Find the Initial Array from given array after range sum queries

Given an array arr[] which is the resultant array when a number of queries are performed on the original array. The queries are of the form [l, r, x] where l is the starting index in the array, r is the ending index in the array and x is the integer elements that has to be added to all the elements in the index range [l, r]. The task is to find the original array.

Examples:

Input: arr[] = {5, 7, 8}, l[] = {0}, r[] = {1}, x[] = {2}
Output: 3 5 8
If query [0, 1, 2] is performed on the array {3, 5, 8}
The resultant array will be {5, 7, 8}

Input: arr[] = {20, 30, 20, 70, 100},
l[] = {0, 1, 3},
r[] = {2, 4, 4},
x[] = {10, 20, 30}
Output: 10 0 -10 20 50

Naive Approach: For each range starting from l to r subtract the corresponding x to get the initial array.



Below is the implementation of the approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to print the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
}
  
// Function to find the original array
void findOrgArr(int arr[], int l[], int r[], int x[],
                int n, int q)
{
    for (int j = 0; j < q; j++) {
        for (int i = l[j]; i <= r[j]; i++) {
  
            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }
  
    printArr(arr, n);
}
  
// Driver code
int main()
{
    // Final array
    int arr[] = { 20, 30, 20, 70, 100 };
  
    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Queries
    int l[] = { 0, 1, 3 };
    int r[] = { 2, 4, 4 };
    int x[] = { 10, 20, 30 };
  
    // Number of queries
    int q = sizeof(l) / sizeof(l[0]);
  
    findOrgArr(arr, l, r, x, n, q);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Utility function to print the contents of an array
static void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        System.out.print(arr[i]+" ");
    }
}
  
// Function to find the original array
static void findOrgArr(int arr[], int l[], int r[], int x[],
                int n, int q)
{
    for (int j = 0; j < q; j++) {
        for (int i = l[j]; i <= r[j]; i++) {
  
            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }
  
    printArr(arr, n);
}
  
// Driver code
public static void  main(String args[])
{
    // Final array
    int arr[] = { 20, 30, 20, 70, 100 };
  
    // Size of the array
    int n =  arr.length;
  
    // Queries
    int l[] = { 0, 1, 3 };
    int r[] = { 2, 4, 4 };
    int x[] = { 10, 20, 30 };
  
    // Number of queries
    int q = l.length;
  
    findOrgArr(arr, l, r, x, n, q);
  
}
}
  
// This code is contributed by
// Shashank_Sharma
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
import math as mt
  
# Utility function to print the 
# contents of an array
def printArr(arr, n):
  
    for i in range(n): 
        print(arr[i], end = " ")
  
# Function to find the original array
def findOrgArr(arr, l, r, x, n, q):
  
    for j in range(q):
        for i in range(l[j], r[j] + 1):
              
            # Decrement elements between
            # l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j]
          
    printArr(arr, n)
  
# Driver code
  
# Final array
arr = [20, 30, 20, 70, 100
  
# Size of the array
n = len(arr)
  
# Queries
l = [0, 1, 3
r = [ 2, 4, 4
x = [ 10, 20, 30 ]
  
# Number of queries
q = len(l)
  
findOrgArr(arr, l, r, x, n, q)
  
# This code is contributed by 
# mohit kumar 29
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
  
// Utility function to print the 
// contents of an array
static void printArr(int[] arr, int n)
{
    for (int i = 0; i < n; i++)
    {
        Console.Write(arr[i] + " ");
    }
}
  
// Function to find the original array
static void findOrgArr(int[] arr, int[] l, 
                       int[] r, int[] x, 
                       int n, int q)
{
    for (int j = 0; j < q; j++) 
    {
        for (int i = l[j]; i <= r[j]; i++)
        {
  
            // Decrement elements between
            // l[j] and r[j] by x[j]
            arr[i] = arr[i] - x[j];
        }
    }
  
    printArr(arr, n);
}
  
// Driver code
public static void Main()
{
    // Final array
    int[] arr = { 20, 30, 20, 70, 100 };
  
    // Size of the array
    int n = arr.Length;
  
    // Queries
    int[] l = { 0, 1, 3 };
    int[] r = { 2, 4, 4 };
    int[] x = { 10, 20, 30 };
  
    // Number of queries
    int q = l.Length;
  
    findOrgArr(arr, l, r, x, n, q);
  
}
}
  
// This code is contributed by
// Akanksha Rai
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach
  
// Utility function to print the contents
// of an array
function printArr(&$arr, $n)
{
    for ($i = 0; $i < $n; $i++)
    {
        echo($arr[$i]);
        echo(" ");
    }
}
  
// Function to find the original array
function findOrgArr(&$arr, &$l, &$r
                        &$x, $n, $q)
{
    for ($j = 0; $j < $q; $j++) 
    {
        for ($i = $l[$j]; $i <= $r[$j]; $i++) 
        {
  
            // Decrement elements between
            // l[j] and r[j] by x[j]
            $arr[$i] = $arr[$i] - $x[$j];
        }
    }
  
    printArr($arr, $n);
}
  
// Driver code
  
// Final array
$arr = array(20, 30, 20, 70, 100);
  
// Size of the array
$n = sizeof($arr);
  
// Queries
$l = array(0, 1, 3 );
$r = array( 2, 4, 4 );
$x = array(10, 20, 30 );
  
// Number of queries
$q = sizeof($l);
  
findOrgArr($arr, $l, $r, $x, $n, $q);
  
// This code is contributed by Shivi_Aggarwal
?>
chevron_right


Output:
10 0 -10 20 50 

Time Complexity: O(n2)

Efficient Approach: Follow the following steps to reach the initial array:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Utility function to print the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
}
  
// Function to find the original array
void findOrgArr(int arr[], int l[], int r[], int x[],
                int n, int q)
{
    int b[n] = { 0 };
  
    for (int i = 0; i < q; i++) {
  
        // Decrement the element at l[i]th index by -x
        b[l[i]] += -x[i];
  
        // Increment the element at (r[i] + 1)th index
        // by x if (r[i] + 1) is a valid index
        if (r[i] + 1 < n)
            b[r[i] + 1] += x[i];
    }
  
    for (int i = 1; i < n; i++)
        // Prefix sum of array b
        b[i] = b[i - 1] + b[i];
  
    // Update the original array
    for (int i = 0; i < n; i++)
        arr[i] = arr[i] + b[i];
  
    printArr(arr, n);
}
  
// Driver code
int main()
{
    // Final array
    int arr[] = { 20, 30, 20, 70, 100 };
  
    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Queries
    int l[] = { 0, 1, 3 };
    int r[] = { 2, 4, 4 };
    int x[] = { 10, 20, 30 };
  
    // Number of queries
    int q = sizeof(l) / sizeof(l[0]);
  
    findOrgArr(arr, l, r, x, n, q);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of above approach 
class GFG{
  
    // Utility function to print the contents of an array 
    static void printArr(int arr[], int n) 
    
        for (int i = 0; i < n; i++) 
        
        System.out.print(arr[i] + " ") ; 
        
    
      
    // Function to find the original array 
    static void findOrgArr(int arr[], int l[], int r[], int x[], 
                    int n, int q) 
    
        int b[] = new int[n] ;
          
        for (int i = 0; i < q; i++)
            b[i] = 0 ;
      
        for (int i = 0; i < q; i++)
        
      
            // Decrement the element at l[i]th index by -x 
            b[l[i]] += -x[i]; 
      
            // Increment the element at (r[i] + 1)th index 
            // by x if (r[i] + 1) is a valid index 
            if (r[i] + 1 < n) 
                b[r[i] + 1] += x[i]; 
        
      
        for (int i = 1; i < n; i++) 
            // Prefix sum of array b 
            b[i] = b[i - 1] + b[i]; 
      
        // Update the original array 
        for (int i = 0; i < n; i++) 
            arr[i] = arr[i] + b[i]; 
      
        printArr(arr, n); 
    
      
    // Driver code 
    public static void main(String []args)
    
        // Final array 
        int arr[] = { 20, 30, 20, 70, 100 }; 
      
        // Size of the array 
        int n = arr.length ;
      
        // Queries 
        int l[] = { 0, 1, 3 }; 
        int r[] = { 2, 4, 4 }; 
        int x[] = { 10, 20, 30 }; 
      
        // Number of queries 
        int q = l.length ;
      
        findOrgArr(arr, l, r, x, n, q); 
        
}
  
// This code is contributed by aishwarya.27
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Utility function to print the contents 
# of an array
def printArr(arr, n):
  
    for i in range(n):
        print(arr[i], end = " ")
  
  
# Function to find the original array
def findOrgArr(arr, l, r, x, n, q):
  
    b = [0 for i in range(n)]
  
    for i in range(q):
  
        # Decrement the element at l[i]th 
        # index by -x
        b[l[i]] += -x[i]
  
        # Increment the element at (r[i] + 1)th 
        # index by x if (r[i] + 1) is a valid index
        if (r[i] + 1 < n):
            b[r[i] + 1] += x[i]
      
    for i in range(n):
          
        # Prefix sum of array b
        b[i] = b[i - 1] + b[i]
  
    # Update the original array
    for i in range(n):
        arr[i] = arr[i] + b[i]
  
    printArr(arr, n)
  
# Driver code
arr = [20, 30, 20, 70, 100]
  
# Size of the array
n = len(arr)
  
# Queries
l = [0, 1, 3 ]
r = [2, 4, 4 ]
x = [10, 20, 30 ]
  
# Number of queries
q = len(l)
  
findOrgArr(arr, l, r, x, n, q)
  
# This code Is contributed by
# Mohit kumar 29
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of above approach 
using System;
  
class GFG
{
  
// Utility function to print the 
// contents of an array 
static void printArr(int[] arr, int n) 
    for (int i = 0; i < n; i++) 
    
        Console.Write(arr[i] + " "); 
    
  
// Function to find the original array 
static void findOrgArr(int[] arr, int[] l, 
                       int[] r, int[] x, 
                       int n, int q) 
    int[] b = new int[n];
      
    for (int i = 0; i < q; i++)
        b[i] = 0 ;
  
    for (int i = 0; i < q; i++)
    
  
        // Decrement the element at l[i]th 
        // index by -x 
        b[l[i]] += -x[i]; 
  
        // Increment the element at (r[i] + 1)th 
        // index by x if (r[i] + 1) is a valid index 
        if (r[i] + 1 < n) 
            b[r[i] + 1] += x[i]; 
    
  
    for (int i = 1; i < n; i++) 
      
        // Prefix sum of array b 
        b[i] = b[i - 1] + b[i]; 
  
    // Update the original array 
    for (int i = 0; i < n; i++) 
        arr[i] = arr[i] + b[i]; 
  
    printArr(arr, n); 
  
// Driver code 
public static void Main()
    // Final array 
    int[] arr = { 20, 30, 20, 70, 100 }; 
  
    // Size of the array 
    int n = arr.Length;
  
    // Queries 
    int[] l = { 0, 1, 3 }; 
    int[] r = { 2, 4, 4 }; 
    int[] x = { 10, 20, 30 }; 
  
    // Number of queries 
    int q = l.Length;
  
    findOrgArr(arr, l, r, x, n, q); 
}
  
// This code is contributed 
// by Akanksha Rai
chevron_right


Output:
10 0 -10 20 50 

Time Complexity: O(n)




Recommended Posts:


Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :