Queries to find the Lower Bound of K from Prefix Sum Array with updates using Fenwick Tree
Last Updated :
27 Sep, 2021
Given an array A[ ] consisting of non-negative integers and a matrix Q[ ][ ] consisting of queries of the following two types:
- (1, l, val): Update A[l] to A[l] + val.
- (2, K): Find the lower_bound of K in the prefix sum array of A[ ]. If the lower_bound does not exist, print -1.
The task for each query of the second type is to print the index of the lower_bound of value K.
Examples:
Input: A[ ] = {1, 2, 3, 5, 8}, Q[ ][ ] = {{1, 0, 2}, {2, 5}, {1, 3, 5}}
Output: 1
Explanation:
Query 1: Update A[0] to A[0] + 2. Now A[ ] = {3, 2, 3, 5, 8}
Query 2: lower_bound of K = 5 in the prefix sum array {3, 5, 8, 13, 21} is 5 and index = 1.
Query 3: Update A[3] to A[3] + 5. Now A[ ] = {3, 2, 3, 10, 8}
Input: A[ ] = {4, 1, 12, 8, 20}, Q[ ] = {{2, 50}, {1, 3, 12}, {2, 50}}
Output: -1
Naive approach:
The simplest approach is to first build a prefix sum array of a given array A[ ], and for queries of Type 1, update values and recalculate the prefix sum. For query of Type 2, perform a Binary Search on the prefix sum array to find the lower bound.
Time Complexity: O(Q*(N*logn))
Auxiliary Space: O(N)
Efficient Approach:
The above approach can be optimized Fenwick Tree. Using this Data Structure, the update queries in the prefix sum array can be performed in logarithmic time.
Follow the steps below to solve the problem:
- Construct the Prefix Sum Array using Fenwick Tree.
- For queries of Type 1, while l > 0, add val to A[l] traverse to the parent node by adding the least significant bit in l.
- For queries of Type 2, perform the Binary Search on the Fenwick Tree to obtain the lower bound.
- Whenever a prefix sum greater than K appears, store that index and traverse the left part of the Fenwick Tree. Otherwise, traverse the right part of the Fenwick Tree now, perform Binary Search.
- Finally, print the required index.
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 calculate and return
// the sum of arr[0..index]
int getSum(int BITree[], int index)
{
int ans = 0;
index += 1;
// Traverse ancestors
// of BITree[index]
while (index > 0)
{
// Update the sum of current
// element of BIT to ans
ans += BITree[index];
// Update index to that
// of the parent node in
// getSum() view by
// subtracting LSB(Least
// Significant Bit)
index -= index & (-index);
}
return ans;
}
// Function to update the Binary Index
// Tree by replacing all ancestors of
// index by their respective sum with val
static void updateBIT(int BITree[], int n,
int index, int val)
{
index = index + 1;
// Traverse all ancestors
// and sum with 'val'.
while (index <= n)
{
// Add 'val' to current
// node of BIT
BITree[index] += val;
// Update index to that
// of the parent node in
// updateBit() view by
// adding LSB(Least
// Significant Bit)
index += index & (-index);
}
}
// Function to construct the Binary
// Indexed Tree for the given array
int* constructBITree(int arr[], int n)
{
// Initialize the
// Binary Indexed Tree
int* BITree = new int[n + 1];
for(int i = 0; i <= n; i++)
BITree[i] = 0;
// Store the actual values in
// BITree[] using update()
for(int i = 0; i < n; i++)
updateBIT(BITree, n, i, arr[i]);
return BITree;
}
// Function to obtain and return
// the index of lower_bound of k
int getLowerBound(int BITree[], int arr[],
int n, int k)
{
int lb = -1;
int l = 0, r = n - 1;
while (l <= r)
{
int mid = l + (r - l) / 2;
if (getSum(BITree, mid) >= k)
{
r = mid - 1;
lb = mid;
}
else
l = mid + 1;
}
return lb;
}
void performQueries(int A[], int n, int q[][3])
{
// Store the Binary Indexed Tree
int* BITree = constructBITree(A, n);
// Solve each query in Q
for(int i = 0;
i < sizeof(q[0]) / sizeof(int);
i++)
{
int id = q[i][0];
if (id == 1)
{
int idx = q[i][1];
int val = q[i][2];
A[idx] += val;
// Update the values of all
// ancestors of idx
updateBIT(BITree, n, idx, val);
}
else
{
int k = q[i][1];
int lb = getLowerBound(BITree,
A, n, k);
cout << lb << endl;
}
}
}
// Driver Code
int main()
{
int A[] = { 1, 2, 3, 5, 8 };
int n = sizeof(A) / sizeof(int);
int q[][3] = { { 1, 0, 2 },
{ 2, 5, 0 },
{ 1, 3, 5 } };
performQueries(A, n, q);
}
// This code is contributed by jrishabh99
Java
// Java program to implement
// the above approach
import java.util.*;
import java.io.*;
class GFG {
// Function to calculate and return
// the sum of arr[0..index]
static int getSum(int BITree[],
int index)
{
int ans = 0;
index += 1;
// Traverse ancestors
// of BITree[index]
while (index > 0) {
// Update the sum of current
// element of BIT to ans
ans += BITree[index];
// Update index to that
// of the parent node in
// getSum() view by
// subtracting LSB(Least
// Significant Bit)
index -= index & (-index);
}
return ans;
}
// Function to update the Binary Index
// Tree by replacing all ancestors of
// index by their respective sum with val
static void updateBIT(int BITree[],
int n, int index, int val)
{
index = index + 1;
// Traverse all ancestors
// and sum with 'val'.
while (index <= n) {
// Add 'val' to current
// node of BIT
BITree[index] += val;
// Update index to that
// of the parent node in
// updateBit() view by
// adding LSB(Least
// Significant Bit)
index += index & (-index);
}
}
// Function to construct the Binary
// Indexed Tree for the given array
static int[] constructBITree(
int arr[], int n)
{
// Initialize the
// Binary Indexed Tree
int[] BITree = new int[n + 1];
for (int i = 0; i <= n; i++)
BITree[i] = 0;
// Store the actual values in
// BITree[] using update()
for (int i = 0; i < n; i++)
updateBIT(BITree, n, i, arr[i]);
return BITree;
}
// Function to obtain and return
// the index of lower_bound of k
static int getLowerBound(int BITree[],
int[] arr, int n, int k)
{
int lb = -1;
int l = 0, r = n - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (getSum(BITree, mid) >= k) {
r = mid - 1;
lb = mid;
}
else
l = mid + 1;
}
return lb;
}
static void performQueries(int A[], int n, int q[][])
{
// Store the Binary Indexed Tree
int[] BITree = constructBITree(A, n);
// Solve each query in Q
for (int i = 0; i < q.length; i++) {
int id = q[i][0];
if (id == 1) {
int idx = q[i][1];
int val = q[i][2];
A[idx] += val;
// Update the values of all
// ancestors of idx
updateBIT(BITree, n, idx, val);
}
else {
int k = q[i][1];
int lb = getLowerBound(
BITree, A, n, k);
System.out.println(lb);
}
}
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 2, 3, 5, 8 };
int n = A.length;
int[][] q = { { 1, 0, 2 },
{ 2, 5 },
{ 1, 3, 5 } };
performQueries(A, n, q);
}
}
Python3
# Python3 program to implement
# the above approach
# Function to calculate and return
# the sum of arr[0..index]
def getSum(BITree, index):
ans = 0
index += 1
# Traverse ancestors
# of BITree[index]
while (index > 0):
# Update the sum of current
# element of BIT to ans
ans += BITree[index]
# Update index to that
# of the parent node in
# getSum() view by
# subtracting LSB(Least
# Significant Bit)
index -= index & (-index)
return ans
# Function to update the
# Binary Index Tree by
# replacing all ancestors
# of index by their respective
# sum with val
def updateBIT(BITree, n,
index, val):
index = index + 1
# Traverse all ancestors
# and sum with 'val'.
while (index <= n):
# Add 'val' to current
# node of BIT
BITree[index] += val
# Update index to that
# of the parent node in
# updateBit() view by
# adding LSB(Least
# Significant Bit)
index += index & (-index)
# Function to construct the Binary
# Indexed Tree for the given array
def constructBITree(arr, n):
# Initialize the
# Binary Indexed Tree
BITree = [0] * (n + 1)
for i in range(n + 1):
BITree[i] = 0
# Store the actual values in
# BITree[] using update()
for i in range(n):
updateBIT(BITree, n, i, arr[i])
return BITree
# Function to obtain and return
# the index of lower_bound of k
def getLowerBound(BITree, arr,
n, k):
lb = -1
l = 0
r = n - 1
while (l <= r):
mid = l + (r - l) // 2
if (getSum(BITree,
mid) >= k):
r = mid - 1
lb = mid
else:
l = mid + 1
return lb
def performQueries(A, n, q):
# Store the Binary Indexed Tree
BITree = constructBITree(A, n)
# Solve each query in Q
for i in range(len(q)):
id = q[i][0]
if (id == 1):
idx = q[i][1]
val = q[i][2]
A[idx] += val
# Update the values of all
# ancestors of idx
updateBIT(BITree, n,
idx, val)
else:
k = q[i][1]
lb = getLowerBound(BITree,
A, n, k)
print(lb)
# Driver Code
if __name__ == "__main__":
A = [1, 2, 3, 5, 8]
n = len(A)
q = [[1, 0, 2],
[2, 5, 0],
[1, 3, 5]]
performQueries(A, n, q)
# This code is contributed by Chitranayal
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to calculate and return
// the sum of arr[0..index]
static int getSum(int []BITree,
int index)
{
int ans = 0;
index += 1;
// Traverse ancestors
// of BITree[index]
while (index > 0)
{
// Update the sum of current
// element of BIT to ans
ans += BITree[index];
// Update index to that
// of the parent node in
// getSum() view by
// subtracting LSB(Least
// Significant Bit)
index -= index & (-index);
}
return ans;
}
// Function to update the Binary Index
// Tree by replacing all ancestors of
// index by their respective sum with val
static void updateBIT(int []BITree,
int n, int index,
int val)
{
index = index + 1;
// Traverse all ancestors
// and sum with 'val'.
while (index <= n)
{
// Add 'val' to current
// node of BIT
BITree[index] += val;
// Update index to that
// of the parent node in
// updateBit() view by
// adding LSB(Least
// Significant Bit)
index += index & (-index);
}
}
// Function to construct the Binary
// Indexed Tree for the given array
static int[] constructBITree(int []arr,
int n)
{
// Initialize the
// Binary Indexed Tree
int[] BITree = new int[n + 1];
for(int i = 0; i <= n; i++)
BITree[i] = 0;
// Store the actual values in
// BITree[] using update()
for(int i = 0; i < n; i++)
updateBIT(BITree, n, i, arr[i]);
return BITree;
}
// Function to obtain and return
// the index of lower_bound of k
static int getLowerBound(int []BITree,
int[] arr, int n,
int k)
{
int lb = -1;
int l = 0, r = n - 1;
while (l <= r)
{
int mid = l + (r - l) / 2;
if (getSum(BITree, mid) >= k)
{
r = mid - 1;
lb = mid;
}
else
l = mid + 1;
}
return lb;
}
static void performQueries(int []A, int n,
int [,]q)
{
// Store the Binary Indexed Tree
int[] BITree = constructBITree(A, n);
// Solve each query in Q
for(int i = 0; i < q.GetLength(0); i++)
{
int id = q[i, 0];
if (id == 1)
{
int idx = q[i, 1];
int val = q[i, 2];
A[idx] += val;
// Update the values of all
// ancestors of idx
updateBIT(BITree, n, idx, val);
}
else
{
int k = q[i, 1];
int lb = getLowerBound(BITree,
A, n, k);
Console.WriteLine(lb);
}
}
}
// Driver Code
public static void Main(String[] args)
{
int []A = { 1, 2, 3, 5, 8 };
int n = A.Length;
int [,]q = { { 1, 0, 2 },
{ 2, 5, 0 },
{ 1, 3, 5 } };
performQueries(A, n, q);
}
}
// This code is contributed by 29AjayKumar
JavaScript
<script>
// Javascript program to implement
// the above approach
// Function to calculate and return
// the sum of arr[0..index]
function getSum(BITree, index) {
let ans = 0;
index += 1;
// Traverse ancestors
// of BITree[index]
while (index > 0) {
// Update the sum of current
// element of BIT to ans
ans += BITree[index];
// Update index to that
// of the parent node in
// getSum() view by
// subtracting LSB(Least
// Significant Bit)
index -= index & (-index);
}
return ans;
}
// Function to update the Binary Index
// Tree by replacing all ancestors of
// index by their respective sum with val
function updateBIT(BITree, n, index, val) {
index = index + 1;
// Traverse all ancestors
// and sum with 'val'.
while (index <= n) {
// Add 'val' to current
// node of BIT
BITree[index] += val;
// Update index to that
// of the parent node in
// updateBit() view by
// adding LSB(Least
// Significant Bit)
index += index & (-index);
}
}
// Function to construct the Binary
// Indexed Tree for the given array
function constructBITree(arr, n) {
// Initialize the
// Binary Indexed Tree
let BITree = new Array(n + 1);
for (let i = 0; i <= n; i++)
BITree[i] = 0;
// Store the actual values in
// BITree[] using update()
for (let i = 0; i < n; i++)
updateBIT(BITree, n, i, arr[i]);
return BITree;
}
// Function to obtain and return
// the index of lower_bound of k
function getLowerBound(BITree, arr, n, k) {
let lb = -1;
let l = 0, r = n - 1;
while (l <= r) {
let mid = Math.floor(l + (r - l) / 2);
if (getSum(BITree, mid) >= k) {
r = mid - 1;
lb = mid;
}
else
l = mid + 1;
}
return lb;
}
function performQueries(A, n, q) {
// Store the Binary Indexed Tree
let BITree = constructBITree(A, n);
// Solve each query in Q
for (let i = 0;
i < q.length;
i++) {
let id = q[i][0];
if (id == 1) {
let idx = q[i][1];
let val = q[i][2];
A[idx] += val;
// Update the values of all
// ancestors of idx
updateBIT(BITree, n, idx, val);
}
else {
let k = q[i][1];
let lb = getLowerBound(BITree,
A, n, k);
document.write(lb + "<br>");
}
}
}
// Driver Code
let A = [1, 2, 3, 5, 8];
let n = A.length;
let q = [[1, 0, 2],
[2, 5, 0],
[1, 3, 5]];
performQueries(A, n, q);
// This code is contributed by gfgking
</script>
Time Complexity: O(Q*(logN)2)
Auxiliary Space: O(N)
Similar Reads
Array range queries to find the number of perfect square elements with updates Given an array arr[] of N integers, the task is to perform the following two queries: query(start, end): Print the number of perfect square numbers in the sub-array from start to endupdate(i, x): Add x to the array element referenced by array index i, that is: arr[i] = x Note: 0 based indexing is fo
15+ min read
Array range queries to count the number of Fibonacci numbers with updates Given an array arr[] of N integers, the task is to perform the following two queries: query(start, end): Print the number of fibonacci numbers in the subarray from start to endupdate(i, x): Add x to the array element referenced by array index i, that is: arr[i] = x Examples: Input: arr = { 1, 2, 3,
15+ min read
Queries to find the first array element exceeding K with updates Given an array arr[] of size N and a 2D array Q[][] consisting of queries of the following two types: 1 X Y: Update the array element at index X with Y.2 K: Print the position of the first array element greater than or equal to K. If there is no such index, then print -1. Examples: Input : arr[] = {
15+ min read
XOR of elements in a given range with updates using Fenwick Tree Given an array A[] of integers and array Q consisting of queries of the following two types: (1, L, R) : Return XOR of all elements present between indices L and R.(2, I, val) : update A[I] to A[I] XOR val. The task is to solve each query and print the XOR for every Query of 1st type, using Fenwick
13 min read
Index of kth set bit in a binary array with update queries Given a binary array arr[] and q queries of following types: k: find the index of the kth set bit i.e. kth 1 in the array.(x, y): Update arr[x] = y where y can either be a 0 or 1. Examples: Input: arr[] = {1, 0, 1, 0, 0, 1, 1, 1}, q = 2 k = 4 (x, y) = (5, 1) Output: Index of 4th set bit: 6 Array aft
15+ min read