Constant time range add operation on an array

3.2

Given an array of size N which is initialized with all zeros. We are given many range add queries, which should be applied to this array. We need to print final updated array as our result.
Examples:

N = 6
Arr = [0, 0, 0, 0, 0, 0]
rangeUpdate1 [0, 2], add 100
Arr = [100, 100, 100, 0, 0, 0]
rangeUpdate1 [1, 5], add 100
Arr = [100, 200, 200, 100, 100, 100]
rangeUpdate1 [2, 3], add 100
Arr = [100, 200, 300, 200, 100, 100]
Which is the final updated array.

This problem can be solved using segment tree with lazy updates in O(log N) time per query but we can do better here, as update operation is not given. We can process each query in constant time using this logic, when a query to add V is given in range [a, b] we will add V to arr[a] and –V to arr[b+1] now if we want to get the actual values of array we will convert the above array into prefix sum array. See below example to understand,

Arr = [0, 0, 0, 0, 0, 0]
rangeUpdate1 [0, 2], add 100
Arr = [100, 0, 0, -100, 0, 0]
rangeUpdate1 [1, 5], add 100
Arr = [100, 100, 0, -100, 0, 0, -100]
rangeUpdate1 [2, 3], add 100
Arr = [100, 100, 100, -100, -100, 0, -100]    
Now we will convert above operation array to prefix sum array as shown below,
Arr = [100, 200, 300, 200, 100, 100]
Which is the final updated array.

So in effect, when we add a value V to specific index of array, It represents adding V to all elements right to this index, that is why we add –V after range to remove its effect after its range of add query.
Please note in below code, if range spans till the last index, the addition of –V is omitted to be in memory limit of the array.

C++

//  C++ program to get updated array after many array range
// add operation
#include <bits/stdc++.h>
using namespace std;

//  Utility method to add value val, to range [lo, hi]
void add(int arr[], int N, int lo, int hi, int val)
{
    arr[lo] += val;
    if (hi != N - 1)
       arr[hi + 1] -= val;
}

//  Utility method to get actual array from operation array
void updateArray(int arr[], int N)
{
    //  convert array into prefix sum array
    for (int i = 1; i < N; i++)
        arr[i] += arr[i - 1];
}

//  method to print final updated array
void printArr(int arr[], int N)
{
    updateArray(arr, N);
    for (int i = 0; i < N; i++)
        cout << arr[i] << " ";
    cout << endl;
}

//  Driver code to test above methods
int main()
{
    int N = 6;

    int arr[N] = {0};

    //  Range add Queries
    add(arr, N, 0, 2, 100);
    add(arr, N, 1, 5, 100);
    add(arr, N, 2, 3, 100);

    printArr(arr, N);
    return 0;
}

Java

// Java program to get updated array after
// many array range add operation
import java.io.*;
 
class GFG 
{
    // Utility method to add value val,
    // to range [lo, hi]
    static void add(int arr[], int N, int lo,
                           int hi, int val)
    {
        arr[lo] += val;
        if (hi != N - 1)
           arr[hi + 1] -= val;
    }
    
    // Utility method to get actual array from 
    // operation array
    static void updateArray(int arr[], int N)
    {
        // convert array into prefix sum array
        for (int i = 1; i < N; i++)
            arr[i] += arr[i - 1];
    }
     
    // method to print final updated array
    static void printArr(int arr[], int N)
    {
        updateArray(arr, N);
        for (int i = 0; i < N; i++)
            System.out.print(""+arr[i]+" ");
        System.out.print("\n");
    }
     
    // Driver code to test above methods
    public static void main (String[] args) 
    {
        int N = 6;
 
        int arr[] = new int[N];
     
        // Range add Queries
        add(arr, N, 0, 2, 100);
        add(arr, N, 1, 5, 100);
        add(arr, N, 2, 3, 100);
     
        printArr(arr, N);
    }
}

// This code is contributed by Prakriti Gupta


Output:

100 200 300 200 100 100

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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