Range Update without using Lazy Propagation and Point Query in a Segment Tree
Given an array arr[] consisting of N 0s and a 2D array Q[][] consisting of queries of the following two types:
- 1 L R X: Increment all the elements in the range [L, R] by X.
- 2 X: Print elements at Xth index of the array.
Input: arr[] = { 0, 0, 0, 0, 0 }, Q[][] = { { 1, 0, 2, 100 }, { 2, 1 }, { 1, 2, 3, 200 }, { 2, 2 }, {
4 } }
Output: 100 300 0
Explanation:
Query1: Incrementing all the array elements in the range [0, 2] by 100 modifies arr[] to { 100, 100, 100, 0, 0 }.
Query2: Print arr[1](=100)
Query3: Incrementing all the array elements in the range [2, 3] by 200 modifies arr[] to { 100, 100, 300, 200, 0 }.
Query4: Print arr[2](=300)
Query5: Print arr[4](=0)Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
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Input: arr[] = { 0, 0 }, Q[][] = { { 1, 0, 1, 100 }, { 2, 1 } }
Output: 100
Approach: The idea is to use the concept of Segment Tree to perform queries of type 2 and the difference array to perform queries of type 1. Follow the steps below to solve the problem:
- For query of type { 1, L, R, X }, update arr[L] += X and arr[R + 1] -= X using Segment tree.
- For query of type { 2, X } Find the sum of array elements in the range [0, X] using Segment Tree and print the sum obtained.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to update the array elements// at idx pos, i.e arr[pos] += Xvoid update(int Tree[], int idx, int s, int e, int pos, int X){ // If current node is a // leaf nodes if (s == e) { // Update Tree[idx] Tree[idx] += X; } else { // Divide segment tree into left // and right subtree int m = (s + e) / 2; // Check if pos lies in left subtree if (pos <= m) { // Search pos in left subtree update(Tree, 2 * idx, s, m, pos, X); } else { // Search pos in right subtree update(Tree, 2 * idx + 1, m + 1, e, pos, X); } // Update Tree[idx] Tree[idx] = Tree[2 * idx] + Tree[2 * idx + 1]; }}// Function to find the sum from// elements in the range [0, X]int sum(int Tree[], int idx, int s, int e, int ql, int qr){ // Check if range[ql, qr] equals // to range [s, e] if (ql == s && qr == e) return Tree[idx]; if (ql > qr) return 0; // Divide segment tree into // left substree and // right substree int m = (s + e) / 2; // Return sum of elements in the range[ql, qr] return sum(Tree, 2 * idx, s, m, ql, min(m, qr)) + sum(Tree, 2 * idx + 1, m + 1, e, max(ql, m + 1), qr);}// Function to find Xth element// in the arrayvoid getElement(int Tree[], int X, int N){ // Print element at index x cout << "Element at " << X << " is " << sum(Tree, 1, 0, N - 1, 0, X) << endl;}// Function to update array elements// in the range [L, R]void range_Update(int Tree[], int L, int R, int X, int N){ // Update arr[l] += X update(Tree, 1, 0, N - 1, L, X); // Update arr[R + 1] += X if (R + 1 < N) update(Tree, 1, 0, N - 1, R + 1, -X);}// Drivers Codeint main(){ // Size of array int N = 5; int Tree[4 * N + 5] = { 0 }; int arr[] = { 0, 0 }; vector<vector<int> > Q = { { 1, 0, 1, 100 }, { 2, 1 } }; int cntQuery = Q.size(); for (int i = 0; i < cntQuery; i++) { if (Q[i][0] == 1) { range_Update(Tree, Q[i][1], Q[i][2], Q[i][3], N); } else { getElement(Tree, Q[i][1], N); } }} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to update the array elements// at idx pos, i.e arr[pos] += Xstatic void update(int Tree[], int idx, int s, int e, int pos, int X){ // If current node is a // leaf nodes if (s == e) { // Update Tree[idx] Tree[idx] += X; } else { // Divide segment tree into left // and right subtree int m = (s + e) / 2; // Check if pos lies in left subtree if (pos <= m) { // Search pos in left subtree update(Tree, 2 * idx, s, m, pos, X); } else { // Search pos in right subtree update(Tree, 2 * idx + 1, m + 1, e, pos, X); } // Update Tree[idx] Tree[idx] = Tree[2 * idx] + Tree[2 * idx + 1]; }}// Function to find the sum from// elements in the range [0, X]static int sum(int Tree[], int idx, int s, int e, int ql, int qr){ // Check if range[ql, qr] equals // to range [s, e] if (ql == s && qr == e) return Tree[idx]; if (ql > qr) return 0; // Divide segment tree into // left substree and // right substree int m = (s + e) / 2; // Return sum of elements in the range[ql, qr] return sum(Tree, 2 * idx, s, m, ql, Math.min(m, qr)) + sum(Tree, 2 * idx + 1, m + 1, e, Math.max(ql, m + 1), qr);}// Function to find Xth element// in the arraystatic void getElement(int Tree[], int X, int N){ // Print element at index x System.out.print("Element at " + X+ " is " + sum(Tree, 1, 0, N - 1, 0, X) +"\n");}// Function to update array elements// in the range [L, R]static void range_Update(int Tree[], int L, int R, int X, int N){ // Update arr[l] += X update(Tree, 1, 0, N - 1, L, X); // Update arr[R + 1] += X if (R + 1 < N) update(Tree, 1, 0, N - 1, R + 1, -X);}// Drivers Codepublic static void main(String[] args){ // Size of array int N = 5; int Tree[] = new int[4 * N + 5]; int arr[] = { 0, 0 }; int[][] Q = { { 1, 0, 1, 100 }, { 2, 1 } }; int cntQuery = Q.length; for (int i = 0; i < cntQuery; i++) { if (Q[i][0] == 1) { range_Update(Tree, Q[i][1], Q[i][2], Q[i][3], N); } else { getElement(Tree, Q[i][1], N); } }}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program to implement# the above approach# Function to update the array elements# at idx pos, i.e arr[pos] += Xdef update(Tree, idx, s, e, pos, X): # If current node is a # leaf nodes if (s == e): # Update Tree[idx] Tree[idx] += X else: # Divide segment tree into left # and right subtree m = (s + e) // 2 # Check if pos lies in left subtree if (pos <= m): # Search pos in left subtree update(Tree, 2 * idx, s, m, pos, X) else: # Search pos in right subtree update(Tree, 2 * idx + 1, m + 1, e,pos, X) # Update Tree[idx] Tree[idx] = Tree[2 * idx] + Tree[2 * idx + 1]# Function to find the sum from# elements in the range [0, X]def sum(Tree, idx, s, e, ql, qr): # Check if range[ql, qr] equals # to range [s, e] if (ql == s and qr == e): return Tree[idx] if (ql > qr): return 0 # Divide segment tree into # left substree and # right substree m = (s + e) // 2 # Return sum of elements in the range[ql, qr] return sum(Tree, 2 * idx, s, m, ql, min(m, qr))+ sum(Tree, 2 * idx + 1, m + 1, e,max(ql, m + 1), qr)# Function to find Xth element# in the arraydef getElement(Tree, X, N): # Prelement at index x print("Element at", X, "is ", sum(Tree, 1, 0, N - 1, 0, X))# Function to update array elements# in the range [L, R]def range_Update(Tree, L, R, X, N): # Update arr[l] += X update(Tree, 1, 0, N - 1, L, X) # Update arr[R + 1] += X if (R + 1 < N): update(Tree, 1, 0, N - 1, R + 1, -X)# Drivers Codeif __name__ == '__main__': # Size of array N = 5 Tree = [0]*(4 * N + 5) arr = [0, 0] Q = [ [1, 0, 1, 100], [2, 1] ] cntQuery = len(Q) for i in range(cntQuery): if (Q[i][0] == 1): range_Update(Tree, Q[i][1], Q[i][2], Q[i][3], N) else: getElement(Tree, Q[i][1], N)# This code is contributed by mohit kumar 29. |
C#
// C# program to implement// the above approachusing System;public class GFG{ // Function to update the array elements // at idx pos, i.e arr[pos] += X static void update(int []Tree, int idx, int s, int e, int pos, int X) { // If current node is a // leaf nodes if (s == e) { // Update Tree[idx] Tree[idx] += X; } else { // Divide segment tree into left // and right subtree int m = (s + e) / 2; // Check if pos lies in left subtree if (pos <= m) { // Search pos in left subtree update(Tree, 2 * idx, s, m, pos, X); } else { // Search pos in right subtree update(Tree, 2 * idx + 1, m + 1, e, pos, X); } // Update Tree[idx] Tree[idx] = Tree[2 * idx] + Tree[2 * idx + 1]; } } // Function to find the sum from // elements in the range [0, X] static int sum(int []Tree, int idx, int s, int e, int ql, int qr) { // Check if range[ql, qr] equals // to range [s, e] if (ql == s && qr == e) return Tree[idx]; if (ql > qr) return 0; // Divide segment tree into // left substree and // right substree int m = (s + e) / 2; // Return sum of elements in the range[ql, qr] return sum(Tree, 2 * idx, s, m, ql, Math.Min(m, qr)) + sum(Tree, 2 * idx + 1, m + 1, e, Math.Max(ql, m + 1), qr); } // Function to find Xth element // in the array static void getElement(int []Tree, int X, int N) { // Print element at index x Console.Write("Element at " + X+ " is " + sum(Tree, 1, 0, N - 1, 0, X) +"\n"); } // Function to update array elements // in the range [L, R] static void range_Update(int []Tree, int L, int R, int X, int N) { // Update arr[l] += X update(Tree, 1, 0, N - 1, L, X); // Update arr[R + 1] += X if (R + 1 < N) update(Tree, 1, 0, N - 1, R + 1, -X); } // Drivers Code public static void Main(String[] args) { // Size of array int N = 5; int []Tree = new int[4 * N + 5]; int []arr = { 0, 0 }; int[,]Q = { { 1, 0, 1, 100 }, { 2, 1, 0, 0 } }; int cntQuery = Q.GetLength(0); for (int i = 0; i < cntQuery; i++) { if (Q[i, 0] == 1) { range_Update(Tree, Q[i, 1], Q[i, 2], Q[i, 3], N); } else { getElement(Tree, Q[i, 1], N); } } }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to implement// the above approach// Function to update the array elements// at idx pos, i.e arr[pos] += Xfunction update(Tree, idx, s, e, pos, X){ // If current node is a // leaf nodes if (s == e) { // Update Tree[idx] Tree[idx] += X; } else { // Divide segment tree into left // and right subtree var m = parseInt((s + e) / 2); // Check if pos lies in left subtree if (pos <= m) { // Search pos in left subtree update(Tree, 2 * idx, s, m, pos, X); } else { // Search pos in right subtree update(Tree, 2 * idx + 1, m + 1, e, pos, X); } // Update Tree[idx] Tree[idx] = Tree[2 * idx] + Tree[2 * idx + 1]; }}// Function to find the sum from// elements in the range [0, X]function sum(Tree, idx, s, e, ql, qr){ // Check if range[ql, qr] equals // to range [s, e] if (ql == s && qr == e) return Tree[idx]; if (ql > qr) return 0; // Divide segment tree into // left substree and // right substree var m =parseInt((s + e) / 2); // Return sum of elements in the range[ql, qr] return sum(Tree, 2 * idx, s, m, ql, Math.min(m, qr)) + sum(Tree, 2 * idx + 1, m + 1, e, Math.max(ql, m + 1), qr);}// Function to find Xth element// in the arrayfunction getElement(Tree, X, N){ // Print element at index x document.write( "Element at " + X + " is " + sum(Tree, 1, 0, N - 1, 0, X) );}// Function to update array elements// in the range [L, R]function range_Update(Tree, L, R, X, N){ // Update arr[l] += X update(Tree, 1, 0, N - 1, L, X); // Update arr[R + 1] += X if (R + 1 < N) update(Tree, 1, 0, N - 1, R + 1, -X);}// Drivers Code// Size of arrayvar N = 5;var Tree = Array(4 * N + 5).fill(0);var arr = [ 0, 0 ];var Q = [[ 1, 0, 1, 100 ], [ 2, 1 ]];var cntQuery = Q.length;for (var i = 0; i < cntQuery; i++) { if (Q[i][0] == 1) { range_Update(Tree, Q[i][1], Q[i][2], Q[i][3], N); } else { getElement(Tree, Q[i][1], N); }}</script> |
Output:
Element at 1 is 100
Time Complexity: O(|Q| * log(N))
Auxiliary Space: O(N)
