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Queries to find the Lower Bound of K from Prefix Sum Array with updates using Fenwick Tree
  • Difficulty Level : Medium
  • Last Updated : 23 Oct, 2020

Given an array A[ ] consisting of non-negative integers and 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 lower_bound of value K.

Examples: 

Input: A[ ] = {1, 2, 3, 5, 8}, Q[ ][ ] = {{1, 0, 2}, {2, 5}, {1, 3, 5}} 
Output:
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 firstly build a prefix sum array of 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 using 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 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 appraoch
#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 ancestores 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 obtian 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 appraoch
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 ancestores 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 obtian 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 appraoch
 
# 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 ancestores
# 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 obtian 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 appraoch
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 ancestores 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 obtian 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
Output: 
1



 

Time Complexity: O(Q*(logN)2) 
Auxiliary Space: O(N)

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