# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // C++ implementation of the approach #include 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; }

## Java

 // 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

## PHP



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.

 // C++ implementation of the approach #include 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; }

Output:

10 0 -10 20 50

Time Complexity: O(n)

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Improved By : Shashank_Sharma