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:

C++

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// 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;
}

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Java

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// 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

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PHP

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<?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
?>

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Output:

10 0 -10 20 50 

Time Complexity: O(n2)

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

  • Take an array b[] of the size of the given array and initialize all of its elements with 0.
  • In array b[], for every query update b[l] = b[l] – x and b[r + 1] = b[r + 1] + x if r + 1 < n. This is because x will cancel out the effect of -x when performed the prefix sum.
  • Take the prefix sum of array b[], and add it to the given array which will produce the initial array.
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// 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;
}

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Output:

10 0 -10 20 50 

Time Complexity: O(n)



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