
    }
    return;
}
void push(int v, int tl, int tr, tree *t)
{
    if( !t[v].offset1 && !t[v].offset2 )
    {
        return;
    }
    t[v].sum = ( t[v].sum + ( t[v].offset1 + t[v].offset2 ) * sp[tr-tl] ) % MOD;
    if( tl != tr )
    {
        int tm = ( tl + tr ) / 2;
        t[v*2].offset1 = ( t[v*2].offset1 + t[v].offset1 ) % MOD;
        t[v*2+1].offset1 = ( t[v*2+1].offset1 + t[v].offset1 * p[tm-tl+1] ) % MOD;
        t[v*2].offset2 = ( t[v*2].offset2 + t[v].offset2 * p[tr-tm] ) % MOD;
        t[v*2+1].offset2 = ( t[v*2+1].offset2 + t[v].offset2 ) % MOD;
    }
    t[v].offset1 = t[v].offset2 = 0;
    return;
}
void range_solve(int x, int y, long long z)
{
    int ca = lca(x, y), ty = y;
    long long cac = z % MOD, cay = 0;
    while( node_chain[x] != node_chain[ca] )
    {
        range_update(1, 0, chain_len[node_chain[x]]-1, 0, node_idx[x], 0, cac, chain[node_chain[x]]);
        cac = cac * p[node_idx[x]+1] % MOD;
        x = DP[0][chain_head[node_chain[x]]];
    }
    range_update(1, 0, chain_len[node_chain[x]]-1, node_idx[ca], node_idx[x], 0, cac, chain[node_chain[x]]);
    cac = cac * p[node_idx[x]-node_idx[ca]] % MOD;
    while( node_chain[ty] != node_chain[ca] )
    {
        cay += node_idx[ty] + 1;
        ty = DP[0][chain_head[node_chain[ty]]];
    }
    cay += node_idx[ty] - node_idx[ca];
    while( node_chain[y] != node_chain[ca] )
    {
        cay -= node_idx[y];
        range_update(1, 0, chain_len[node_chain[y]]-1, 0, node_idx[y], cac*p[cay]%MOD, 0, chain[node_chain[y]]);
        cay--;
        y = DP[0][chain_head[node_chain[y]]];
    }
    if( node_idx[y] != node_idx[ca] )
    {
        range_update(1, 0, chain_len[node_chain[y]]-1, node_idx[ca]+1, node_idx[y], cac*p[1]%MOD, 0, chain[node_chain[y]]);
    }
    return;
}
int min(int x, int y)
{
    return ( x < y ) ? x : y ;
}
int max(int x, int y>
{
    return ( x > y ) ? x : y;
}
long long solve(int x, int ancestor)
{
    long long ans = 0;
    while( node_chain[x] != node_chain[ancestor] )
    {
        ans = ( ans + sum(1, 0, chain_len[node_chain[x]]-1, 0, node_idx[x], chain[node_chain[x]]) ) % MOD;
        x = DP[0][chain_head[node_chain[x]]];
    }
    ans = ( ans + sum(1, 0, chain_len[node_chain[x]]-1, node_idx[ancestor], node_idx[x], chain[node_chain[x]]) ) % MOD;
    return ans;
}
int main()
{
    int U, Q, x, y, i;
    long long R, t;
    scanf("%d%lld", &N, &R);
    R %= MOD;
    p[0] = sp[0] = 1;
    for( i = 1 ; i  < = N ; i++ )
    {
        p[i] = R * p[i-1] % MOD;
        sp[i] = ( sp[i-1] + p[i] ) % MOD;
    }
    for( i = 0 ; i  <  N - 1 ; i++ )
    {
        scanf("%d%d", &x, &y);
        insert_edge(x-1, y-1, 1);
    }
    preprocess();
    scanf("%d%d", &U, &Q);
    while(U--)
    {
        scanf("%d%d%lld", &x, &y, &t);
        range_solve(x-1, y-1, t);
    }
    while(Q--)
    {
        scanf("%d%d", &x, &y);
        i = lca(x-1, y-1);
        printf("%lld\n", ( solve(x-1, i) + solve(y-1, i) - sum(1, 0, chain_len[node_chain[i]]-1, node_idx[i], node_idx[i], chain[node_chain[i]]) + MOD ) % MOD);
    }
    return 0;
}

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#2 Code Example with Python Programming

Code - Python Programming


import sys

sys.setrecursionlimit(10000000)

n, r = input().strip().split(' ')
n, r = [int(n), int(r)]

g = {}
gv = {}
edge = []

for c in range(1, n):
    i, j = input().strip().split(' ')
    if i not in g.keys():
        g[i] = []
        gv[i] = 0
    if j not in g.keys():
        g[j] = []
        gv[j] = 0

    g[i].append(j)
    g[j].append(i)


def find_path(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return path
    if start not in graph.keys():
        return None
    for node in graph[start]:
        if node not in path:
            newpath = find_path(graph, node, end, path)
            if newpath: return newpath
    return None


u, p = input().strip().split(' ')
u, p = [int(u), int(p)]

for c in range(u):
    i, j, x = input().strip().split(' ')
    i, j, x = [str(i), str(j), int(x)]
    path = find_path(g, i, j)
    for pa in path:
        gv[pa] = gv[pa] + x
        x *= r
for c in range(p):
    i, j = input().strip().split(' ')
    path = find_path(g, i, j)
    s = 0
    for p in path:
        s += gv[p]
    print(s % 100711433)

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#3 Code Example with Java Programming

Code - Java Programming


import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileInputStream;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;
import java.util.StringTokenizer;

public class Solution2 {

  static final int LOGN = 17;
  static final int MOD = 100_711_433;

  static long power(long a, int n) {
    long r = 1;
    for (; n > 0; n >>= 1, a = a*a%MOD) {
      if ((n & 1) > 0) {
        r = r*a%MOD;
      }
    }
    return r;
  }

  static long add(long x, long y) {
    return (x+y)%MOD;
  }

  static int inv(int x) {
    long r = 1;
    for (; x > 1; x = MOD%x) {
      r = MOD/x * -r % MOD;
    }
    return (int) r;
  }

  static int[] dep;
  static int[][] par;
  static List < Integer>[] e;
  
  static class Node {
    int d;
    int v;
    int p;
    Node(int d, int v, int p) {
      this.d = d;
      this.v = v;
      this.p = p;
    }
  }
  
  static void dfs(int d, int v, int p) {
    Deque < Node> queue = new LinkedList<>();
    queue.add(new Node(d, v, p));
    while (!queue.isEmpty()) {
      Node node = queue.poll();
      dep[node.v] = node.d;
      par[0][node.v] = node.p;
      for (int u: e[node.v]) {
        if (u != node.p) {
          queue.add(new Node(node.d+1, u, node.v));
        }
      }
    }
  }

  static int r;
  static int invr;
  static long[] d;
  static long[] dd;
  
  static class Node2 {
    int v;
    int p;
    boolean start = true;
    Node2(int v, int p) {
      this.v = v;
      this.p = p;
    }
  }
  
  static void dfs2(int v, int p) {
    Deque < Node2> queue = new LinkedList<>();
    queue.add(new Node2(v, p));
    while (!queue.isEmpty()) {
      Node2 node = queue.peek();
      if (node.start) {
        for (int u: e[node.v]) {
          if (u != node.p) {
            queue.push(new Node2(u, node.v));
          }
        }
        node.start = false;
      } else {
        if (node.p >= 0) {
          d[node.p] = add(d[node.p], d[node.v] * r);
          dd[node.p] = add(dd[node.p], dd[node.v] * invr);
        }
        queue.remove();
      }
    }
  }

  static long[] path;

  static class Node3 {
    int v;
    int p;
    long s;
    Node3(int v, int p, long s) {
      this.v = v;
      this.p = p;
      this.s = s;
    }
  }
  
  static void dfs3(int v, int p, long s) {
    Deque < Node3> queue = new LinkedList<>();
    queue.add(new Node3(v, p, s));
    while (!queue.isEmpty()) {
      Node3 node = queue.poll();
      path[node.v] = (node.s+d[node.v]+dd[node.v])%MOD;
      for (int u: e[node.v]) {
        if (u != node.p) {
          queue.push(new Node3(u, node.v, (node.s+d[node.v]+dd[node.v])%MOD));
        }
      }
    }    
  }

  static int lca(int v, int u) {
    if (dep[v]  <  dep[u]) {
      int temp = v;
      v = u;
      u = temp;
    }
    for (int i = LOGN; --i >= 0; ) {
      if (1 << i  < = dep[v]-dep[u]) {
        v = par[i][v];
      }
    }
    if (v == u) {
      return v;
    }
    for (int i = LOGN; --i >= 0; ) {
      if (par[i][v] != par[i][u]) {
        v = par[i][v];
        u = par[i][u];
      }
    }
    return par[0][v];
  }

  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    BufferedWriter bw = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

    StringTokenizer st = new StringTokenizer(br.readLine());
    int n = Integer.parseInt(st.nextToken());
    r = Integer.parseInt(st.nextToken()) % MOD;
    invr = r > 0 ? inv(r) : 0;
    e = new List[n];
    for (int i = 0; i  <  n; i++) {
      e[i] = new LinkedList<>();
    }
    for (int i = 0; i  <  n-1; i++) {
      st = new StringTokenizer(br.readLine());
      int u = Integer.parseInt(st.nextToken())-1;
      int v = Integer.parseInt(st.nextToken())-1;
      e[v].add(u);
      e[u].add(v);
    }
    dep = new int[n];
    par = new int[LOGN][n];
    dfs(0, 0, -1);    
    for (int j = 1; j  <  LOGN; j++) {
      for (int i = 0; i  <  n; i++) {
        par[j][i] = par[j-1][i] < 0 ? -1 : par[j-1][par[j-1][i]];
      }
    }
    st = new StringTokenizer(br.readLine());
    int upd = Integer.parseInt(st.nextToken());
    int q = Integer.parseInt(st.nextToken());
    d = new long[n];
    dd = new long[n];
    for (int i=0; i  <  upd; i++) {
      st = new StringTokenizer(br.readLine());
      int v = Integer.parseInt(st.nextToken()) - 1;
      int u = Integer.parseInt(st.nextToken()) - 1;
      int w = lca(v, u);
      int x = Integer.parseInt(st.nextToken());

      if (r > 0) {
        long xl = power(r, dep[v]-dep[w]), xr = power(r, dep[u]-dep[w]);
        d[v] = add(d[v], x);
        if (w > 0) {
          d[par[0][w]] = add(d[par[0][w]], - (x * xl % MOD * r));
        }
        dd[u] = add(dd[u], x * xl % MOD * xr);
        dd[w] = add(dd[w], - x * xl);
      } else {
        d[v] = add(d[v], x);
      }
    }
    dfs2(0, -1);
    path = new long[n];
    dfs3(0, -1, 0);
    for (int i=0; i  <  q; i++) {
      st = new StringTokenizer(br.readLine());
      int v = Integer.parseInt(st.nextToken()) - 1;
      int u = Integer.parseInt(st.nextToken()) - 1;
      int w = lca(v, u);
      long result = ((path[v]+path[u]-path[w]-(w > 0 ? path[par[0][w]] : 0)) % MOD + MOD) % MOD;
      bw.write(String.valueOf(result) + "\n");
    }
    
    bw.newLine();
    bw.close();
    br.close();
  }  
}

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